#linear-algebra | Mathematics | Page 111

ocean sequoia Jul 21, 2020, 1:49 AM

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idk how else you would answer that its kinda yes or no XD

half storm Jul 21, 2020, 1:50 AM

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Yea lol. I don't think I could have given a different answer.

ocean sequoia Jul 21, 2020, 1:50 AM

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i was trying to use a property maybe to solve something dont worry thats not the 'real'question

half storm Jul 21, 2020, 1:50 AM

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Got you.

ocean sequoia Jul 21, 2020, 1:50 AM

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thanks

half storm Jul 21, 2020, 1:50 AM

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No problem. As a side note, a vector space is finite dimensional if there exists a finite set of vectors that spans the space.

ocean sequoia Jul 21, 2020, 1:51 AM

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ah that makes sense so its only infinite if there is not a finite set

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and in this case m vectors would span the space

half storm Jul 21, 2020, 1:51 AM

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Yup.

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The standard basis with 1 in the first coordinate then all zeros, 1 in the second coordinate all zeroes etc.

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Any vector space that is a finite cartesian product of a field is a finite dimensional vector space.

median forum Jul 21, 2020, 1:52 AM

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wait

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what is F here?

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a finite field?

half storm Jul 21, 2020, 1:53 AM

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I assumed that it was an infinite field. He didn't specify.

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I was wondering that.

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I don't think it would though.

pallid rampart Jul 21, 2020, 1:54 AM

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Well it doesn’t matter

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As long as F is a field

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F^n is an n dimensional vector space over F for any field F

half storm Jul 21, 2020, 1:55 AM

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Yea because if you're assuming that vector addition and scalar multiplication are defined in the way that you would usually define them, then I was thinking it shouldnt

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yea

torpid horizon Jul 21, 2020, 2:40 AM

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hi

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can someone help me here

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on a question

half storm Jul 21, 2020, 2:41 AM

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I can try shoot.

torpid horizon Jul 21, 2020, 2:41 AM

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im not sure how to do numbers 10 and 11 but for 10 I got the answer B

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thisis the given

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📎 unknown.png

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here are questions 10 and 11

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📎 unknown.png

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it's multiple choice

half storm Jul 21, 2020, 2:43 AM

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You're gonna have to type some of this out

wintry steppe Jul 21, 2020, 2:43 AM

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um do you have the pic of the original problem

half storm Jul 21, 2020, 2:43 AM

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Or that

median forum Jul 21, 2020, 2:44 AM

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this writing is as bad as mine

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yikes

wintry steppe Jul 21, 2020, 2:45 AM

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ok look I'll try to translate

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nevermind, bad idea

median forum Jul 21, 2020, 2:51 AM

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Find the change of basis matrix $C = C_b$, B_2$ that changes $B_1$ coordinates into $B_2$ coordinates :

stoic pythonBOT Jul 21, 2020, 2:51 AM

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Fractal:
Compile Error! Click the errors reaction for details. (You may edit your message)

📎 241386462174511106.png

median forum Jul 21, 2020, 2:51 AM

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something like that

torpid horizon Jul 21, 2020, 2:53 AM

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yes

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that what it says

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so it says....

wintry steppe Jul 21, 2020, 2:54 AM

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@torpid horizon what is B

torpid horizon Jul 21, 2020, 2:54 AM

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for option b?

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on 10?

wintry steppe Jul 21, 2020, 2:54 AM

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no

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i mean the problem

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oh nvm

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didnt see the other pic my bad

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well think about it

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how do you turn (3,1) into (1,1) and (2,1) into (1,2)

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essentially, you are mapping 3x to x, y to y for the first vector

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and 2x to x and y to 2y for the second vector

torpid horizon Jul 21, 2020, 2:55 AM

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so the instructionssay B sub 1 = (3 1), (2 1) ) and b sub 2 = .....and let h = R ^2 map onto R^2 be linear such that h (3 1) = ( 1 ) and h (2 1 ) = ( 1 2 )

wintry steppe Jul 21, 2020, 2:55 AM

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and thats exactly what a matriix represents

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so we want to find a matrix A such that (3,1) maps to (1,1)

torpid horizon Jul 21, 2020, 2:56 AM

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so is the answer to 10 letter B?

wintry steppe Jul 21, 2020, 2:56 AM

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and also maps (2,1) to (1,2)

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why are you rushing

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you asked for help so im trying to explain

torpid horizon Jul 21, 2020, 2:57 AM

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yes go ahead

wintry steppe Jul 21, 2020, 2:57 AM

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so lets say A is a 2x2 matrix containing column vectors (a,b) and (c,d)

torpid horizon Jul 21, 2020, 2:58 AM

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ahuh

wintry steppe Jul 21, 2020, 2:58 AM

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we know A * (3,1) = (1,1)

torpid horizon Jul 21, 2020, 2:58 AM

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ok

wintry steppe Jul 21, 2020, 2:58 AM

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just a note: (3,1) and (1,1) are column vectors here

torpid horizon Jul 21, 2020, 2:59 AM

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ok

wintry steppe Jul 21, 2020, 2:59 AM

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so compute the matrix multiplication

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and tell me what u get

torpid horizon Jul 21, 2020, 2:59 AM

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so which am i multiplying?

wintry steppe Jul 21, 2020, 2:59 AM

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we have A * (3,1) = (1,1)

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where (3,1) and (1,1) are vectors in R^2

torpid horizon Jul 21, 2020, 3:00 AM

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ahuh

wintry steppe Jul 21, 2020, 3:00 AM

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and A is made up of the column vectors (a,b) and (c,d)

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so multiply that out

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our goal is to find a, b, c and d

torpid horizon Jul 21, 2020, 3:02 AM

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so do i multipl both?

wintry steppe Jul 21, 2020, 3:02 AM

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just multiply

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A * (3,1)

torpid horizon Jul 21, 2020, 3:02 AM

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what is A here?

wintry steppe Jul 21, 2020, 3:03 AM

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its [(a,b), (c,d)]

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(a,b) is a column vectors

torpid horizon Jul 21, 2020, 3:03 AM

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( 3 a + b 3c + d)

wintry steppe Jul 21, 2020, 3:04 AM

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and that = (1,1)

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so 3a + b = 1

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3c + d = 1

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now lets do it for the other vector

torpid horizon Jul 21, 2020, 3:04 AM

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ahuh

wintry steppe Jul 21, 2020, 3:04 AM

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so A * (2,1) = (1,2)

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same idea

torpid horizon Jul 21, 2020, 3:05 AM

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so after i write ......3a + b = 1

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3c+ d = 1

wintry steppe Jul 21, 2020, 3:05 AM

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multiply A * (2,1) = (1,2)

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we r gonna have 4 equations after this

torpid horizon Jul 21, 2020, 3:06 AM

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ok let me see

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i got...

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2a + b = 1

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2c + d = 2

wintry steppe Jul 21, 2020, 3:06 AM

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great

torpid horizon Jul 21, 2020, 3:06 AM

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so what would i do next

wintry steppe Jul 21, 2020, 3:07 AM

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so 1 = 2a + b = 3a + b

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so b = 1-2a

torpid horizon Jul 21, 2020, 3:07 AM

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ahuh

wintry steppe Jul 21, 2020, 3:07 AM

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=> 3a + 1-2a = 1

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=>a = 0

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=> b = 1

torpid horizon Jul 21, 2020, 3:08 AM

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ahuh

wintry steppe Jul 21, 2020, 3:08 AM

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now for c and d

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2c + d = 1 + 1

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= 1 + 3c + d

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so 2c + d = 1 + 3c + d

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=> d = 1 + c + d

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=> 1 + c = 0

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=> c = -1

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now for d, we know 2c + d = 2

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so d = 2 - 2c = 2(1-c) = 2(1-(-1)) = 2 * 2 = 4

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so we have (a,b) = (0,1)

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and (c,d) = (-1,4)

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so our transformation matrix A = [(0,1), (-1,4)]

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where each vector in () is a column vector

torpid horizon Jul 21, 2020, 3:11 AM

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ahuh

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so that's the answer to 11 which is letter C?

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did i do #10 correctly i got the answer was B

wintry steppe Jul 21, 2020, 3:13 AM

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well what did u do for 10?

torpid horizon Jul 21, 2020, 3:13 AM

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i did this....

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📎 unknown.png

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i got B?

soft tundra Jul 21, 2020, 3:19 AM

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📎 2018_practice_problem.JPG

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Is the answer to c that the error vector must be orthogonal in order to be the shortest distance to b?

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📎 asdfasdfasdf.jpg

torpid horizon Jul 21, 2020, 3:20 AM

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how would i find the lipshitz for #14?

half storm Jul 21, 2020, 3:21 AM

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@soft tundra Yea

soft tundra Jul 21, 2020, 3:22 AM

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ty 🙂

torpid horizon Jul 21, 2020, 3:23 AM

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can someone please help me

rigid cypress Jul 21, 2020, 3:23 AM

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gimme a sec

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where's the question exactly

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@torpid horizon

torpid horizon Jul 21, 2020, 3:25 AM

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📎 unknown.png

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im stuck on these last two

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it says if S = <v1 , v2, etc) is a linearly dependent subset of V then V sub j element of S such tht span S = span (S, (vsubj))

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true or flase

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i said True

rigid cypress Jul 21, 2020, 3:28 AM

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think about what linear dependence means

torpid horizon Jul 21, 2020, 3:30 AM

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ahuh

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so is it false then

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idk

rigid cypress Jul 21, 2020, 3:30 AM

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here

torpid horizon Jul 21, 2020, 3:30 AM

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means it equals 0

rigid cypress Jul 21, 2020, 3:30 AM

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it could be true or false

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so think about it like this

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if i gave you a 3d vector space V

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and then a subset S = {(1, 0, 0), (2, 0, 0)}

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the values of S are linearly dependent right

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but is there a vector you could add to S to make it span V

torpid horizon Jul 21, 2020, 3:32 AM

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so it doesnt mattter?>

rigid cypress Jul 21, 2020, 3:33 AM

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?

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what im saying by "it could be true or false" is that i wont give you the answer

torpid horizon Jul 21, 2020, 3:35 AM

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so meaningless?

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how about for 20

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#20

rigid cypress Jul 21, 2020, 3:36 AM

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???

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i didnt say it's meaningless

torpid horizon Jul 21, 2020, 3:36 AM

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i mean question #20

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i just said true

half storm Jul 21, 2020, 3:36 AM

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is that $ S \textbackslash {v_j } $

torpid horizon Jul 21, 2020, 3:36 AM

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yes

rigid cypress Jul 21, 2020, 3:36 AM

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i was talking about 19

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i cant read 20 tbqh

half storm Jul 21, 2020, 3:37 AM

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fuck it I can't Tex it but Im' assuming it's S take away vj

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But @rigid cypress is right. You have to think about what it means for a set to linearly independent and what exactly is the "span of a set of vectors".

ocean sequoia Jul 21, 2020, 3:39 AM

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do you know what a direct sum is?

torpid horizon Jul 21, 2020, 3:40 AM

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not really

ocean sequoia Jul 21, 2020, 3:41 AM

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what does it mean to span a space?

torpid horizon Jul 21, 2020, 3:41 AM

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tht all vectors fall within that space

ocean sequoia Jul 21, 2020, 3:41 AM

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also i cant read the question so im going off the help

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can you be a bit more specific

half storm Jul 21, 2020, 3:41 AM

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That second problem that you listed, the most I can make out of that is that f is a linear map from V into W and that B_1 is a basis for V and that B_2 is a basis for W

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I can't figure out what the rest is trying to say.

torpid horizon Jul 21, 2020, 3:42 AM

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yeah my teaches writing kinda bad

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which part was it

ocean sequoia Jul 21, 2020, 3:45 AM

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is that it?

torpid horizon Jul 21, 2020, 3:45 AM

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yes

cursive narwhal Jul 21, 2020, 3:47 AM

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Let $S$ be a linearly dependent subset of $V$. Since it is linearly dependent, there exists a vector in it, call it $v_j$, such that $v_j$ can be written as a linear combination of the other vectors in $S$. So, it is the case that:

$$span(S) = span(S \setminus {v_j})$$

stoic pythonBOT Jul 21, 2020, 3:48 AM

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Abhijeet Vats:

📎 340191992636243968.png

cursive narwhal Jul 21, 2020, 3:49 AM

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@torpid horizon

half storm Jul 21, 2020, 3:49 AM

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\setminus cool.

torpid horizon Jul 21, 2020, 3:51 AM

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yes

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thats the question

cursive narwhal Jul 21, 2020, 3:51 AM

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I've given you the answer

ocean sequoia Jul 21, 2020, 3:51 AM

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i dont even understand what the question is

cursive narwhal Jul 21, 2020, 3:51 AM

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The statement given to you is true.

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I mean, as I understand it, it holds.

ocean sequoia Jul 21, 2020, 3:53 AM

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is it just asking if you can make S linearly independent?

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or that if you remove a vector V_j it doesnt change the span?

torpid horizon Jul 21, 2020, 3:54 AM

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yes

half storm Jul 21, 2020, 3:54 AM

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the second one.

torpid horizon Jul 21, 2020, 3:54 AM

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yes

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i said true as well

half storm Jul 21, 2020, 3:54 AM

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That is the correct answer.

torpid horizon Jul 21, 2020, 3:54 AM

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number 20 im not sure what the answer is

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but i wrote

ocean sequoia Jul 21, 2020, 3:55 AM

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that is true do you understand why?

rigid cypress Jul 21, 2020, 3:55 AM

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do you think 19 is true

torpid horizon Jul 21, 2020, 3:55 AM

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C

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i kind of understood it

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and for 19 i wrote true

rigid cypress Jul 21, 2020, 3:55 AM

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we're not going to give you any direct solutions

torpid horizon Jul 21, 2020, 3:56 AM

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im just trying to understand the problems

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because it talks about spans

ocean sequoia Jul 21, 2020, 3:56 AM

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@torpid horizon hang on lets do one at a time

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what is a span

torpid horizon Jul 21, 2020, 3:56 AM

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okie

ocean sequoia Jul 21, 2020, 3:56 AM

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it seems likeyour def is weak

torpid horizon Jul 21, 2020, 3:56 AM

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span is where you have vectors that fall within a coordinate plane

ocean sequoia Jul 21, 2020, 3:56 AM

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no

torpid horizon Jul 21, 2020, 3:56 AM

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no?

ocean sequoia Jul 21, 2020, 3:56 AM

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no

torpid horizon Jul 21, 2020, 3:56 AM

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i saw a vid maybe i misunderstood

ocean sequoia Jul 21, 2020, 3:56 AM

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i think you did

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it happens this stuff can be confusing the first time you go over it

torpid horizon Jul 21, 2020, 3:57 AM

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omg yes

half storm Jul 21, 2020, 3:58 AM

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The definition you gave is a sufficnet way of viewing the span of two vectors in R^2; But there is a more general definition, and it is the one that you will almost always want to refer to.

dire thunder Jul 21, 2020, 3:58 AM

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span is set of all linear combinations of vectors

half storm Jul 21, 2020, 3:58 AM

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^

dire thunder Jul 21, 2020, 3:59 AM

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i mean in R^2 it is plane

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but generally it is not

ocean sequoia Jul 21, 2020, 3:59 AM

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do you know what a linear combination is

cursive narwhal Jul 21, 2020, 3:59 AM

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there are a grand total of 3 people helping out and it's not very useful

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two of you need to back off

dire thunder Jul 21, 2020, 3:59 AM

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correctly would say if n >= 2 and you have exactly 2 independet vectors you will get plane

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hi vats

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catSad

half storm Jul 21, 2020, 4:00 AM

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Yea I'll leave this to you guys.

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I like answering questions because it helps me know that I understand the material.

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But I don't mind.

ocean sequoia Jul 21, 2020, 4:00 AM

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same but ill back off for the same reason

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you are right vats

torpid horizon Jul 21, 2020, 4:00 AM

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ahuh

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every feedback helps

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since im lost lol

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so dont worry if u guys know a lot

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enlighten me

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lol

wintry steppe Jul 21, 2020, 4:01 AM

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@soft tundra yes

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must be normal to the plane by definition

cursive narwhal Jul 21, 2020, 4:02 AM

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pixie, do you know what a vector space is? Can I assume a basic level of knowledge from you?

torpid horizon Jul 21, 2020, 4:03 AM

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yeah i just started the course

cursive narwhal Jul 21, 2020, 4:04 AM

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Because right now, I'm just going to give you the abstraction straight up and you have to grapple with it.

Let $(V,+,\cdot)$ be a vector space over $\mathbb{F}$. Let $S$ be a list of vectors $S = (v_1,v_2,\ldots,v_n)$. So, this is a list of $n$ vectors from the vector space. We're not saying anything more about them. Then, I will define the following set:

$$span(S) \coloneqq {\sum_{k=1}^{n} \alpha_k v_k: \alpha_k \in \mathbb{F} }$$

This is the definition of the span of $S$. It has geometric meaning when you're talking about $\bR^3$ or $\bR^2$ or even $\bR$. But here's the abstraction.

stoic pythonBOT Jul 21, 2020, 4:04 AM

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Abhijeet Vats:

📎 340191992636243968.png

cursive narwhal Jul 21, 2020, 4:05 AM

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Simply put: the span of a list of vectors is the set of all possible linear combinations of those vectors.

torpid horizon Jul 21, 2020, 4:08 AM

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ah

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ok

cursive narwhal Jul 21, 2020, 4:08 AM

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Does that make sense? So, for instance, suppose that $V = \bR^2$ and I'll define $S \coloneqq ((1,2),(1,3))$. If this is a vector space over $\bR$, then $span(S)$ consists of vectors of the form:

$$\alpha \cdot (1,2) + \beta \cdot (1,3) = (\alpha+\beta,2\alpha+3\beta)$$

with $\alpha,\beta \in \bR$. That's the general form of the vectors that will lie in my span.

stoic pythonBOT Jul 21, 2020, 4:08 AM

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Abhijeet Vats:

📎 340191992636243968.png

cursive narwhal Jul 21, 2020, 4:09 AM

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Linear Algebra can be a bit tricky if you're just starting out with it and haven't been exposed to proofs before. I don't recommend using videos to learn it. Use books instead.

dire thunder Jul 21, 2020, 4:10 AM

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any math can be a bit tricky

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megathink

torpid horizon Jul 21, 2020, 4:10 AM

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yeh i feel like LA is hard or me

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diff eqns not so much

wintry steppe Jul 21, 2020, 4:10 AM

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i personally enjoyed LA more than diff eq

torpid horizon Jul 21, 2020, 4:10 AM

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im the opposite

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lol

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thank u guys for ur help

cursive narwhal Jul 21, 2020, 4:11 AM

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Have a look at Klaus Janich's Linear Algebra textbook. You can use the linear algebra lecture notes by E. Kowalski as a supplement

torpid horizon Jul 21, 2020, 4:11 AM

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oh ok

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is it free

dire thunder Jul 21, 2020, 4:11 AM

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lib

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gen

cursive narwhal Jul 21, 2020, 4:12 AM

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The lecture notes are freely available (Just search up Linear Algebra lecture notes by E. Kowalski). Getting a copy of Janich's text will cost money but you can use libgen

torpid horizon Jul 21, 2020, 4:12 AM

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ah ok ill look right now

half storm Jul 21, 2020, 4:13 AM

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Linear Algebra is definitely a class that introduces you to the idea of abstraction. A Diffy Q course will generally give you some nice physical examples to see how the occur and get some good intuition of the information they convey.

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Also, DE taught at the undergrad level is fairly algorithmic.

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You recognize a certain type of problem and there is a procedure that you follow to solve it.

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Linear algebra is more about thinking about the structures are and how they are related to one another.

median forum Jul 21, 2020, 4:16 AM

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not pdes

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odes are

half storm Jul 21, 2020, 4:16 AM

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Yea PDEs delves more into the theory

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I kind of wish my professor delved more into the theory about DE. DE and linear algebra probably should be taken in the same year, or you should try to take them concurrently.

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There is obviously an extensive amount of theory for ODE's but generally it is not discussed.

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not at undergrad anyways.

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You do get some of it though. Most of my ODE's class was filled with theorem statements that could be pretty hard to grapple with and most of the theorems are generally not proved because you need a course in analysis and LA to really understand them.

cursive narwhal Jul 21, 2020, 4:19 AM

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John, where did you do your undergrad?

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(Just curious haha)

half storm Jul 21, 2020, 4:19 AM

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At a liberal arts college.

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I can't really tell you cause it's a small school and I might dox myself. Sorry 😦

cursive narwhal Jul 21, 2020, 4:19 AM

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It's fine, it's fine

median forum Jul 21, 2020, 4:19 AM

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gl proving frobenius LULW

half storm Jul 21, 2020, 4:20 AM

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I'm mostly speaking from my own experience, but I've heard similar stories from my friends who went to other schools.

cursive narwhal Jul 21, 2020, 4:21 AM

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it's a bit different where i'm studying but maybe that's because the culture for math in my school is a bit different

half storm Jul 21, 2020, 4:21 AM

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Yea, I kind of wish I went to a university. I think it would helped me in some ways. I wasn't a great student. At least not my junior and senior years of college. I did pretty well first two years.

torpid horizon Jul 21, 2020, 4:55 AM

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im looking at the

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resources u posted

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ty

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i appreciate everyones help here

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idk if u guys are still online

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but ty

ivory basin Jul 21, 2020, 4:56 AM

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📎 c7338a01f0d97ace2e08b2ca5d5c1b11.png

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i need help on part c of this problem

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i know axiom 7 or K(u+v)=Ku+kv is the axiom that fails but i dont get why

half storm Jul 21, 2020, 5:04 AM

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@ivory basin In order for V to not be a vector space, it has to be the case that there exists a k in our field of scalars - which is implied to be the real numbers in this case - such that Axiom 7 doesn't hold.

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What I'd recommend you do, is consider both expressions on each sides of the equality seperately, and consider what k has to be so that they are not equal.

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Does that make sense?

ivory basin Jul 21, 2020, 5:08 AM

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but when i think of both sides as seperate expressions i still cant think of any k scalars that break the axiom

half storm Jul 21, 2020, 5:12 AM

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Alright, I'll actually work through it.

half storm Jul 21, 2020, 5:36 AM

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Hmmm, I checked out Axiom 7 and it seems to work out 😮

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Of course, we could both be doing it incorrectly

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How do you "know" that it is Axiom 7 that is the issue?

ivory basin Jul 21, 2020, 6:12 AM

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im not sure either my friend said he asked his tutor and the tutor said axiom 7 failed

dusky epoch Jul 21, 2020, 6:14 AM

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which axiom do y'all number 7

half storm Jul 21, 2020, 6:15 AM

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scalars distirbute over vector addition. If k is a scalar taken from a the field F and u and v are vectors in V, then K( u + v ) = ku + kv.

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That's the one her is referring to.

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But it seems to check out to me.

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There is another axiom that deals with distributivity. Basically it's distributivity over scalar addition

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so if k and l are scalars from a field F and v a vector in V , then (k + l) v = kv + lv

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Did you consider checking that property?

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That could possibly be the axiom that isn't satisfied.

ivory basin Jul 21, 2020, 6:21 AM

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📎 2f04bbeb069b5e289289efd0e7503293.png

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this is what my friend sent me and said this is the right answer

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but it doesnt look right to me

half storm Jul 21, 2020, 6:23 AM

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Well they kind of just skipped to saying $ ku \oplus kv \neq k ( u \oplus v ) $ but they didn't verify that.

stoic pythonBOT Jul 21, 2020, 6:23 AM

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JohntheDon:

📎 671939363071983616.png

half storm Jul 21, 2020, 6:23 AM

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When you went to verify it yourself. You saw that this is actually a true statement. I verified for myself as well. It seems to hold - thought of course we could both be doing it wrong.

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I have a feelling though that the other distributive property may not hold.

dusky epoch Jul 21, 2020, 6:24 AM

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ku + kv = (ku1 + k - 1, ku2 + k - 1) + (kv1 + k - 1, kv2 + k - 1) = (ku1 + kv1 + 2k - 2 + 1, ku2 + kv2 + 2k - 2 + 1)

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it does hold

half storm Jul 21, 2020, 6:24 AM

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That's what we both got.

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So their tutor is wrong.

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Give me a sec to check the other distributive property.

ivory basin Jul 21, 2020, 6:25 AM

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so then which axiom is the one that fails because i tried them all

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unless i did one wrong

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i did all the axioms and it looks like it is a vector space but the question says prove that V isnt a vector space so idk what to do

half storm Jul 21, 2020, 6:36 AM

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I'm checking and its looking like a vector space to me lol

ivory basin Jul 21, 2020, 6:37 AM

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thats what i think but idk why my teacher phrases the question as prove its not a vector space

dusky epoch Jul 21, 2020, 6:41 AM

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okay so you know what might be helpful

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expressing the operations in terms of the "classical"/standard addition and scaling on R^2

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letting z = (1,1) and (+) and (*) be the new operations, we have
u (+) v = u+v+z
k (*) u = k(u+z) - z

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(k + l) (*) u = (k+l)(u+z) - z
k (*) u (+) l (*) u = k(u+z) - z + l(u+z) - z + z = ku + kz + lu + lz - 2z + z = (k+l)u + (k+l-1)z

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hm

noble hemlock Jul 21, 2020, 7:38 AM

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Over a commutative ring, why is the trace of a square matrix the sum of its characteristic eigenvectors?

dusky epoch Jul 21, 2020, 8:01 AM

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it's not

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did you mean to ask why the trace of a matrix equals the sum of its eigenvalues

noble hemlock Jul 21, 2020, 9:17 AM

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yeah that's what I meant sorry @dusky epoch

dusky epoch Jul 21, 2020, 9:18 AM

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i believe you might wanna assume its charpoly actually factors over your ring

noble hemlock Jul 21, 2020, 9:22 AM

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yes

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Assuming that, how do I go about proving it? @dusky epoch

dusky epoch Jul 21, 2020, 9:28 AM

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leibniz formula for the determinant i guess

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and also vieta's formulas

noble hemlock Jul 21, 2020, 9:29 AM

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how

zealous widget Jul 21, 2020, 11:10 AM

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how can I show that if $n$ projection matrices $P_{1}$ up to $P_{n}$ of dimension $n \times n$ sum to $I$ then (assuming these projection matrices are for projecting onto lines), then these independent line vectors $a_{1}$ through to $a_{n}$ are orthogonal?

dusky epoch Jul 21, 2020, 11:15 AM

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so $P_i$ is the projector onto $\ang{a_i}$?

stoic pythonBOT Jul 21, 2020, 11:15 AM

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Ann:

📎 183668144404037632.png

zealous widget Jul 21, 2020, 11:15 AM

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Yes

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correct

dusky epoch Jul 21, 2020, 11:15 AM

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(btw it's $n \times n$)

stoic pythonBOT Jul 21, 2020, 11:15 AM

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Ann:

📎 183668144404037632.png

noble hemlock Jul 21, 2020, 11:15 AM

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Is the adjoint of an orthogonal matrix also orthogonal?

zealous widget Jul 21, 2020, 11:15 AM

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oh OK thanks

stoic pythonBOT Jul 21, 2020, 11:15 AM

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catfood:

📎 467744230010519552.png

zealous widget Jul 21, 2020, 11:15 AM

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question edited

dusky epoch Jul 21, 2020, 11:16 AM

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right so
it seems to me like if we assume the $a_i$ are of unit length (normalizing otherwise to make it the case), then $P_i = a_ia_i^T$

stoic pythonBOT Jul 21, 2020, 11:16 AM

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Ann:

📎 183668144404037632.png

zealous widget Jul 21, 2020, 11:16 AM

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hmmm

dusky epoch Jul 21, 2020, 11:16 AM

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(@noble hemlock yes)

zealous widget Jul 21, 2020, 11:17 AM

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my algebra textbook taught me that it should be $P_i = \frac{a_ia_i^T} {a_i^Ta_i}$

stoic pythonBOT Jul 21, 2020, 11:17 AM

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catfood:

📎 467744230010519552.png

noble hemlock Jul 21, 2020, 11:17 AM

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can you show why? @dusky epoch

dusky epoch Jul 21, 2020, 11:17 AM

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@zealous widget yeah i normalized the a_i to have length 1

zealous widget Jul 21, 2020, 11:17 AM

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oh yeah good point

dusky epoch Jul 21, 2020, 11:17 AM

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@noble hemlock A^T A = I implies A^T (A^T)^T = I

noble hemlock Jul 21, 2020, 11:18 AM

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what's A^T?

zealous widget Jul 21, 2020, 11:18 AM

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transpose

gray dust Jul 21, 2020, 11:18 AM

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transpose of A

noble hemlock Jul 21, 2020, 11:18 AM

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ah okay that makes sense

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thanks

dusky epoch Jul 21, 2020, 11:18 AM

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hm ok so

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$\sum_{i=1}^n a_ia_i^T = I$

stoic pythonBOT Jul 21, 2020, 11:18 AM

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Ann:

📎 183668144404037632.png

dusky epoch Jul 21, 2020, 11:18 AM

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hmm

zealous widget Jul 21, 2020, 11:19 AM

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I tried a clunky method of adding all the fractions

dusky epoch Jul 21, 2020, 11:19 AM

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hold up.

zealous widget Jul 21, 2020, 11:19 AM

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tryna find some cancellations

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didnt work

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take your time

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no rush lmao

dusky epoch Jul 21, 2020, 11:19 AM

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ok no nvm.

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i might need to rest for a bit

zealous widget Jul 21, 2020, 11:19 AM

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so we'll do this later then?

dusky epoch Jul 21, 2020, 11:19 AM

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maybe?

zealous widget Jul 21, 2020, 11:20 AM

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sure thing

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rest is more imporrtant than my q

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so go and do whatever :D

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@dusky epoch nevermind did it myself with my clunky fraction method and some tweaking

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thanks for the help though

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was greatly appreciated!

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actually

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wait

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no

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i didn't do it

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nvm

onyx comet Jul 21, 2020, 1:30 PM

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does ayone have advanced knowledge for la?

real stirrupBOT Jul 21, 2020, 1:41 PM

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From

#❓how-to-get-help

Rule 1

The help channels are solely for help with math, so feel free to post your question. Asking whether you can ask a question or if anyone knows about some specific topic is unnecessary, so please try to avoid questions of that nature.

modern palm Jul 21, 2020, 2:30 PM

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how do you prove that any line through the origin in R^3 is a subspace?

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specifically that it is closed under addition

half storm Jul 21, 2020, 2:31 PM

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Well a line is given by a vector equation right?

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something to this effect

modern palm Jul 21, 2020, 2:31 PM

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yes

half storm Jul 21, 2020, 2:32 PM

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$$ L := r(t) = <a_1 , a_2, a_3> + t<b_1,b_2,b_3> where t \in \mathbb{R}$$

stoic pythonBOT Jul 21, 2020, 2:33 PM

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JohntheDon:

📎 671939363071983616.png

half storm Jul 21, 2020, 2:33 PM

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In order to show something is a subspace, you need to show that 0 is contained in the set

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that for any two points in the set that their sum is in the set.

cursive narwhal Jul 21, 2020, 2:33 PM

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\langle, \rangle

modern palm Jul 21, 2020, 2:33 PM

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yep

half storm Jul 21, 2020, 2:33 PM

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and that it's closed under scalar multiplication.

modern palm Jul 21, 2020, 2:34 PM

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thats what I learned, makes sense

half storm Jul 21, 2020, 2:35 PM

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o.k. so we'ere considering that the line is passing through the origin it follows there exists a t_0 such that r(t_0) = 0

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i.e. the zero vector.

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This is by assumption of our supposed line.

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Does that make sense?

modern palm Jul 21, 2020, 2:35 PM

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yep, makes sense

half storm Jul 21, 2020, 2:36 PM

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Cool, alright now take any two points lying on the line L, then it has to be the case that such a point in R^3 satisfies the aformentioned equation.

modern palm Jul 21, 2020, 2:36 PM

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ok

half storm Jul 21, 2020, 2:39 PM

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So I'm saying suppose that <c_1,c_2,c_3> and <d_1,d_2,d_3> \text{ are on the line, then this statment is true. That there exists t_1 and t_2 such that }
$$ \langle c_1, c_2, c_3 \rangle = \langle a_1, a_2, a_3 \rangle + t_1 \langle b_1, b_2, b_3 \rangle$$

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Ugh.

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Well I'm trying to say is <c_1, c_2, c_3 > = <a_1, a_2, a_3> + t_1 <b_1, b_2, b_3> and <d_1, d_2, d_3> = <a_1, a_2, a_3> + t_2 < b_1, b_2, b_3>

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So we sum them.

modern palm Jul 21, 2020, 2:43 PM

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so.. <c_1, c_2, c_3 > and <d_1, d_2, d_3> are two lines?

half storm Jul 21, 2020, 2:43 PM

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They are points on the line.

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We want to show that for any two point on the line that their sum is also contained on the line.

zealous widget Jul 21, 2020, 2:44 PM

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hey
would anyone here mind explaining why $P_1P_2=0$ ($P_1$ and $P_2$ being projection matrices of size $n \times n$) means that the vectors $a_1$ and $a_2$ are orthogonal? (the vectors $a_1$ and $a_2$ are the vectors that the projection matrices project onto)

stoic pythonBOT Jul 21, 2020, 2:44 PM

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catfood:

📎 467744230010519552.png

modern palm Jul 21, 2020, 2:45 PM

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what are <a_1, a_2, a_3> and < b_1, b_2, b_3>

half storm Jul 21, 2020, 2:45 PM

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<a_1,a_2,a_3> is the initial point on the line and <b_1, b_2, b_3> is the direction vector.

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<b_1, b_2, b_3> is the vector that gives you direction in which the line points.

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Does that make sense?

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Did you go over the vector equation of a line in R^3 in class?

modern palm Jul 21, 2020, 2:47 PM

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if <b_1, b_2, b_3> is a direction vector, then isnt the equation you gave in the form of l = p_0 + td, where p0 is a point and d = <b_1, b_2, b_3>. So isnt those two equations of two lines?

half storm Jul 21, 2020, 2:48 PM

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They are the same line because the two equations I gave you have the same direction vector and the same initial point on the line.

modern palm Jul 21, 2020, 2:48 PM

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ok

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oh ok i see

half storm Jul 21, 2020, 2:49 PM

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The vector equation for a line is completely determined by a point on the line and it's direction vector. It's defined uniquely in this way.

modern palm Jul 21, 2020, 2:49 PM

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they are using the same points as well as the same direction vector, its just that the two lines have different t

half storm Jul 21, 2020, 2:49 PM

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Yup

modern palm Jul 21, 2020, 2:50 PM

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so we add them?

half storm Jul 21, 2020, 2:50 PM

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because the t is the number that determines where on the line you are.

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yup

modern palm Jul 21, 2020, 2:53 PM

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ohh i think i know where this is going

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that smart lol

half storm Jul 21, 2020, 2:54 PM

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so if you add them you get <c_0, c_1, c_2> + <d_1, d_2, d_3> = <a_1, a_2, a_3> + t_1<d_1, d_2, d_2> + <a_1, a_2, a_3> + t_2<d_1,d_2,d_3>
= 2<a_1,a_2,a_3> + (t_1 + t_2) <b_1, b_2, b_3>

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I hope you can see that this is a point on the line because we have the same direction vector only scaled by a different amount and the point 2<a_1,a_2,a_3> is merely scaled by a factor 2.

modern palm Jul 21, 2020, 2:56 PM

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yep

half storm Jul 21, 2020, 2:59 PM

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There's probably a clear way of writing this in order to get rid of the factor 2 and make it explicitly clear that it's a point on the line. Maybe @cursive narwhal can answer.

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oh i know what it is

modern palm Jul 21, 2020, 3:00 PM

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I understand what you mean

cursive narwhal Jul 21, 2020, 3:00 PM

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wot

half storm Jul 21, 2020, 3:00 PM

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by our assumption it's a line passing through the origin.

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so <a_1,a_2,a_3> is necessarily zero.

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the zero vector.

modern palm Jul 21, 2020, 3:00 PM

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yep

half storm Jul 21, 2020, 3:01 PM

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So, if we had noted this in the beginning, the equation would have reduced instead from <a_1,a_2,a_3> + t<b_1,b_2,b_3> to simply t<b_1,b_2,b_3>

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O.k. now we need only to show that it's closed under scalar multiplication.

modern palm Jul 21, 2020, 3:02 PM

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yep

half storm Jul 21, 2020, 3:03 PM

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So we let k be any element of the real numbers and we need to show that for any point on the line, i.e. a point <c_1,c_2,c_3> = t_0 <b_1,b_2,b_3> for some t_0 when scaled by k k<c_1,c_2,c_3> is also a point on a line.

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so k<c_1,c_2,c_3> = k t_0 <b_1,b_2,b_3> where k * t_0 is a real number.

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and since we have the same direction vecotr, it also is point on the line.

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so L is closed under scalar multiplication

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So L is a subspace.

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@cursive narwhal I was trying to see why the factor 2 didn't go away in order to make it clear that we had another point on the line. But then I realized that this statement doesn't hold for lines in general and I had the general equation of a line in R^3.

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@cursive narwhal but then after using the information we had by assumption that it was a line through the origin it's clearly a point on the line.

cursive narwhal Jul 21, 2020, 3:06 PM

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oh lol sorry i couldn't answer

half storm Jul 21, 2020, 3:06 PM

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@cursive narwhal In short I thought the standard method of proof didn't work and was going to need some help, but I see now I didn't lol.

cursive narwhal Jul 21, 2020, 3:06 PM

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didn't know the context

half storm Jul 21, 2020, 3:06 PM

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All good.

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Does that make sense @modern palm ?

modern palm Jul 21, 2020, 3:08 PM

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yep, Im just gonna read over again so it all comes together

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thanks

half storm Jul 21, 2020, 3:08 PM

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No problem

modern palm Jul 21, 2020, 3:15 PM

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@half storm Wait, for the proof of closed under scalar multiplication, you wrote "a point <c_1,c_2,c_3> = t_0 <b_1,b_2,b_3> for some t_0"

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isnt that the equation for a line?

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why is that a point

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unless I didnt know you can write points like this

half storm Jul 21, 2020, 3:16 PM

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The equation for a line is given for r(t) = t<b_1,b_2,b_3> for any t in the reals. That is, a line is the set of all points for any t in the reals.

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when you say that a point lies on the line, you are saying that there specifically exists a t_0 in the reals such that equation is satisfied.

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A line is the set of all points that satfisfy that equation, a point on line the is just one of those points that is given by taking a specific real number and multiplying the direciton vector by that.

modern palm Jul 21, 2020, 3:17 PM

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so, t_0 would be a real number

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ok

half storm Jul 21, 2020, 3:17 PM

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Yea it's a specific real number.

modern palm Jul 21, 2020, 3:18 PM

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makes sense, thanks again

half storm Jul 21, 2020, 3:18 PM

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No problem.

zealous widget Jul 21, 2020, 3:25 PM

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hey
would anyone here mind explaining why $P_1P_2=0$ ($P_1$ and $P_2$ being projection matrices of size $n \times n$) means that the vectors $a_1$ and $a_2$ are orthogonal? (the vectors $a_1$ and $a_2$ are the vectors that the projection matrices project onto)

stoic pythonBOT Jul 21, 2020, 3:25 PM

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catfood:

📎 467744230010519552.png

half storm Jul 21, 2020, 3:27 PM

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It's a consequence of the definition of matrix multiplication.

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So when you say that a_1 and a_2 are the vectors that the matrices P1 and P2 project onto, you're saying that for any x in F^n that the P 1 projects x onto a_1, i.e. gives you how much it projects in the direction of a_1 right?

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@zealous widget Here's a pretty informal proof, I'm assuming that we are working in Euclidean space e.g. $\mathbb{R}^n$ for some natural number $n$. $ Let x \in \mathbb{R}^n$ then $(P_1P_2)(x) = 0x \implies P_1(P_2)(x) = 0$. Now $P_2(x)$ gives you the projection of $x$ onto $a_2$. Note that the projection o $x$ onto $a_2$ points in the same direction as $a_2$. Now consider $(P_2)((P_1)(x)). (P_2)((P_1)(x))$ gives you the direction of $(P_1)(x)$ in the direction of $a_2$. But we have that $(P_2)((P_1)(x)) = 0.$ Note that $(P_2)((P_1)(x))$ points in the same direction as $a_2$. To say $(P_2)((P_1)(x))$ = 0. Means that $(P_1)(x)$ is orthogonal to $a_2$ i.e. that the inner product / dot product of $(P_1)(x) \cdot a_2 = 0$ . and since $(P_1)(x)$ points in the same direction as $a_1$, we can conclude that $a_1 \cdot a_2 = 0.$

stoic pythonBOT Jul 21, 2020, 3:46 PM

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JohntheDon:

📎 671939363071983616.png

hazy gull Jul 21, 2020, 4:02 PM

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for #6 (a) in this picture, can someone tell me why you multiply the basis s by the resulting vector? It seems like you would divide not multiply, becase p(t)*s=[1,2,3] right?

📎 Screenshot_from_2020-07-21_10-47-44.png

cursive narwhal Jul 21, 2020, 4:08 PM

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?

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[1,2,3] are supposed to be the coefficients of p(t) under the basis S.

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@hazy gull

zealous widget Jul 21, 2020, 4:13 PM

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@half storm How does $P_2P_1x=0$ imply that $P_1x$ is orthogonal to $a_2$?

stoic pythonBOT Jul 21, 2020, 4:13 PM

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catfood:

📎 467744230010519552.png

hazy gull Jul 21, 2020, 4:14 PM

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think of the dot product as how much one vector is in the direction of the other

zealous widget Jul 21, 2020, 4:14 PM

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hmm

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i've never thought of that before

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thanks

hazy gull Jul 21, 2020, 4:14 PM

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np

zealous widget Jul 21, 2020, 4:15 PM

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@hazy gull wait

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what I don't understand here though

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is how $P_2$ represents $a_2$ in some way

stoic pythonBOT Jul 21, 2020, 4:15 PM

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catfood:

📎 467744230010519552.png

hazy gull Jul 21, 2020, 4:16 PM

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what is a2

half storm Jul 21, 2020, 4:17 PM

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@zealous widget P_2 is the projection of matrix onto a_2. Meaning that any vector that you take in x R^n, (P_2)(x) gives you a new vector that is in the same direction of a_2. i.e. it is a scalar multiple of a_2. This is what a projection matrix does

zealous widget Jul 21, 2020, 4:18 PM

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hmm

half storm Jul 21, 2020, 4:19 PM

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If you've taken multivariable calculus, then this is taught there and there is actually a formula for the projection that is given. What you are doing in LA is giving that linear opeartor in the form of a matrix.

zealous widget Jul 21, 2020, 4:20 PM

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but $P_2P_1x=0$ still doesn't imply that $P_1x$ is orthogonal to $a_2$
as in I don't get where how $P_2^T$ represents a vector on $a_2$

stoic pythonBOT Jul 21, 2020, 4:20 PM

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catfood:

📎 467744230010519552.png

half storm Jul 21, 2020, 4:20 PM

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P_2T is irrelevant to this proof.

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You can probably use something about it for different proof but I'm using a different way of proving it.

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(P_2)(x) is a vector right, because P_2 is a matrix that takes vectors in R^n and maps them to vectors in R^n right?

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Does that make sense?

zealous widget Jul 21, 2020, 4:22 PM

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Yes

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but there is only 1 vector in $P_2P_1x=0$

stoic pythonBOT Jul 21, 2020, 4:22 PM

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catfood:

📎 467744230010519552.png

zealous widget Jul 21, 2020, 4:23 PM

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in fact the end product of that equation should be a vector

half storm Jul 21, 2020, 4:23 PM

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Right

zealous widget Jul 21, 2020, 4:23 PM

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not trying to disagree with you or anything

half storm Jul 21, 2020, 4:23 PM

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It is.

zealous widget Jul 21, 2020, 4:23 PM

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i'm just having a hard time understanding

half storm Jul 21, 2020, 4:23 PM

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I know.

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so x is a vector in R^n so (P_1)(x) is also a vector in R^n

zealous widget Jul 21, 2020, 4:24 PM

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Yes

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understood

half storm Jul 21, 2020, 4:25 PM

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and more importantly it is a vector that points in the same direction as a e.g. If a is a vector in R^3 <a_1,a_2,a_3> , then (P_1)(x) = k<a_1,a_,2_<a_3> for some k in reals

zealous widget Jul 21, 2020, 4:25 PM

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yep

half storm Jul 21, 2020, 4:27 PM

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So $P_2(P_1(x))$ is giving giving you how much of the vector (P_1)(x) is in the direction of a_2.

zealous widget Jul 21, 2020, 4:27 PM

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correct

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but the result is a vector

half storm Jul 21, 2020, 4:28 PM

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Oh yes, I see what you're saying.

zealous widget Jul 21, 2020, 4:32 PM

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@half storm very bad timing, i gotta go in a few min for martial art training lol

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finish your message

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and we can do it again in 40 min lol

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maybe in DMs

half storm Jul 21, 2020, 4:37 PM

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So we know that (P_2)(P_1(x)) gives us a vector that is in the direction of the a_2. e.g. if a_2 = <a_1,a_2,a_3> , then P_2((P_1)(x)) = k <a_1,a_2,a_3>.
But we also know by assumption that P_2((P_1)(x)) = 0. So k <a_1,a_2,a_3>. = (0,0,0). which means that it has to always be that k = 0.

zealous widget Jul 21, 2020, 4:38 PM

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right i really gtg now lol

half storm Jul 21, 2020, 4:38 PM

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lol go ahead.

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I'll just @ you.

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@zealous widget O.k. I just thought of a better proof in my head that is more streamlined. It requires a little bit of knowledge of the definition of multivariable calculus. I see that you are familiar with physics. So you should know that the general definition of the projection of a vector ,say a, onto another vector b in $R^n$ is given by $$ \frac{a \cdot b}{|b|^2} b$$

stoic pythonBOT Jul 21, 2020, 4:47 PM

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JohntheDon:

📎 671939363071983616.png

half storm Jul 21, 2020, 4:48 PM

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@zealous widget Consider the vector $a_1$, we know that $(P_1)(a_1) = a_1 $

stoic pythonBOT Jul 21, 2020, 4:48 PM

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JohntheDon:

📎 671939363071983616.png

half storm Jul 21, 2020, 4:49 PM

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@zealous widget So $P_2(P_1(a_1)) = P_2(a_1)$

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@zealous widget Now $P_2(a_1) = ka_2 = 0 $ $k \in \mathbb{R}$ by definition of vector projection, and where 0 is the zero vector And more specifically, we know that k is defined by the aforementioned formula $$ k = \frac{a_1 \cdot a_2}{|a_2|^2} a_2$$

cursive narwhal Jul 21, 2020, 4:55 PM

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\text{by definition of vector projection......}

dusky epoch Jul 21, 2020, 4:55 PM

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you do realize you can use dollars to delimit math mode so you won't have to spam \text everywhere right

half storm Jul 21, 2020, 4:55 PM

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yea

dusky epoch Jul 21, 2020, 4:56 PM

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Now $P_2(a_1) = ka_2 = 0$ for some $k \in \bR$ by definition of vector projection...

stoic pythonBOT Jul 21, 2020, 4:56 PM

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Ann:

📎 183668144404037632.png

dusky epoch Jul 21, 2020, 4:56 PM

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etc

half storm Jul 21, 2020, 4:56 PM

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oh nvm.

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That is different than what I saw thinking

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@zealous widget So $ k = 0 \implies \frac{a_1 \cdot a_2}{|a_2|^2} a_2 = 0 \implies a_1 \cdot a_2 = 0$ ; Therefore $a_1$ and $a_2$ are orthogonal.

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lol the struggle.

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I gotta read a TeX primer or something

stoic pythonBOT Jul 21, 2020, 5:03 PM

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JohntheDon:

📎 671939363071983616.png

#

JohntheDon:

📎 671939363071983616.png

#

JohntheDon:

📎 671939363071983616.png

#

JohntheDon:

📎 671939363071983616.png

#

JohntheDon:

📎 671939363071983616.png

zealous widget Jul 21, 2020, 5:17 PM

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aight

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@half storm session finished

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lemme just assimilate this stuff first

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crap i gotta shower

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another 15 min i guess

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goddamn sweat

delicate zealot Jul 21, 2020, 5:39 PM

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So I just learned about cofactor expansion along rows and columns to compute determinants and it got me thinking: in the case where you want to compute the determinant and in row reductions there is a row of zeros, what's stopping me from doing cofactor expansion along that row? The determinant would then obviously be zero, and I realize that doing row operations would scale the determinant and change the sign depending on the row reductions you did, but I just have a "bad feeling" about this I guess and that it would change the determinant significantly if I did this.

zealous widget Jul 21, 2020, 5:42 PM

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right

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fully back now

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i'll admit

half storm Jul 21, 2020, 5:42 PM

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pretty sure there are theorems that the determinant is unique and that the cofactor expansion along any row is the same.

zealous widget Jul 21, 2020, 5:42 PM

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i have no fucking clue what multivariable calculus is

half storm Jul 21, 2020, 5:43 PM

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and that if there exist a row of zeroes, then the determinant is zero.

zealous widget Jul 21, 2020, 5:43 PM

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but

half storm Jul 21, 2020, 5:43 PM

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so I think it's o.k.

zealous widget Jul 21, 2020, 5:43 PM

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i understand the proof partially

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havent finished reading ir yt

delicate zealot Jul 21, 2020, 5:43 PM

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ok I knew about the theorems I just didn't want to believe it lol

zealous widget Jul 21, 2020, 5:43 PM

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*it yet

half storm Jul 21, 2020, 5:43 PM

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:LUL:

zealous widget Jul 21, 2020, 5:44 PM

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the multivariable calculus formula you described is exactly the same as the one in my linear algebra textbook

half storm Jul 21, 2020, 5:44 PM

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lol i understand @delicate zealot seems too good to be true.

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Oh o.k.

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so that's good

zealous widget Jul 21, 2020, 5:44 PM

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except yours doesnt use transposes lmao

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right lemme finish reading le holy proof

half storm Jul 21, 2020, 5:45 PM

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Are you a CS student?

zealous widget Jul 21, 2020, 5:45 PM

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@half storm your proof follows

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actually no

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i was hoping nobody would ask

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so i wouldn't seem arrogant and braggy

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but I'm a 13 year old asian kid

half storm Jul 21, 2020, 5:46 PM

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That's great lol

zealous widget Jul 21, 2020, 5:46 PM

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i mean

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not to disclose my personal story but

half storm Jul 21, 2020, 5:46 PM

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If you're doing LA now, then you're way ahead of the game.

zealous widget Jul 21, 2020, 5:47 PM

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i was an arrogant dick 1 - 2 years ago

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being ahead pumped a sense of superiority into me

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now i dont want to return to that phase lol

delicate zealot Jul 21, 2020, 5:47 PM

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@zealous widget you very much are not the only one that's struggled with this, I was an ass in calc 1

zealous widget Jul 21, 2020, 5:47 PM

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If you're doing LA now, then you're way ahead of the game.
@half storm true

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@zealous widget you very much are not the only one that's struggled with this, I was an ass in calc 1
@delicate zealot haha lmao

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where was i

half storm Jul 21, 2020, 5:48 PM

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Yea, I think every person who goes through this phase at some point. You feel very accomplished. Eventually you learn to stop being this because it's going to be hard to make friends or you get humbled when the material gets more difficult.

zealous widget Jul 21, 2020, 5:49 PM

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@half storm your proof follows
but its a bit disjointed to read in its current form

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i'll have to write it out again

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when i get back to my pc

half storm Jul 21, 2020, 5:49 PM

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Yea It's not pretty.

zealous widget Jul 21, 2020, 5:53 PM

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Yea, I think every person who goes through this phase at some point. You feel very accomplished. Eventually you learn to stop being this because it's going to be hard to make friends or you get humbled when the material gets more difficult.
@half storm all kids go through an arrogant attention/validation seeking phase, though this lasts different times for different people. I tend to think it lasts longer for those who are spoiled more in some way - such as "cool" kiddos in secondary school (high school for the non british)

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altho this isnt a psychology server

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so we should stop here lol

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@half storm thanks for the proof man

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greatly appreciated

half storm Jul 21, 2020, 5:55 PM

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No problem 😄

zealous widget Jul 21, 2020, 5:55 PM

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love ya <3

half storm Jul 21, 2020, 5:56 PM

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❤️

zealous widget Jul 21, 2020, 5:56 PM

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💝

delicate zealot Jul 21, 2020, 6:01 PM

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@half storm dude I still couldn't shake it even with the theorems so I messed around with a small 3x3 matrix and found the determinant before and after row reductions and whaddya know the determinant stayed 0 the whole time 😄

📎 unknown.png

half storm Jul 21, 2020, 6:02 PM

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@delicate zealot lol yea, Sometimes the only way you're really gonna convince yourself is by doing what you're doing i.e. working some examples. And then also looking at the proofs.

ocean sequoia Jul 21, 2020, 6:03 PM

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Determinants suck

delicate zealot Jul 21, 2020, 6:03 PM

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my professor gave me a giant 6x6 matrix and asked if it was invertible and I found that I could take a row and add it to another to get a row of zeros and I was like wait can I just say that the determinant is 0?

half storm Jul 21, 2020, 6:04 PM

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Whenever, there is a proof that gives an amazing result that makes the whole problem way easier I literally have to go check the proof and do some examples because it's just so convenient and I'm like "Thank the lord 🙏" that this works. Lol gotta check it instead of taking it for granted.

delicate zealot Jul 21, 2020, 6:04 PM

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cause this looks nasty otherwise

📎 unknown.png

umbral smelt Jul 21, 2020, 6:05 PM

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I think that's not invertible

hazy gull Jul 21, 2020, 6:05 PM

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so I still dont get the anwer to #6 a. shouldnt it be the S^(-1)*[1,2,3]? because I'm translating something with respect to S back to the normal basis

📎 Screenshot_from_2020-07-21_10-47-44.png

umbral smelt Jul 21, 2020, 6:05 PM

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take a look at row 1 and row 3

ocean sequoia Jul 21, 2020, 6:05 PM

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Pretty positive if all the rows are independent it’s invertable

umbral smelt Jul 21, 2020, 6:08 PM

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When you apply r1+r3, you'll get zeroes on the 1st row, and when you get full zeroes on a row or a column, you'll get that the determinant is 0

hazy gull Jul 21, 2020, 6:08 PM

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^^

half storm Jul 21, 2020, 6:08 PM

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Oh yea definitely non invertible then.

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@delicate zealot You can do what you did and show that the determinant is zero and thus the matrix is non-invertible.

delicate zealot Jul 21, 2020, 6:09 PM

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Yeah that's what I'm in the process of doing right now 🙂

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well actually I'm doing this:
$\det B = (-1)^1 \det A$

stoic pythonBOT Jul 21, 2020, 6:10 PM

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super64guy:

📎 275399578965114881.png

delicate zealot Jul 21, 2020, 6:11 PM

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because it took 1 row reduction to get to a new matrix and then I compute the determinant of B and divide by (-1)^1 to get the determinant of A (which will become 0)

hazy gull Jul 21, 2020, 6:28 PM

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would [p(x)]_s be p(x) from the normal basis written in the s basis or p(x) in the s basis written using the normal basis?

gray dust Jul 21, 2020, 6:30 PM

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not sure where normal basis comes in. p(x) is just p(x), [p(x)]_s is likely the coords of p(x) wrt the s basis

half storm Jul 21, 2020, 6:30 PM

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^

hazy gull Jul 21, 2020, 6:34 PM

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so [p(x)]_s should be p(x) * s^(-1)

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?

gray dust Jul 21, 2020, 6:34 PM

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idk what that means

hazy gull Jul 21, 2020, 6:36 PM

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vector p with respect to s is the vector p multiplied by the inverse of s?

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because im translating p to s

gray dust Jul 21, 2020, 6:38 PM

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s is a basis, no? idk what you're thinking of in saying the "inverse" of s

hazy gull Jul 21, 2020, 6:38 PM

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yes s would be a basis

gray dust Jul 21, 2020, 6:38 PM

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a basis is simply a linearly independent list of vectors that spans the vector space you're looking at, idk how you talk of an "inverse" of that

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just to make sure everything's clear, take any vector v in the space. let B=(b_1,...,b_n) be a basis, so there exist unique scalars c_1,...,c_n where v=c_1b_1+...+c_nb_n. this is alternatively said as, the coords of v wrt B are [v]_B=(c_1,...,c_n)

half storm Jul 21, 2020, 6:51 PM

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@hazy gull I think I understand your question. That notation can refer to two things, If T is a linear operator on a vector space V and $\Beta$ is a basis for V, then the matrix representation is denoted by ${[T]}{\Beta} $ If you are talking about a coordinate vector of a vector say $x \in V$ with respect to a basis $\Beta$ for V, then the notation is given as ${[v]}{\Beta}$

stoic pythonBOT Jul 21, 2020, 6:53 PM

#

JohntheDon:
Compile Error! Click the errors reaction for details. (You may edit your message)

📎 671939363071983616.png

half storm Jul 21, 2020, 6:54 PM

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I can't get it to TeX correctly but what you're notation was suggesting woudl have to refer to what @gray dust is talking about because a poynomial is not a linear operator on a vector space.

hazy gull Jul 21, 2020, 7:01 PM

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O ya sorry, I'm just trying to figure out the notation for a vector with respect to a basis and also for the change of basis because I'm confused by what []_s means exactly

gray dust Jul 21, 2020, 7:04 PM

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before going into change of basis, make sure you have this down

just to make sure everything's clear, take any vector v in the space. let B=(b_1,...,b_n) be a basis, so there exist unique scalars c_1,...,c_n where v=c_1b_1+...+c_nb_n. this is alternatively said as, the coords of v wrt B are [v]_B=(c_1,...,c_n)
see how taking a particular linear combo of the b's produces v, you take the scalars in front of the b's and put them into a list, then you have the coordinates of v wrt B

supple hemlock Jul 21, 2020, 7:39 PM

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Hey, I need some help with this problem

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https://prnt.sc/tm4n10

Lightshot

Screenshot

Captured with Lightshot

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📎 8i4kLm3XFGYAAAAASUVORK5CYII.png

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I know that det(A*A-1) = det(I) = 1

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Why + or - 1?

limber sierra Jul 21, 2020, 7:41 PM

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first since the entries of C and C^-1 are integers, det(C) and det(C^-1) should also be integers (can you see why? you'll need to prove this if you havent already)

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now since det(A*A^-1) = 1

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you can apply the fact that det(XY) = det(X)det(Y)

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to get det(A) * det(A^-1) = 1

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but note that both det(A) and det(A^-1) are integers

zealous widget Jul 21, 2020, 7:41 PM

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@half storm just an update to say your proof worked :D

limber sierra Jul 21, 2020, 7:41 PM

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so for this fact to be true

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either they're both 1, or they're both -1

half storm Jul 21, 2020, 7:42 PM

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Great 😄

supple hemlock Jul 21, 2020, 7:42 PM

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aij = aji/det(A) for all ij for A^-1

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if det(A) = 1, aij = aji

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so they're related by an integer

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OHH, I SEE IT NOW

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THANK YOU @limber sierra

hazy gull Jul 21, 2020, 11:58 PM

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@gray dust ok that makes sense so far, so can you give an example of what a change of basis will look like

gray dust Jul 22, 2020, 12:05 AM

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change of basis is useful but there's not much material to cover to learn them. first you should cover the matrix representation of a linear map wrt bases of its domain & codomain. once you got that, then all you need to know next is that a change of basis matrix from bases A to B is the matrix of the identity map wrt A & B, ie if P is that change of basis matrix, then for any vector x, P acts upon the coords of x wrt A to produce the coords of x wrt B

half storm Jul 22, 2020, 12:10 AM

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Need some help proving this one. I'm pretty sure I already know how to prove that this is linearly independent. I need to show that it generates V and that's where I'm struggling.

📎 Capture.PNG

hollow finch Jul 22, 2020, 12:22 AM

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you might be able to use the fact that n linearly independent n-dimensional vectors will always span an n-dimensional space.

half storm Jul 22, 2020, 12:25 AM

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god damn it.

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🤦‍♂️

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That would certainly do it.

spice ruin Jul 22, 2020, 1:25 AM

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📎 unknown.png

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Can someone explain how I back substitute for the last step? I already got it to echlon form on my own and couldnt figure out how the free variable interacts with back substitution just from looking at the answer

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📎 unknown.png

hollow finch Jul 22, 2020, 1:28 AM

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thats a great question

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reversing the matrix notation, the first two lines of your matrix say
$$x_1+x_3=4$$
$$-x_2+x_3=1$$

stoic pythonBOT Jul 22, 2020, 1:29 AM

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nix:

📎 196803594576592896.png

hollow finch Jul 22, 2020, 1:30 AM

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since x_3 is in both equations, it makes it the best choice for your free variable.

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so solve for the others in terms of x_3 and then youll be able to split the vectors

shy mango Jul 22, 2020, 1:32 AM

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hi all, sorry to interrupt. i was wondering if i have a matrix A with nullity that is 0, how do i prove that every b in C(A) has a unique solution for the equation Ax=b? thanks in advance

hollow finch Jul 22, 2020, 1:32 AM

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by using
$$\begin{pmatrix}x_1\x_2\x_3 \end{pmatrix}$$ as your solution

stoic pythonBOT Jul 22, 2020, 1:32 AM

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nix:

📎 196803594576592896.png

hollow finch Jul 22, 2020, 1:32 AM

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@shy mango channel is in use right now, try one of the questions channels #❓how-to-get-help

slow scroll Jul 22, 2020, 1:33 AM

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404, lets go to #help-5

shy mango Jul 22, 2020, 1:34 AM

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@hollow finch thanks for the tip lol, ill read it

hollow finch Jul 22, 2020, 1:34 AM

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np, we all start somewhere catthumbsup

spice ruin Jul 22, 2020, 1:35 AM

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oh dang im dumb. i was interpreting "in terms on x_3" as "solve for x_3"

hollow finch Jul 22, 2020, 1:36 AM

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nah youre not dumb. this stuff isnt easy

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probably one of the most confusing parts of the beginning of a LA course imo

spice ruin Jul 22, 2020, 1:37 AM

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I dont think its too bad yet, just trying to get ahead a little since i have a proofs + linear combo starting in a month

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and proofs are scary

hollow finch Jul 22, 2020, 1:38 AM

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good for you for getting ahead. yeah LA is a really beautiful subject, but many people struggle with the proofs.

spice ruin Jul 22, 2020, 1:39 AM

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LA seems pretty cool thus far, I was watching some higher level stuff that I thought was rad and everyone is telling me i need to learn LA before i make an actual attempt at some of the other stuff

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epic math time has some coolvideos about representation theory but its pretty incomplete as a learning source

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i took multivar already and really loved the vector calc portion too

half storm Jul 22, 2020, 1:47 AM

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I'm trying to refine my knowledge of LA so I can actually get to doing some of the more fun stuff and really understand it i.e. multivariable analysis, measure theory stats and some comp sci stuff

spice ruin Jul 22, 2020, 1:52 AM

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rad

dire bough Jul 22, 2020, 2:02 AM

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LA is powerful stuff.

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For measure theory though, you'll want analysis more than anything else.

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But yeah, if you're doing anything applied or computational, LA is an absolute must.

cobalt tartan Jul 22, 2020, 3:21 AM

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Quick question: For some Ax = b, where A is some matrix, if I perform elementary row operations to it to obtain A', does A'x = b? Or does it equal instead some b'?

half storm Jul 22, 2020, 3:26 AM

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The solution set of a row reduced echelon matrix has the same solution set as the original matrix. So I'm positive thats a no.

cobalt tartan Jul 22, 2020, 3:27 AM

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A no to?

half storm Jul 22, 2020, 3:28 AM

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I'm saying that b = b'.

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You can probably prove it like this.

cobalt tartan Jul 22, 2020, 3:29 AM

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Ah yea

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Ok that's what I thought

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Naw it's for a non LA course, they want me to prove something else and I was like "hey wait the deterimnat doesn't change if I go and do adding/subtracting of rows"

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LA showing up in other courses epically fatpepedab

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saving my ass

wintry steppe Jul 22, 2020, 3:32 AM

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to answer your original question: performing elementary row operations corresponds to left-multiplying by an elementary matrix. so if Ax = b and E is an elementary matrix corresponding to some row operation, then applying that row operation gives you A'x = Eb, so it will, as you say, instead be some b'

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(assuming A' is the matrix EA, which is obtained from A by performing the row operation that E corresponds to)

half storm Jul 22, 2020, 3:33 AM

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Suppose that A' is the row reduced echelon form of A, so we have $Ax = b$ and $A' = (E_1E_2 \dots E_N)A \implies A'x = b' \implies (E_1E_2 \dots E_N)A x = b' \implies (E_1E_2 \dots E_N) b = b'$

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@cobalt tartan I got you I'm proving it for myself to see if my intution is correct

cobalt tartan Jul 22, 2020, 3:34 AM

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Ah

half storm Jul 22, 2020, 3:34 AM

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Just to see if my intution is correct.

stoic pythonBOT Jul 22, 2020, 3:35 AM

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JohntheDon:

📎 671939363071983616.png

half storm Jul 22, 2020, 3:35 AM

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So i was wrong. 😦

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So just because they have the same solution set doesn't mean that they necessarily map each x to the same thing.

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It's just the images of the matrices are equal.

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@vague canyon glad you came to say the right shit.

wintry steppe Jul 22, 2020, 3:37 AM

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wrong @ ?

half storm Jul 22, 2020, 3:38 AM

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I said that they have the same solution set, but I thought that for any x that $Ax = A'x$ basically.

stoic pythonBOT Jul 22, 2020, 3:38 AM

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JohntheDon:

📎 671939363071983616.png

half storm Jul 22, 2020, 3:38 AM

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which isn't true.

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But that's what I told her.

wintry steppe Jul 22, 2020, 3:40 AM

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as long as they get the correct solution and as long as everyone involved works towards the right answer

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no big deal

wintry steppe Jul 22, 2020, 4:44 AM

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@dusky epoch can you explain permutation matrices to me

dusky epoch Jul 22, 2020, 4:45 AM

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what about 'em

wintry steppe Jul 22, 2020, 4:45 AM

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yeah exactly

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what is the point of them

dusky epoch Jul 22, 2020, 4:47 AM

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well, for every permutation on n points there's a corresponding linear map on R^n which permutes the coordinates accordingly

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its matrix in the standard basis is the corresponding permutation matrix

wintry steppe Jul 22, 2020, 4:47 AM

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but what is the point? use?

dusky epoch Jul 22, 2020, 4:48 AM

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i don't know how to answer that lmao

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i'm really not sure what you're looking for

wintry steppe Jul 22, 2020, 4:48 AM

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i am not sure either

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my book spends a whole subchapter on them

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but i don't know why

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i am just so confused

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like i don't understand why i had to read that

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i don't feel like i know anything new or that i should know?

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are there any applications of them..

limber sierra Jul 22, 2020, 4:52 AM

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the fact that the elementary row operations correspond to left-multiplication by elementary matrices is quite useful

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permutation matrices of course corresponding to permuting rows

dire thunder Jul 22, 2020, 4:53 AM

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some linear transformations are literally just permutation of basis vectors

dusky epoch Jul 22, 2020, 4:53 AM

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"bUt WhAt ArE tHe ApPlIcAtIoNs"

wintry steppe Jul 22, 2020, 4:54 AM

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ann

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i don't know why you're doing that

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there's a clear, to say, the determinant

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or even row echelon form

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i don't see how that is the case here lol

limber sierra Jul 22, 2020, 4:54 AM

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is the ability to work algebraically with permutations not of value to you?

dire thunder Jul 22, 2020, 4:55 AM

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sully

limber sierra Jul 22, 2020, 4:55 AM

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like literally every proof involving row reduction can be made far simpler with elementary matrices

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since it can be understood entirely in terms of multiplication by (invertible!) matrices

dusky epoch Jul 22, 2020, 4:55 AM

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nah namington who needs simple proofs? real proofs take up 100 pages each and require dozens of mathematicians to pore over /j

limber sierra Jul 22, 2020, 4:55 AM

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(with nice det properties!)

wintry steppe Jul 22, 2020, 4:56 AM

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sigh ann

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@limber sierra i don't think i was told about this....

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so i don't know how to use them for that lol

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broke

limber sierra Jul 22, 2020, 4:56 AM

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what the fuck did they teach then

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if they spent a subchapter on them

wintry steppe Jul 22, 2020, 4:56 AM

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lol

limber sierra Jul 22, 2020, 4:56 AM

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without giving their motivating theorem

quartz compass Jul 22, 2020, 4:56 AM

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at some point, you're gonna have to come up with your own context for why things matter to you

wintry steppe Jul 22, 2020, 4:57 AM

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that's a good question namington

limber sierra Jul 22, 2020, 4:57 AM

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in any case, permutation matrices are also an easy way to think of permutations, well, linearly

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this is naturally useful in group theory

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(and combinatorics kinda although usually combinatorial techniques are more high-tech)

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reducing mathematics problems to linear algebra problems is very useful in practice

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since we understand linear algebra very very well

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the existence of permutation matrices, and the fact that they obey fairly "nice" properties, gives us an easy way to discuss permutations (and hence the symmetric group and etcetc) through LA

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which again, we understand quite well

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this is the usual way we formulate the representation theory of S_n for instance

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but thats probably beyond your scope

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for now the best way to think of them is "they give us ways to talk about gaussian elimination via left-multiplication by matrices with nice properties (invertibility, determinant one of two values, etc)"

wintry steppe Jul 22, 2020, 5:01 AM

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ok

limber sierra Jul 22, 2020, 5:01 AM

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"and matrix multiplication is fairly 'easy' to reason about"

wintry steppe Jul 22, 2020, 5:01 AM

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mechanical

compact basin Jul 22, 2020, 1:18 PM

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Hello. I’ve been studying DFT

https://youtu.be/ds0cmAV-Yek

To implement this in python language, I would like to understand how this work. But I could not find a good lecture notes or resources that is directly related to this video. Anyone has any recommendations? Thank you in advance

YouTube

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Doga's a super smart dude who writes a Turkish blog "Bi Lim Ne Güzel ...

▶ Play video

#

What I am highly interested in is what is the input, how does it be deal with, then what is the output. I reckon that this must be in linear algebra, thus I asked here

polar cloud Jul 22, 2020, 1:29 PM

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If it’s not a bother, ping me when this question is answered! I have an urgent question but don’t want to impose

zealous widget Jul 22, 2020, 3:44 PM

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is it possible to have a matrix satisfying $Ax=x$, where x is a vector of dimensions $n \times 1$ and where A is a matrix of dimension $n \times n$

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where's texit?

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is it possible to have a matrix satisfying $Ax=x$, where x is a vector of dimensions $n \times 1$ and where A is a matrix of dimension $n \times n$ and where $A \neq I$

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$A$

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,help

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,help

stoic pythonBOT Jul 22, 2020, 3:45 PM

#

A brief description and guide on how to use me was sent to your DMs! Please use ,list to see a list of all my commands, and ,help cmd to get detailed help on a command!

zealous widget Jul 22, 2020, 3:46 PM

#

i'll copy a previous working message to see if that works]

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I said that they have the same solution set, but I thought that for any x that $Ax = A'x$ basically.

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what

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is it possible to have a matrix satisfying $Ax=x$, where x is a vector of dimensions $n \times 1$ and where A is a matrix of dimension $n \times n$ and where $A \neq I$

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why it no work

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oh nvm

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im stupid

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answer is yes

dusky epoch Jul 22, 2020, 3:53 PM

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are you asking whether it's possible to have $(\forall x \in \bR^n)(Ax = x)$ but $A \neq I$?

zealous widget Jul 22, 2020, 3:53 PM

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not for all x

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i meant for some specific x

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i'm stupid]

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it's true

dusky epoch Jul 22, 2020, 3:53 PM

#

so then you're asking if it's possible for a matrix to have 1 as eigenvalue

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fkgjlsdfkhsf

zealous widget Jul 22, 2020, 3:54 PM

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i haven't learnt about eigenvalues yet

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that's 2 - 3 chapters away in my textbook

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but yes its easily provable to be true

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i'm an idiot

#

sorry

steady fiber Jul 22, 2020, 6:09 PM

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also look into Fast Fourier Transforms

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implementing a DFT in code naively is pretty slow (like really slow, dont do it that way)

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the linear algebra behind it is quite simple, you simply multiply your input by a dft matrix

viscid kernel Jul 22, 2020, 9:20 PM

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Anyone know what the notation in red means ? Btw, “stel” means “suppose”

📎 image0.jpg

half storm Jul 22, 2020, 9:22 PM

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I'm guessing the family of functions that are continuous on the set [a,b]

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Yea that has to be it.

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Because they tell you what the set is equal to on the right.

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Not sure what language that is but "f continu" seems very much like " f continuous"

gray dust Jul 22, 2020, 9:23 PM

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yes V is the vector space of continuous real valued functions on the interval [a,b]

viscid kernel Jul 22, 2020, 9:25 PM

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Thanks but, whats the meaning of notation which comes before [a,b]

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@half storm continu means continuous xD its in Dutch

latent ledge Jul 22, 2020, 9:25 PM

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Let $A$ and $B$ be nonsimilar $n\times n$ complex matrices with both the same minimal polynomial and the same characteristic polynomial. Show that $n\geq4$ and that the common minimal polynomial does not equal the common characteristic polynomials.

stoic pythonBOT Jul 22, 2020, 9:25 PM

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Konoha:

📎 628829193270198283.png

gray dust Jul 22, 2020, 9:26 PM

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if S is a set, C(S) sometimes denotes a set of continuous functions on S. its definition is provided on the right anyway

viscid kernel Jul 22, 2020, 9:27 PM

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Thanks

gray dust Jul 22, 2020, 9:27 PM

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no prob

latent ledge Jul 22, 2020, 9:30 PM

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for my question, suppose that A and B are 4 by 4, and their minimal poly is like p(x)=x^4

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they have different jordan forms, I suppose

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<@&286206848099549185>

compact basin Jul 22, 2020, 10:17 PM

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@gleaming adder ok. thank you very much!

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@steady fiber I dunno what the input is. But multiplication ok. I’ll google it

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@gleaming adder I know such useful libraries, but as I said, what I want to know is that the mathematics to visualize any picture with DFT/FFT.

steady fiber Jul 22, 2020, 10:24 PM

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input is function you want a DFT of

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sampled at discrete, regular intervals

median forum Jul 22, 2020, 10:26 PM

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start by knowing the basics of a fourier transform
then a discrete fourier transform
then finally, how the fft makes the operations faster by conveniently changing the matrix multiplications

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but really basic stuff
like what does it input, what are you supposed to get
what is the expected frequency space distribution given an input
and so on

compact basin Jul 22, 2020, 10:37 PM

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But when and how will the actual image dataset (pixel) be used by FFD ? I supposed that it would be the input

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Also I have no idea. How to create each circle.....

📎 image0.png

steady fiber Jul 22, 2020, 10:40 PM

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you want to create the animation?

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because that's different from just doing the DFT

compact basin Jul 22, 2020, 10:49 PM

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If I make you wrong, sorry. I do want to make an animation with python language. Input is any image

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The image is exactly what I wanna implement

steady fiber Jul 22, 2020, 10:51 PM

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for that you need to use complex numbers, if you want to animate it for a 2D image

compact basin Jul 22, 2020, 10:52 PM

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Yeah I want to animate 2D

ocean sequoia Jul 23, 2020, 1:27 AM

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📎 Screen_Shot_2020-07-22_at_9.27.29_PM.png

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im actually kinda frustrated i cant figure this one out

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im assuming it has to do with the rank nullity theorem

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because if range T(V) was zero entries it would just take it to the null space but i cant figure out how to state it rigorously

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i think thats why it is atleast or did you mean me?

stiff frost Jul 23, 2020, 1:31 AM

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Ignore what I deleted, it was a misunderstanding

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What could you say about the range of T if you only had a few nonzero entries (less than dim W, say)?

ocean sequoia Jul 23, 2020, 1:32 AM

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it wouldnt be able to form a basis for W

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because dim T(v) < dim(W)

stiff frost Jul 23, 2020, 1:33 AM

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The range of T usually isn't a basis for anything

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T doesn't have a dimension

ocean sequoia Jul 23, 2020, 1:33 AM

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err excuse me the image here

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so dim T(v)

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in LADR they do refer to dim range T

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📎 Screen_Shot_2020-07-22_at_9.34.34_PM.png

stiff frost Jul 23, 2020, 1:35 AM

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So it seems you already have an understanding of why the matrix of T has at least dim range T entries

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when you made a claim about "because dim T(V)"

ocean sequoia Jul 23, 2020, 1:37 AM

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well if dim T(V) < dim(W) then it cant be surjective for W which means it cant form a basis for W

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could just something simple

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like

stiff frost Jul 23, 2020, 1:38 AM

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We don't really care about a basis for W

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I regret bringing up dimW

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because the problem doesn't refer to it

ocean sequoia Jul 23, 2020, 1:39 AM

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it does say with respect to basis

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wasnt sure if maybe there was like a contridiction in there

stiff frost Jul 23, 2020, 1:39 AM

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Every matrix for a linear transformation is "with respect to a basis pair"

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the problem was just emphasizing that the choice of basis pair doesn't matter

ocean sequoia Jul 23, 2020, 1:40 AM

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oh

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ok i forgot all about dim(W)

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doesnt exist any more

stiff frost Jul 23, 2020, 1:41 AM

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There are few ways to think about this. We could start by thinking about the extreme case: like what if for some basis the matrix had no nonzero entries? What would that tell you about T?

ocean sequoia Jul 23, 2020, 1:42 AM

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huh let me think hang on

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honestly im not sure

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my gut wants me to say something like every vector in domain

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is needed

stiff frost Jul 23, 2020, 1:44 AM

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Ok, let's forget this problem for a bit. What is "the matrix for a transformation T in L(V,W) with respect to bases B and C"?

ocean sequoia Jul 23, 2020, 1:45 AM

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its the linear transformation that takes a B a basis of a subspace V to the basis C of the subspace W

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that is uniquely determined on B and C

stiff frost Jul 23, 2020, 1:46 AM

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False

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For a couple reasons

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Firstly, not every transformation T in L(V,W) takes a basis to a basis

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Secondly, and maybe more importantly, a matrix isn't a linear transformation

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Do you have a book where you can review the definition of L(V,W) and "matrix for a transformation with repsect to a pair of bases"?

ocean sequoia Jul 23, 2020, 1:47 AM

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i do but it refers to matricies

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as linear maps

stiff frost Jul 23, 2020, 1:48 AM

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maybe it does, but I bet it doesn't refer to "matrices with respect to a pair of bases" as linear maps

ocean sequoia Jul 23, 2020, 1:49 AM

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📎 Screen_Shot_2020-07-22_at_9.48.55_PM.png

stiff frost Jul 23, 2020, 1:49 AM

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I have no objection to those sentences. (and those sentences don't refer to matrices as linear maps, but maybe other sentences in your book do)

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But anyway, you need to know the definition of the matrix of a linear map with respect to a pair of bases

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we can't solve this problem without it, since the problem is about those matrices

ocean sequoia Jul 23, 2020, 1:53 AM

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the definition of a matrix of a linear map with respect to a pair of bases is the linear map $T(v_1,...,v_2)$ to $A_1,_k w_1,...A_m,_k w_m)$

stoic pythonBOT Jul 23, 2020, 1:54 AM

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brzig:

📎 571557137118003201.png

ocean sequoia Jul 23, 2020, 1:54 AM

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where v1,...vn and w1,..,wm are basis of V and W

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why is that definition wrong

half storm Jul 23, 2020, 1:54 AM

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A matrix is a linear transformation

ocean sequoia Jul 23, 2020, 1:54 AM

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Thats what i thought...

stiff frost Jul 23, 2020, 1:55 AM

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It's confusing to call a matrix a linear transformation when we're in a context where a linear transformation will have many different matrices

half storm Jul 23, 2020, 1:55 AM

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It's not a linear transformation from V into W but it's a linear transformation yea?

stiff frost Jul 23, 2020, 1:55 AM

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I would recommend not saying so

cursive narwhal Jul 23, 2020, 1:55 AM

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no

stiff frost Jul 23, 2020, 1:55 AM

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but some books would say so, especially in the beginning

cursive narwhal Jul 23, 2020, 1:56 AM

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there's a bijection between the set of m x n matrices and the set of vector space homomorphisms. they're not the same thing, however

gray dust Jul 23, 2020, 1:56 AM

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please listen to dirib's point on the distinction between a linear map vs any of its matrix representations (given there exist finite bases of its domain & codomain)

half storm Jul 23, 2020, 1:57 AM

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Yea a matrix isn't a linear transformation, the left-multiplication transformation of a matrix A is a linear transformation

stiff frost Jul 23, 2020, 1:57 AM

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"is the linear map $T(v_1,...,v_2)$ to $A_1,_k w_1,...A_m,_k w_m)$" this isn't really clear to me. $T(v_1,...,v_2)$ wouldn't normally be defined

stoic pythonBOT Jul 23, 2020, 1:57 AM

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dirib:

📎 224701279736299520.png

half storm Jul 23, 2020, 1:57 AM

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I guess that's the nuance that I'm missing.

ocean sequoia Jul 23, 2020, 1:57 AM

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honestly im pulling that straight from LADR

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here

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📎 Screen_Shot_2020-07-22_at_9.57.51_PM.png

stiff frost Jul 23, 2020, 1:58 AM

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That's not what it says though

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What it says is fine

ocean sequoia Jul 23, 2020, 1:58 AM

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could you maybe elaborate on why what i said was different?

stiff frost Jul 23, 2020, 1:58 AM

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If $T\in\mathcal L(V,W)$ and $v_7\in V$ then $T(v_7)$ is defined. But "$T(v_1,...,v_2)$ isn't.

stoic pythonBOT Jul 23, 2020, 1:58 AM

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dirib:

📎 224701279736299520.png

ocean sequoia Jul 23, 2020, 1:59 AM

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oh

stiff frost Jul 23, 2020, 1:59 AM

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Like T takes in a single vector from V at a time

ocean sequoia Jul 23, 2020, 1:59 AM

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i typoed

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im so sorry

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i meant T(v1,..,vm) where v1,...,vm are a basis for V

stiff frost Jul 23, 2020, 1:59 AM

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That's still not defined

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T takes in one vector at a time

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T(v1) and T(v2) are defined

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but T(v1,...,vm) doesn't make sense really

ocean sequoia Jul 23, 2020, 2:00 AM

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ok im now very confused are you maybe able to spell it out for me?

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hang on let me

stiff frost Jul 23, 2020, 2:01 AM

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Spell out what I was just saying about T, or spell out the definition of the matrix given in LADR?

ocean sequoia Jul 23, 2020, 2:01 AM

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give you my understanding and you can maybe tell me why im wrong

stiff frost Jul 23, 2020, 2:01 AM

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ok

ocean sequoia Jul 23, 2020, 2:01 AM

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if you dont mind

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Ok i assumed that if $T /in L(V,W)$ and if $v_1,...,v_m$ $w1,...,w_n$are a basis of V,W then the matrix is a representation of the linear transformation T from V to W uniquely determined on the basis of V and W(which means that any transformation from vk to wk will be equal to each other)

stiff frost Jul 23, 2020, 2:05 AM

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I don't understand what you mean by "any transformation from vk to wk will be equal to each other". And "a representation" is vague in a way that the definition (say, from LADR) makes concrete for us.

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Also w_1,... better not be a basis of V unless W=V or something

stoic pythonBOT Jul 23, 2020, 2:06 AM

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brzig:

📎 571557137118003201.png

ocean sequoia Jul 23, 2020, 2:08 AM

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"any transformation from vk to wk will be equal to each other" I mean that if a transformation T takes takes v1 to w1 and a transformation S takes v1 to w1 then S(v1) = T(v1)

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honestly unquiely determined has always been a bit shaky but the way i understood it is that there is only one way to take v1 to w1

stiff frost Jul 23, 2020, 2:08 AM

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That's true I guess, but most transformations won't take v1 to w1 so that doesn't really matter for this disucssion

ocean sequoia Jul 23, 2020, 2:09 AM

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of course

stiff frost Jul 23, 2020, 2:09 AM

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Rather than drilling down into what you had in mind with the "any transformation from vk to wk will be equal to eachother" stuff, I think it might be good to walk through the definition in LADR fresh. Then look at an example (does LADR have examples for you?). And then maybe come back to the problem you started with

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Let's walk through the definition together

ocean sequoia Jul 23, 2020, 2:09 AM

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ok let me repost

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📎 Screen_Shot_2020-07-22_at_10.10.01_PM.png

stiff frost Jul 23, 2020, 2:10 AM

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I think you're okay with the first sentence "Suppose...basis of W."

latent ledge Jul 23, 2020, 2:10 AM

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there is a problem about this definition in 3a

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you can take v1=(1,0,0,...,0)

ocean sequoia Jul 23, 2020, 2:11 AM

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hang on

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can i finish walking through the definition? I really appreciate the problem ill check it out after

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is that ok?

latent ledge Jul 23, 2020, 2:11 AM

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sure

ocean sequoia Jul 23, 2020, 2:12 AM

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so is the difference between a matrix and a linear transformation that a matrix takes in every vector at once

stiff frost Jul 23, 2020, 2:12 AM

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Maybe this sounds like a silly question, but it's critical to the second sentence. What is a matrix?

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That is not a matrix

ocean sequoia Jul 23, 2020, 2:12 AM

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no worries

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a matrix is a rectangular array of elements

stiff frost Jul 23, 2020, 2:13 AM

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Yes

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and those elements are numbers/scalars/field elements (as opposed to linear maps or vectors or soemthing else)

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So to define an "m-by-n matrix" here, we just need to know what all the numbers are

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Well, they are the $A_{j,k}$ that appear in the equations that the definition says the matrix is "defined by"

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And this is a little tricky because of all the indices.

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n is dim V. m is dim W. j and k are arbitrary

stoic pythonBOT Jul 23, 2020, 2:15 AM

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dirib:

📎 224701279736299520.png

ocean sequoia Jul 23, 2020, 2:15 AM

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and j,k < n

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right?

stiff frost Jul 23, 2020, 2:16 AM

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wrong (in general)

ocean sequoia Jul 23, 2020, 2:16 AM

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ok

stiff frost Jul 23, 2020, 2:16 AM

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the definition says the matrix is "m-by-n"

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so j<=m and k<=n

ocean sequoia Jul 23, 2020, 2:16 AM

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ok

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im with you

stiff frost Jul 23, 2020, 2:17 AM

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j and k

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ok

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so What is, say, A_{1,1}?

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Well, the equation in the defintion has A_{1,k} in it

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so we set k=1 to find out.

ocean sequoia Jul 23, 2020, 2:17 AM

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A 1 is the first element in the first row and column

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A1,1

stiff frost Jul 23, 2020, 2:18 AM

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right