#linear-algebra | Mathematics | Page 32

serene axle Sep 25, 2019, 3:23 AM

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but this is false

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no mathematician sits down and just stares at a proof and suddenly figure out the steps in an instant

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they do it by trial and error

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there's a reason why all these stuff took thousands of years to build upon cause these patterns weren't easy to recognize in the first place.

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i know professors and some of your fellow classmates makes you feel this way

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but in reality they 100% did not sit down and just understood everything over night.

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if they do then they're most likely exaggerating. the human mind is not made for these type of things.

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that's why books exists to contain information.

wintry steppe Sep 25, 2019, 3:27 AM

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I just don't understand why force everyone to learn this if it doesn't pertain to their field of study; This causes more stress and anxiety I ever felt before and trying to figure this out with a textbook that isn't very useful makes it even more difficult. I'm reading the chapter but it doesn't make it easier to understand how to find the answer to the math problem

serene axle Sep 25, 2019, 3:27 AM

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universities just wants your money

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there's a reason why people call it garbage

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but to make garbage into gold you have to own up to it

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make the most out of it by enjoying it.

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i know it doesn't sit right but if you can enjoy the process it'll make things easier.

jagged saffron Sep 25, 2019, 3:32 AM

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📎 unknown.png

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Would someone be able to help with showing well definedness for 2?

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I think i have to show that if

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$v_1 + W = v_1'+W$ then $<v_1+W,v_2+W> = <v_1'+W,v_2+W>$

stoic pythonBOT Sep 25, 2019, 3:33 AM

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Victoria:

📎 480327108606820383.png

jagged saffron Sep 25, 2019, 3:34 AM

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To show this is well defined, consider $\langle v_1+W,v_2+W \rangle'$ and $\langle v_1'+W,v_2+W \rangle'$ such that $v_1 + W = v_1'+W$ . Then, we have to show that $\langle v_1+W,v_2+W \rangle' = \langle v_1'+W,v_2+W \rangle'$ But since by definition of equal cosets, we have that $v_1 - v_1' \in W$, so we have that $\langle v_1-v_1',v_2 \rangle = 0 = \langle v_1,v_2 \rangle - \langle v_1',v_2\rangle = \langle v_1+W,v_2+W \rangle' - \langle v_1'+W,v_2+W \rangle'$

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does this work?

stoic pythonBOT Sep 25, 2019, 3:36 AM

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Victoria:

📎 480327108606820383.png

jagged saffron Sep 25, 2019, 3:55 AM

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also I'm kinda stuck on proving that the new inner product has the property of positive definiteness where <w,w> = 0 iff w = 0

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Positive definiteness: We just have to show that $\langle v_1+W,v_2+W \rangle' = 0$ iff $v_1+W = v_2+W$ as nonnegative is already given to us by our original function.
$\xrightarrow{}$ direction:
$\langle v_1+W,v_1+W\rangle' = 0 \xrightarrow{} \langle v_1,v_1 \rangle = 0$ Let $w_1 \in W$ then $\langle v_1+w_1,v_1 \rangle =\langle v_1+w_1+W,v_1+W\rangle'$ However, we showed in well definedness that this holds iff $v_1 +W = w_1 + W \xrightarrow{} v_1 + W = W \xrightarrow{} v_1 = 0$.

stoic pythonBOT Sep 25, 2019, 4:00 AM

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Victoria:

📎 480327108606820383.png

jagged saffron Sep 25, 2019, 4:00 AM

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i think this is right? Im honestly not too sure

zealous kettle Sep 25, 2019, 1:57 PM

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can I get help understanding this bit regarding elementary row operations? thanks

📎 image0.png

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I think one of the symbols is called kronecker delta? it was mentioned earlier but I dont know what it means

dusky epoch Sep 25, 2019, 1:58 PM

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$\delta_{ij} = \begin{cases} 1 & i=j \ 0 & i\neq j \end{cases}$

stoic pythonBOT Sep 25, 2019, 1:58 PM

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Ann:

📎 183668144404037632.png

zealous kettle Sep 25, 2019, 1:59 PM

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huh, alright, thanks

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so delta_(ik_i) is 1 if i = k_i and 0 if i =/= k_i?

dusky epoch Sep 25, 2019, 2:04 PM

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yes

zealous kettle Sep 25, 2019, 2:05 PM

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thanks

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what does k mean?

dusty hare Sep 25, 2019, 4:53 PM

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📎 unknown.png

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How does that factor to (x+1)(x+1)

winter reef Sep 25, 2019, 4:53 PM

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(a+b)^2 = a^2 + 2ab + b^2

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and ask such questions in #prealg-and-algebra

clever cedar Sep 25, 2019, 6:46 PM

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is the inverse of a matrix A the same thing as the product of the elementary operations needed to get a matrix A to RREF

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in other words does U = A(inverse)

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also is the identity of a matrix the same thing as the RREF of a matrix

cobalt tartan Sep 25, 2019, 6:52 PM

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I can see why this exists but don't know how to get it?

📎 unknown.png

gray dust Sep 25, 2019, 6:54 PM

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@clever cedar not always, if square matrix A is full rank, then yes rref(A) = I

cobalt tartan Sep 25, 2019, 6:54 PM

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Like I think that it has something to do with the definition of a matrix fo a linear mapping?

clever cedar Sep 25, 2019, 6:56 PM

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my textbook explained the form of B = UA and said that B is the reduced for echelon form of A, and U is the product of all elementary matricies needed to get A to RREF. Therefore A * U = B can only be possible if U is the inverse, correct?

quaint heart Sep 25, 2019, 7:08 PM

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Full rank isn't enough

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You need invertible

clever cedar Sep 25, 2019, 7:09 PM

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i should've stated that i assume the matrix is invertible

north sierra Sep 25, 2019, 7:10 PM

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A = (1,2,3),(5,8,11)

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this would not span R^3 but is there a possibility of it spanning a plane in R^3 ?

clever cedar Sep 25, 2019, 8:18 PM

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is the product of elementary row operations of A to A(inverse) the same thing as the inverse of A

clear otter Sep 25, 2019, 8:24 PM

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@clever cedar if you take the product and multiply it by A, do you get hte identity matrix?

clever cedar Sep 25, 2019, 8:24 PM

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funny u mention that, i actually just found the product of elementary row operations of to A(inverse) and it was A

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wait no

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that cant be true

clear otter Sep 25, 2019, 8:26 PM

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@clever cedar if you think something is the inverse of another thing, say u and v, then it should satisfy u*v = identity

clever cedar Sep 25, 2019, 8:26 PM

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ffs

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thats exactly why this is confusing

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for me

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i cant understand why we care about elementary row operations

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when we have that formula

clear otter Sep 25, 2019, 8:26 PM

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So multiply out all of your elementary matrices to get a single matrix

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If it satisfies the above, then you've found an inverse.

clever cedar Sep 25, 2019, 8:28 PM

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ty, i need some time to think over and reread

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thanks for ur time

half ice Sep 25, 2019, 8:31 PM

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@north sierra
Those two vectors span a plane in R3, yes

north sierra Sep 25, 2019, 8:31 PM

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okay

half ice Sep 25, 2019, 8:31 PM

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Two linearly independent vectors span a 2D space

north sierra Sep 25, 2019, 8:31 PM

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but it wouldnt span all of R^3 right

half ice Sep 25, 2019, 8:31 PM

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Nop

north sierra Sep 25, 2019, 8:32 PM

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okay

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i have a test tomorrow Kaynex

half ice Sep 25, 2019, 8:32 PM

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There's many points in R3 that cannot be made with those vectors

north sierra Sep 25, 2019, 8:32 PM

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thank you for the help these past few days

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also to everyone else that helped

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okay i see

half ice Sep 25, 2019, 8:32 PM

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Np, anything else you want to ask? You seem pretty confident

north sierra Sep 25, 2019, 8:32 PM

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yeah

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im good for now

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yes im confident now

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just hope i do good

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tomorrow

cobalt tartan Sep 25, 2019, 9:16 PM

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I'm not sure how to approach this question

📎 unknown.png

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So far my ideas are that uh

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R^n -> R^n are isomorphic

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And change of basis is linear

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But that's all I got

thorn robin Sep 25, 2019, 9:27 PM

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try and draw a diagram

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you have B[L]C going from (R^n,B) to (R^n,C)

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and on a separate line draw C[L]B going from (R^n,C) to (R^n,B)

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connect the two rows of the diagram with vertical arrows P : (R^n,C) -> (R^n,B) which is the change of basis and P^{-1} on the other side

cobalt tartan Sep 25, 2019, 9:30 PM

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Wait there's no P^{-1} though ?

thorn robin Sep 25, 2019, 9:31 PM

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instead of P^{-1} you can just draw P going in the other direction

cobalt tartan Sep 25, 2019, 9:31 PM

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Wait

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I'm confused

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📎 MVIMG_20190925_173146.jpg

thorn robin Sep 25, 2019, 9:32 PM

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one sec

cobalt tartan Sep 25, 2019, 9:33 PM

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More like this?

📎 MVIMG_20190925_173316.jpg

thorn robin Sep 25, 2019, 9:34 PM

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one of the top/bottom arrows points the wrong way

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depending on what B[L]C means

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otherwise yes

cobalt tartan Sep 25, 2019, 9:35 PM

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Um

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In my book B[L]C means from a basis C to a basis B

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📎 unknown.png

thorn robin Sep 25, 2019, 9:35 PM

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ok then the top arrow should point left

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wait I think that's wrong

cobalt tartan Sep 25, 2019, 9:36 PM

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Wait it's cLb = PbLcP

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The arrows I think are in the right direction?

thorn robin Sep 25, 2019, 9:36 PM

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I'm not sure I understand what you drew

cobalt tartan Sep 25, 2019, 9:36 PM

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Um

thorn robin Sep 25, 2019, 9:36 PM

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you have P going C to C

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and the other P going D(?) to B

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P should go C to B

cobalt tartan Sep 25, 2019, 9:37 PM

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cLb goes from basis B to basis C

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Oh

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So P should be diagonal?

thorn robin Sep 25, 2019, 9:37 PM

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here's what I have

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top row: C[L]B going B to C

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bottom row: B[L]C going C to B

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left column: P going up from C to B

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right column: P going down from C to B

cobalt tartan Sep 25, 2019, 9:40 PM

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📎 MVIMG_20190925_173954.jpg

thorn robin Sep 25, 2019, 9:40 PM

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right

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the point of drawing this is just to organize our thoughts and see what is happening
it also suggests that the P you want to use to solve the problem is the change of basis matrix from C to B, just because that's what makes the diagram make sense

cobalt tartan Sep 25, 2019, 9:41 PM

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Ah

thorn robin Sep 25, 2019, 9:41 PM

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now to prove it you just have to take a vector in R^n and write it in C basis

cobalt tartan Sep 25, 2019, 9:41 PM

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Wait I think that this diagram is slightly wrong

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Should be other way lol

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I want to show that (P)(bLc)(P) = (cLb)

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This diagram shows that (P)(cLb)(P) = (bLc)

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It's the same thing though

thorn robin Sep 25, 2019, 9:42 PM

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ah yeah you're right
I thought the convention for the B and C subscripts was the opposite when I first looked at the question

cobalt tartan Sep 25, 2019, 9:43 PM

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Yea

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Idk why they write it like that

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Is confusing

thorn robin Sep 25, 2019, 9:43 PM

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so then you should take P to be the change of basis matrix from B to C

cobalt tartan Sep 25, 2019, 9:44 PM

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Yea

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So if we have some change of basis matrix from B to C, multiplied by a change of basis matrix from C to B, multiplied by a change of abasis matrix B to C, we should end up with a change of basis matrix from B to C lol

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Now how do I state that mathematicalyy...

thorn robin Sep 25, 2019, 9:45 PM

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what you said isn't quite what you want

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because you referred to all 4 matrices as change of bases

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but L is anything

cobalt tartan Sep 25, 2019, 9:46 PM

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Ah right

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L is just some linear mapping

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Right

thorn robin Sep 25, 2019, 9:46 PM

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stating it mathematically is just the equation given in the problem

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you just need to calculate what happens to a vector written in B basis when you apply the composition of the 3 matrices

cobalt tartan Sep 25, 2019, 9:47 PM

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Hrm

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Wait

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Hrm

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It says to prove that there exists a matrix P

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Hrm

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Like that wouldn't prove that a matrix P exists, would it?

thorn robin Sep 25, 2019, 9:50 PM

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we know there exists a matrix P that changes basis B into basis C

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I assume we can take that for granted, yes?

cobalt tartan Sep 25, 2019, 9:51 PM

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Hrmm

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I mean

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THe question is basically saying to prove that such a matrix exists isn't it?

clear otter Sep 25, 2019, 9:51 PM

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I think that's what they're trying to prove.

cobalt tartan Sep 25, 2019, 9:51 PM

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📎 unknown.png

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"Prove that there exists a matrix P such that"

thorn robin Sep 25, 2019, 9:54 PM

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oh hmm ok

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well then let's figure out how to write down P explictly

cobalt tartan Sep 25, 2019, 9:55 PM

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Well the book also gives that we have

📎 unknown.png

thorn robin Sep 25, 2019, 9:55 PM

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how about using that definition on the identity map

cobalt tartan Sep 25, 2019, 9:56 PM

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By the identity map you mean L:R^n->R^n, L((x1 ... x_n)) = (x_1 ... x_n) right?

thorn robin Sep 25, 2019, 9:57 PM

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yup

cobalt tartan Sep 25, 2019, 9:57 PM

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If we use the standard basis with the identity map then we'd just get the identity matrix

thorn robin Sep 25, 2019, 9:57 PM

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then the property from the picture you posted is [v]C = C[I]B [v]B

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so take P = C[I]B

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yes that's true but we have two arbitrary bases, not the standard basis

cobalt tartan Sep 25, 2019, 9:58 PM

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WHat is C[I]B?

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Is that identity?

thorn robin Sep 25, 2019, 9:58 PM

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it's the matrix of the identity map in the bases B, C

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same notation as your picture

cobalt tartan Sep 25, 2019, 9:59 PM

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Ah ok

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So if we have P = c[I]b, then (P)(bLc)(P) = (c[I]b)(bLc)(c[I]b)

thorn robin Sep 25, 2019, 10:00 PM

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yes

cobalt tartan Sep 25, 2019, 10:00 PM

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And if it's R^n->R^n, c[I]b must exist yea?

thorn robin Sep 25, 2019, 10:01 PM

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any linear map has a matrix in any basis

cobalt tartan Sep 25, 2019, 10:01 PM

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Yes

thorn robin Sep 25, 2019, 10:01 PM

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regardless of whether it's R^n or the identity map

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so yes

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here are the two things you know: $P[v]_B = [v]_C$ and ${}_C[L]_B[v]_B = [Lv]_C$

stoic pythonBOT Sep 25, 2019, 10:02 PM

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Seoin:

📎 181597954883321856.png

thorn robin Sep 25, 2019, 10:03 PM

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now you just need to use them to show that $P{}_B[L]_CP[v]_B = {}_C[L]_B[v]_B$

stoic pythonBOT Sep 25, 2019, 10:03 PM

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Seoin:

📎 181597954883321856.png

thorn robin Sep 25, 2019, 10:04 PM

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which will give the statement the problem asks for since you will verify that this is true for every v

cobalt tartan Sep 25, 2019, 10:04 PM

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Ah

thorn robin Sep 25, 2019, 10:05 PM

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everything make sense?

cobalt tartan Sep 25, 2019, 10:06 PM

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Um

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Kinda

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STill a bit muddlede

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But uh gonna take a few minutes to try to think through it

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And see if it makes more sense once I put it down on paper

thorn robin Sep 25, 2019, 10:06 PM

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if you understand everything I wrote
then to prove it you just need to do the calculation I mentioned

cobalt tartan Sep 25, 2019, 10:07 PM

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By the calculation you mentioned you mean $P{}_B[L]_CP[v]_B = {}_C[L]_B[v]_B$

stoic pythonBOT Sep 25, 2019, 10:07 PM

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Liria ^(;,;)^:

📎 288883123863683072.png

cobalt tartan Sep 25, 2019, 10:07 PM

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right?

thorn robin Sep 25, 2019, 10:07 PM

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yes

cobalt tartan Sep 25, 2019, 10:08 PM

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Ah, and (cLb)(v_b) = (Lv)_c because of linearity...?

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Ah, no, it's a definition

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Ok

thorn robin Sep 25, 2019, 10:10 PM

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yeah I mean
it couldn't really be any other way

cobalt tartan Sep 25, 2019, 10:19 PM

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Ah

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Ok I think I see now

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Like the actual math is kinda sketchy but like

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It makes sense logically now?

thorn robin Sep 25, 2019, 10:20 PM

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what is sketchy?

cobalt tartan Sep 25, 2019, 10:20 PM

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Mmmm

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I wrote it out like this

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📎 unknown.png

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Like

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Oh wait brackets are missing

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But uh

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I can't concievably see myself coming up with that in a short period of time haha

thorn robin Sep 25, 2019, 10:22 PM

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yup looks good

cobalt tartan Sep 25, 2019, 10:42 PM

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Thank you!

north sierra Sep 25, 2019, 10:43 PM

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📎 unknown.png

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i dont understand of how this is a subset of the xz plane

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i know you explained it before @half ice but i still dont really get it

gray dust Sep 25, 2019, 10:47 PM

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what's the y component of those vectors?

north sierra Sep 25, 2019, 10:48 PM

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0

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so any scalar times v1 and v2 would = 0 in y

gray dust Sep 25, 2019, 10:48 PM

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mhmm

north sierra Sep 25, 2019, 10:49 PM

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so i get that part

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sooo

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like

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the plane would still be in R^3 tho right

steady fiber Sep 25, 2019, 10:51 PM

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does the plane cover all of R^3

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if I give you any point in R^3 with a non-zero y component

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would it be in the plane

gray dust Sep 25, 2019, 10:51 PM

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~~these are 3d vectors... they exist in 3d aka R^3~~

north sierra Sep 25, 2019, 10:52 PM

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no it would not cover all of R^3

steady fiber Sep 25, 2019, 10:52 PM

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so it's a subset of R^3

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if it was not a strict subset of R^3, then it would cover all of R^3

north sierra Sep 25, 2019, 10:53 PM

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what does xz-plane refer to?

steady fiber Sep 25, 2019, 10:53 PM

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the plane that goes through the x and z axes

gray dust Sep 25, 2019, 10:53 PM

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all points (x,y,z) where y=0

north sierra Sep 25, 2019, 10:53 PM

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oh

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so would you call (1,2) (5,7) an xz plane

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i mean xy

gray dust Sep 25, 2019, 10:54 PM

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those aren't points in R^3

north sierra Sep 25, 2019, 10:54 PM

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yeah ik

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i meant to put only 2

steady fiber Sep 25, 2019, 10:54 PM

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I mean it's still the xy plane, but it's just the xy plane in R^2

gray dust Sep 25, 2019, 10:54 PM

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rather they're in R^2 where you have (x,y) coords

steady fiber Sep 25, 2019, 10:55 PM

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the xy plane in R^2, is all of R^2

north sierra Sep 25, 2019, 10:56 PM

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always?

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oh nvm

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so xz plane in R^3 would be not be all of R^3 then right

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itd be a subset of R^3

gray dust Sep 25, 2019, 10:57 PM

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yes

north sierra Sep 25, 2019, 10:57 PM

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ok

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got it

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thx !

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to be more specific

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wouldn't you say xz plane is a proper subet of R^3?

gray dust Sep 25, 2019, 11:03 PM

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you could

north sierra Sep 25, 2019, 11:03 PM

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ok

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thx again

scarlet hamlet Sep 25, 2019, 11:10 PM

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hey i'm not really getting the intuition behind this

north sierra Sep 25, 2019, 11:10 PM

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behind what exactly?

scarlet hamlet Sep 25, 2019, 11:10 PM

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"If L:V->W is a linear mapping, L is one to one iff Ker(L) = {0}"

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like yea the proof makes sense

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but its not really intuitive to me?

sonic osprey Sep 25, 2019, 11:11 PM

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Well if Ker(L) isn't just {0} then it clearly wouldn't be one-to-one

scarlet hamlet Sep 25, 2019, 11:12 PM

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its not 'clearly' to me yet so thats why im asking

sonic osprey Sep 25, 2019, 11:12 PM

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Well the definition of one-to-one is that no two elements get sent to the same element

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0 is always sent to 0 in a linear map

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So if anything else is also sent to 0, then your map isn't one-to-one

scarlet hamlet Sep 25, 2019, 11:13 PM

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ohh ok i see

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yep ok thanks

half ice Sep 25, 2019, 11:14 PM

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The other direction is harder. It turns out that if any output has multiple inputs, then the zero output also has multiple inputs

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"iff" means this statement is really two statements

scarlet hamlet Sep 25, 2019, 11:15 PM

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yeah

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i'm just asking cuz i've been using the lemma

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like so often

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but i dont think i actually get it

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like i get rank nullity for example

half ice Sep 25, 2019, 11:16 PM

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It's very useful yeah. You can check one-to-one-ness just by checking the zero output

scarlet hamlet Sep 25, 2019, 11:16 PM

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for me, rank nullity makes sense but this doesnt really yet if that makes sense

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its like

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i know how to use it but it still feels weird?

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idk if its relatable at all

half ice Sep 25, 2019, 11:17 PM

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It definitely is relatable. In this case, I would suggest trying to prove it yourself! It's not a very hard proof. You need the fact that the function is linear, and it's over pretty quick

scarlet hamlet Sep 25, 2019, 11:17 PM

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ok thanks 🙂

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i tend to try and just hammer out examples but i think thats why i dont do well on the proof sections of tests lol

zealous kettle Sep 25, 2019, 11:51 PM

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https://cdn.discordapp.com/attachments/622430759088816129/626421170102599689/image0.jpg Hi! I don't understand the simplification from

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$E_{ik}A_{kj} = A_{rj} + cA_{sj} , i = r$

stoic pythonBOT Sep 25, 2019, 11:52 PM

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Spes:

📎 179701226995318785.png

zealous kettle Sep 25, 2019, 11:52 PM

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I get that E_{ik} = delta_{rk} + cdelta_{sk}

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which makes it

stoic pythonBOT Sep 25, 2019, 11:53 PM

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Spes:

📎 179701226995318785.png

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Spes:

📎 179701226995318785.png

cloud cedar Sep 26, 2019, 12:42 AM

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when you sum over something with the kronecker delta it acts to "replace" the index that gets summed over with the other: $\sum_j\delta_{ij}A_{jk} = A_{ik}$; it's like we took the $j$ index on $A$ and replaced it with an $i$ using the delta

stoic pythonBOT Sep 26, 2019, 12:42 AM

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📎 128822128072982528.png

cloud cedar Sep 26, 2019, 12:44 AM

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in that step, youre summing over k, so the first Akj becomes Aij and the second turns into Asj (then since i=r you can replace the subscript i with an r)

toxic pendant Sep 26, 2019, 1:04 AM

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I'd like to understand why when a determinant for a homogenous matrix is 0, there are an infinite amount of solutions

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Could someone explain it to me like I'm 5

steady fiber Sep 26, 2019, 1:07 AM

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a 5 year old wouldn't understand matrices

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or determinants

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or homogenity

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or infinite solutions

toxic pendant Sep 26, 2019, 1:07 AM

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Says the one with a pfp of someone who definitely would at 5

steady fiber Sep 26, 2019, 1:08 AM

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Pi_thonk

toxic pendant Sep 26, 2019, 1:08 AM

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Wait what these emotes

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Mathematicians are weebs

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📎 Screenshot_20190925-210852.png

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Qed

grave plank Sep 26, 2019, 1:12 AM

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When I show the existence of the zero vector in a subspace, do I need to show that it is the result of the additive inverse property?

steady fiber Sep 26, 2019, 1:12 AM

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when the determinant is 0, at least one row is not linearly independent from the other rows. If you row reduce such a matrix, you will have an all zero row at the bottom. If you find the solutions, since at least one variable is dependent on the other variables, there ceases to be a unique solution

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and you can vary the dependent variable(s) around to get infinite solutions

toxic pendant Sep 26, 2019, 1:13 AM

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Oh

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That makes sense

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This makes too much sense

steady fiber Sep 26, 2019, 1:15 AM

#

I mean if it's a zero vector in the main vector space

#

then it's still going to be the zero vector in the sub space

#

you just have to show it exists in the subspace

#

you already know it is the result of the additive inverse property in the main vector space, no need to re-prove it

grave plank Sep 26, 2019, 1:16 AM

#

so would i just need to prove the additive identity?

#

well

#

i shouldnt have to

steady fiber Sep 26, 2019, 1:17 AM

#

you just need to prove the zero vector exists in the subspace (and you have to prove the subspace is closed under addition and scalar multiplication)

#

that's all

grave plank Sep 26, 2019, 1:17 AM

#

ah

#

ok

toxic pendant Sep 26, 2019, 1:17 AM

#

~~btw ishigami best girl~~

steady fiber Sep 26, 2019, 1:17 AM

#

technically proving addition and scalar multiplication is closed already proves the existence of the zero vector in the subspace

#

but I like to independently check for the zero vector first

grave plank Sep 26, 2019, 1:19 AM

#

ok thank you

zealous kettle Sep 26, 2019, 1:21 AM

#

@cloud cedar why does summing over the kronecker delta "replace" the index that gets summed over?

cloud cedar Sep 26, 2019, 1:21 AM

#

write out the terms of $\sum_j\delta_{ij}A_{jk}$

stoic pythonBOT Sep 26, 2019, 1:22 AM

#

📎 128822128072982528.png

cloud cedar Sep 26, 2019, 1:22 AM

#

and youll see that youre only left with Aik

steady fiber Sep 26, 2019, 1:22 AM

#

since only the term where i and j are equal are kept, the kronecker delta function multiplies every other term with 0

cloud cedar Sep 26, 2019, 1:22 AM

#

yup

steady fiber Sep 26, 2019, 1:23 AM

#

it's the same concept as integrating a function multiplied by a shifted dirac delta function

#

just discretized

paper egret Sep 26, 2019, 1:24 AM

#

📎 unknown.png

#

what does inconsistennt mean

zealous kettle Sep 26, 2019, 1:24 AM

#

I see, thanks

steady fiber Sep 26, 2019, 1:24 AM

#

there is no solution

paper egret Sep 26, 2019, 1:24 AM

#

ah ty

steady fiber Sep 26, 2019, 1:24 AM

#

the constraints on the solutions

#

don't match up

#

so the constraints are "inconsistent" (with each other)

zealous kettle Sep 26, 2019, 1:26 AM

#

another question: in future scenarios when I have trouble understanding a particular theorem, how should I visualize it to help me understand it?

steady fiber Sep 26, 2019, 1:27 AM

#

depends from theorem to theorem

#

might not even be possible to visualize it fully

#

if it gets abstract enough

#

that's when you have to just trust the math checks out

zealous kettle Sep 26, 2019, 1:27 AM

#

hmm

#

what about for this theorem specifically? what could I do?

steady fiber Sep 26, 2019, 1:28 AM

#

which theorem

#

oh I see it

#

for this one you can just try writing out some simple 2x2 or 3x3 elementary row operation matrices

#

and see what happens when they multiply random matrices

#

get a feel for it and convince yourself it's true

zealous kettle Sep 26, 2019, 1:32 AM

#

hmm ok

steady fiber Sep 26, 2019, 1:32 AM

#

for most matrix proofs, just try doing some examples with 2x2 and 3x3 matrices (or like 2x3/3x2/2x1/1x2/3x1/1x3)

#

to get a feel for it

#

bigger sizes get a bit unreasonable to do by hand

#

1x1 is just scalars

zealous kettle Sep 26, 2019, 1:33 AM

#

somewhat related question: in matrix multiplication AB, is the product a linear combination of B?

#

for example

#

take an elementary matrix E
1 1
0 1

#

and another matrix A
2 3
4 5

#

I guess what I'm really asking is

steady fiber Sep 26, 2019, 1:37 AM

#

let C be a matrix such that AB = C, for two matrices A and B
The rows of C are in the row space of B
The columns of C are in the column space of A

zealous kettle Sep 26, 2019, 1:37 AM

#

is there a way to think of E as the identity matrix with a 1 in the 1,2 position, and how it affects the product?

steady fiber Sep 26, 2019, 1:37 AM

#

does that answer your question

zealous kettle Sep 26, 2019, 1:38 AM

#

what is row space and column space?

steady fiber Sep 26, 2019, 1:38 AM

#

a linear algebra textbook can probably explain it better than me

#

or someone else here

zealous kettle Sep 26, 2019, 1:38 AM

#

is it like all possible combinations of the rows of B?

steady fiber Sep 26, 2019, 1:39 AM

#

yes

#

the rows of the product EA would be a combination of the rows of A, so that's why it's a row operation

#

you can indeed look at E and predict what row operation it would be

#

the elementary matrices are quite simple

#

all 3 elementary matrices are simple changes to the identity:
if E looks like the identity with 2 rows swapped -> row swap operation for those two rows
if E looks like the identity with one non-1 element along diagonal -> multiplies that row by that non-1 number
if E looks like the identity with one non-0 element outside a diagonal, then you're adding a multiple of a row to another

#

there's also column operations if you want to look into it

#

where it would instead be AE, and the product would do column operations instead

zealous kettle Sep 26, 2019, 1:43 AM

#

so to clarify

#

row space is like all linear combinations of rows as vectors

steady fiber Sep 26, 2019, 1:45 AM

#

yes

#

it's the span of the rows

zealous kettle Sep 26, 2019, 1:46 AM

#

The rows of C are in the row space of B
The columns of C are in the column space of A

#

this is still a little fuzzy to me

steady fiber Sep 26, 2019, 1:47 AM

#

you'll get to it sometime

zealous kettle Sep 26, 2019, 1:47 AM

#

alright

#

also wont the span of the rows cover the entire ... um... space?

#

like it could reach everywhere, right?

steady fiber Sep 26, 2019, 1:48 AM

#

if the rows are in R^n

#

they can span up to R^n

#

but not necessarily

zealous kettle Sep 26, 2019, 1:50 AM

#

oh because they could be all missing a x_k unknown where 1 <= k <= n

steady fiber Sep 26, 2019, 1:50 AM

#

there is a possibility there aren't enough linearly independent rows

#

to span R^n

#

or that there aren't even enough rows

#

like in the matrix:
1 1 1
1 2 3

#

the rows are in R^3

north sierra Sep 26, 2019, 1:51 AM

#

@paper egret p sure Hx = 0 will always be consistent becuase the augmented column will always be 0 in RREF

steady fiber Sep 26, 2019, 1:51 AM

#

but there's only 2 rows

#

2 rows cannot possibly span all of R^3

zealous kettle Sep 26, 2019, 1:51 AM

#

why not?

steady fiber Sep 26, 2019, 1:52 AM

#

you haven't learned why?

#

but you're learning matrices?

zealous kettle Sep 26, 2019, 1:52 AM

#

I probably have hang on

steady fiber Sep 26, 2019, 1:52 AM

#

that's a pretty big theorem to skip

zealous kettle Sep 26, 2019, 1:52 AM

#

I know for sure I didn't skip anything

steady fiber Sep 26, 2019, 1:52 AM

#

you need n linearly independent vectors to span a vector space with dimension n

#

that's a theorem

zealous kettle Sep 26, 2019, 1:53 AM

#

the first mention of "linearly independent" is ~15 pages down

#

but that makes sense though, to some degree

#

since if you have
1 2
2 4
or any two rows that are colinear

north sierra Sep 26, 2019, 1:55 AM

#

yeah

zealous kettle Sep 26, 2019, 1:55 AM

#

ah the "missing a row part" also makes sense now

#

its like you can't cover all of 2d space with only <1, 2>

steady fiber Sep 26, 2019, 1:55 AM

#

yes

north sierra Sep 26, 2019, 1:56 AM

#

yeah

#

is origin the same as 0 vector?

zealous kettle Sep 26, 2019, 1:57 AM

#

finally

#

are you asking me?

north sierra Sep 26, 2019, 1:57 AM

#

anyone who knows

steady fiber Sep 26, 2019, 1:57 AM

#

I mean depends on how you want to define origin

#

if you define origin as the point that's (0,0,0,...0) (all coordinates 0)

#

then no

north sierra Sep 26, 2019, 1:58 AM

#

like what if you say span(v1,v2) is a line through the origin and v1

#

would that mean 0 vector and v1

steady fiber Sep 26, 2019, 1:59 AM

#

also by zero vector do you mean "vector with 0 magnitude" or the vector with the zero vector property

north sierra Sep 26, 2019, 1:59 AM

#

um im not sure lol

steady fiber Sep 26, 2019, 1:59 AM

#

I'll go with the latter

north sierra Sep 26, 2019, 1:59 AM

#

lol

#

yeah probably

steady fiber Sep 26, 2019, 2:00 AM

#

the span automatically contain the 0 vector

#

it has to

north sierra Sep 26, 2019, 2:00 AM

#

oh ok

steady fiber Sep 26, 2019, 2:00 AM

#

since it says it's through the origin, it contains the origin too

#

doesn't mean the origin and zero vector would be the same

north sierra Sep 26, 2019, 2:01 AM

#

oh

#

so for my answer

#

i should just wrtie oigin

#

origin

#

origin + v1 or whatever

#

if it asked me to provide a geometric description of span(v1,v2) where they where v1 and v2 were independent

steady fiber Sep 26, 2019, 2:09 AM

#

Define Vector space: $V=\left{\left(x_{1},x_{2}\right)| x_{1},x_{2}\in \mathbb{R} \right}$

Define vector addition: $\left( x_1,x_2 \right)+\left( y_1,y_2 \right) = (x_1+y_1+1, x_2+y_2+1)$

Define scalar multiplication: $c\left( x_1,x_2 \right) = \left( cx_1+c-1,cx_2+c-1 \right)$

#

the zero vector in this vector space is (-1,-1) if you want to verify it, so this is an example where the zero vector isn't the origin

stoic pythonBOT Sep 26, 2019, 2:10 AM

#

PorosInMyAshe:

📎 300027066861223979.png

steady fiber Sep 26, 2019, 2:10 AM

#

A geometric description would the plane that contains both vectors v1 and v2

#

that's what span(v1, v2) is

zealous kettle Sep 26, 2019, 2:34 AM

#

I got through half a page in hk today 😫

#

I'll never finish linalg at this rate

steady fiber Sep 26, 2019, 2:35 AM

#

no one "finishes" lin alg

zealous kettle Sep 26, 2019, 2:35 AM

#

bad word choice

#

I'll never finish this book

#

Which is like a intro to linalg

#

ashe if you don't mind me asking how did you learn lin alg?

serene axle Sep 26, 2019, 6:09 AM

#

he breathed it probably

#

like most people in this server does.

#

opencry

#

then there's me and you doing it with sweat and blood

#

no pain; no gain

jagged forum Sep 26, 2019, 7:33 AM

#

Does anyone know how to compute the basis of $W \cap V$ for subspaces W, V in R^5?

stoic pythonBOT Sep 26, 2019, 7:33 AM

#

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

📎 213026416759930881.png

jagged forum Sep 26, 2019, 7:34 AM

#

I forgot how to do any linear algebra computation, and my professor in my LAII course decided to give us a computation quesiton

dusky epoch Sep 26, 2019, 8:00 AM

#

how are your subspaces given

jagged forum Sep 26, 2019, 8:06 AM

#

📎 unknown.png

steady fiber Sep 26, 2019, 12:03 PM

#

I learned introductory Lin alg in a university from a professor

scarlet hamlet Sep 26, 2019, 4:17 PM

#

📎 unknown.png

#

all i did was write out c[L]b and b[L]c

#

and uh i have no idea where to go from there

#

theres this one too

#

📎 unknown.png

#

i literally dont have a clue yikes

#

any insights would help ❤

jagged saffron Sep 26, 2019, 4:26 PM

#

wait i misread nvm

scarlet hamlet Sep 26, 2019, 4:29 PM

#

haha itsok

#

ive literally just stared at the thing for 15 mins

half ice Sep 26, 2019, 4:39 PM

#

In any general algebraic thingy, if Ax = A, then x is a "do nothing" element

#

Here, you have TST = T

#

ST is a "do nothing" which means S undoes T.

#

@scarlet hamlet

#

Mind you, this is probably not the only answer, but it's the easiest one to get

#

Oh wait, the kernel is non trivial so T isn't invertible. Huh.

distant gate Sep 26, 2019, 4:46 PM

#

I forgot some linear algebra, does the null space and range of a linear transformation change with respect to different basis?

scarlet hamlet Sep 26, 2019, 5:23 PM

#

@half ice i feel like im going in circles LOL

#

ive been writing stuff down

#

but its not leading anywhere

thorn robin Sep 26, 2019, 5:25 PM

#

@scarlet hamlet do you know @cobalt tartan?
you two had the exact same question, in the exact same notation, one day apart lol

#

you can scroll up to see our discussion about the problem

cobalt tartan Sep 26, 2019, 5:25 PM

#

Oh

thorn robin Sep 26, 2019, 5:25 PM

#

(referring to your Q4)

cobalt tartan Sep 26, 2019, 5:25 PM

#

Maybe same class

#

LOL

#

Yea that is uh same class as me

thorn robin Sep 26, 2019, 5:26 PM

#

😂

cobalt tartan Sep 26, 2019, 5:27 PM

#

@scarlet hamlet q5 is uh you're almost definitely overthinking it

#

When u see it you'll go wtf did I spend 3 hours on this for

#

Q4 I'm still not 100% sure I understand lol

scarlet hamlet Sep 26, 2019, 5:37 PM

#

LOL wait

#

do u go to UW

#

@cobalt tartan

cobalt tartan Sep 26, 2019, 5:37 PM

#

Yea

#

LOL

#

MATH235?

scarlet hamlet Sep 26, 2019, 5:37 PM

#

YEAH LOOOOOOOOOOOOL

cobalt tartan Sep 26, 2019, 5:38 PM

#

What section

scarlet hamlet Sep 26, 2019, 5:38 PM

#

i knew id find another loo person here

#

uh

#

i have no idea

#

i have satriano at 1030

cobalt tartan Sep 26, 2019, 5:38 PM

#

Oh

#

Ok

#

Different section

#

Uhm

#

Anyways q5 uh

sonic osprey Sep 26, 2019, 5:38 PM

#

I thought UW meant University of Washington rip

half ice Sep 26, 2019, 5:38 PM

#

Waterloo?

cobalt tartan Sep 26, 2019, 5:38 PM

#

Is easier than it seems

#

Yea

scarlet hamlet Sep 26, 2019, 5:38 PM

#

yeah waterloo

#

waterpoo

#

same thing

half ice Sep 26, 2019, 5:39 PM

#

Cool cool. I was Western over here

#

Nice to see Canadian rep

#

You guys are the better school

cobalt tartan Sep 26, 2019, 5:39 PM

#

Haha

scarlet hamlet Sep 26, 2019, 5:40 PM

#

my gpa is trash x.x

cobalt tartan Sep 26, 2019, 5:40 PM

#

@scarlet hamlet Q5 uh what mapping takes T(x) and produces T(x)

scarlet hamlet Sep 26, 2019, 5:41 PM

#

T(x) = x

#

??

#

wait what

#

idk i have 0 confidence in anything math

cobalt tartan Sep 26, 2019, 5:42 PM

#

Uh let's go with an easier example

#

Say R2 -> R2

scarlet hamlet Sep 26, 2019, 5:46 PM

#

uh

#

im literally so confused lol

cobalt tartan Sep 26, 2019, 5:54 PM

#

Say that you have T:R^2-> R^2

#

Such that T((x, y)) = (y, x)

#

What is a linear mapping S:R2-> R^2

#

Such that S(T((x, y))) = (y, x)

#

@scarlet hamlet?

scarlet hamlet Sep 26, 2019, 5:57 PM

#

S(y,x) = (y,x)?

cobalt tartan Sep 26, 2019, 5:57 PM

#

Yes

#

So the mapping S((x, y)) is...?

scarlet hamlet Sep 26, 2019, 5:59 PM

#

one to one? onto? isomorphic?

#

id say so lol

cobalt tartan Sep 26, 2019, 6:00 PM

#

Yep

#

Does that give you a hint lol

scarlet hamlet Sep 26, 2019, 6:01 PM

#

ya a bit

#

i think?

cobalt tartan Sep 26, 2019, 6:02 PM

#

Q5 is big brain

#

Lol

scarlet hamlet Sep 26, 2019, 6:02 PM

#

waterloo is too big brain for me

cobalt tartan Sep 26, 2019, 6:02 PM

#

Same

scarlet hamlet Sep 26, 2019, 6:02 PM

#

naw u seem like an intelligent being

#

i have the iq of a donkey

cobalt tartan Sep 26, 2019, 6:03 PM

#

Uh I'm low IQ wannabe mathie 😂

scarlet hamlet Sep 26, 2019, 6:03 PM

#

i lowkey celebrated when i did q1 and q2 on my own

cobalt tartan Sep 26, 2019, 6:05 PM

#

Oh

#

Same

#

A4 is easier

scarlet hamlet Sep 26, 2019, 6:05 PM

#

did u already do it

#

whoa

#

i spent the whole weekend applying to jobs so im SoOo behind

cobalt tartan Sep 26, 2019, 6:07 PM

#

Uh I looked at it, gonna start tonight

#

Have uh stat midterm to catch up + study for first

scarlet hamlet Sep 26, 2019, 6:07 PM

#

oh same

#

stat230?

#

LOL

#

ok i feel like we should take this convo to dm

cobalt tartan Sep 26, 2019, 6:09 PM

#

Ya

thorn robin Sep 26, 2019, 6:22 PM

#

and they lived happily ever after

north sierra Sep 26, 2019, 6:23 PM

#

a plane in R^2 would have to span all of R^2 right

#

wait i said that wrong

north sierra Sep 26, 2019, 7:32 PM

#

[105]
[014]
[003]

#

this is an augmented matrix btw

dusky epoch Sep 26, 2019, 7:32 PM

#

do you mean

#

[ 1, 0, 5 ]
[ 0, 1, 4 ]
[ 0, 0, 3 ]

north sierra Sep 26, 2019, 7:33 PM

#

yeah

dusky epoch Sep 26, 2019, 7:33 PM

#

don't write matrices like what you just did. matrix entries don't need to be single-digit integers

north sierra Sep 26, 2019, 7:33 PM

#

ok sorry

#

so x1=5 x2=4 and is there an x3?

half ice Sep 26, 2019, 8:44 PM

#

There's no solution to your system, as the bottom line implies 0 = 3
@north sierra

junior shard Sep 26, 2019, 8:54 PM

#

📎 20190926_135442.jpg

#

I got the derivatives but im not sure how to show that theyre linearly independent

subtle walrus Sep 26, 2019, 8:55 PM

#

by using the definition of linear independence

junior shard Sep 26, 2019, 8:56 PM

#

How do i put them into matix

#

So i set cf+cf'+cf''= 0

subtle walrus Sep 26, 2019, 8:58 PM

#

you use different coefficients for each, but yeah

junior shard Sep 26, 2019, 9:00 PM

#

Do what would i do from here?

📎 20190926_145923.jpg

subtle walrus Sep 26, 2019, 9:02 PM

#

yeah

#

now you have to show that c_0=c_1=c_2=0

junior shard Sep 26, 2019, 9:06 PM

#

How do i set up the matrix for this?

junior shard Sep 26, 2019, 9:33 PM

#

Is it correct to say c_2 must be 0 since its the only one that contains x^2

stoic pythonBOT Sep 27, 2019, 12:52 AM

#

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

📎 323837464982716416.png

grave plank Sep 27, 2019, 12:52 AM

#

Is my proof also enough to show that span(u,v,w) = span(u,v)?

steady fiber Sep 27, 2019, 1:09 AM

#

pog I saw university of waterloo in chat

scarlet hamlet Sep 27, 2019, 1:11 AM

#

@steady fiber r u a fellow looser too

steady fiber Sep 27, 2019, 1:11 AM

#

no

#

I'm at the much better university a few hundred kilometers away

#

but every single one of my high school friend is at waterloo

scarlet hamlet Sep 27, 2019, 1:12 AM

#

lol uoft?

steady fiber Sep 27, 2019, 1:12 AM

#

u of tears

scarlet hamlet Sep 27, 2019, 1:12 AM

#

sorry but we won the meme war

steady fiber Sep 27, 2019, 1:12 AM

#

you did

#

we were completely destroyed in the meme war

#

no one denies it

scarlet hamlet Sep 27, 2019, 1:13 AM

#

💪

#

musclethink

grave plank Sep 27, 2019, 1:32 AM

#

<@&286206848099549185>

tranquil junco Sep 27, 2019, 1:33 AM

#

fam

#

https://media.discordapp.net/attachments/490536027400962080/507304636693217321/24ecc97000.png

grave plank Sep 27, 2019, 1:33 AM

#

yah

#

7:52

#

idk what youre on

tranquil junco Sep 27, 2019, 1:33 AM

#

thonkzoom

#

repost ur question

#

or send screencap or smth

grave plank Sep 27, 2019, 1:33 AM

#

https://cdn.discordapp.com/attachments/540211747613704221/626944189216849961/323837464982716416.png

tranquil junco Sep 27, 2019, 1:33 AM

#

like 95% of the time its people ponging helpers randomly and the other 5% its reposti

grave plank Sep 27, 2019, 1:34 AM

#

can my proof generalize for span(u,v,w) = span(u,v)

sonic osprey Sep 27, 2019, 1:35 AM

#

Do you think you can

grave plank Sep 27, 2019, 1:36 AM

#

oh wait

#

i proved span(u,v) is a subset of span(u,v,w)

#

so by the definition of the subset

half ice Sep 27, 2019, 1:39 AM

#

No, you proved what you said you did

grave plank Sep 27, 2019, 1:40 AM

#

i mean earlier

#

in another part of the problem

#

my bad

#

but why cant i say span(u,v,w) = span(u,v) for my proof?

half ice Sep 27, 2019, 1:43 AM

#

You showed that any element in span(u,v,w) is also an element in span(u,v). This proves that span(u,v,w) is a subset of span(u,v)

#

With that recent picture

grave plank Sep 27, 2019, 1:43 AM

#

oh yeah, because if i think of it in terms of the logical form of the subset

#

an element of span(u,v,w) -> span(u,v)

#

i mean also just intuitively

half ice Sep 27, 2019, 1:44 AM

#

It's a very easy proof to show that span(u,v) is a subset of span(u,v,w)

grave plank Sep 27, 2019, 1:44 AM

#

yeah

half ice Sep 27, 2019, 1:44 AM

#

So you should be able to show they're equal sets

grave plank Sep 27, 2019, 1:45 AM

#

yeah

#

ty

cedar solar Sep 27, 2019, 2:35 AM

#

What is a good interpretation of pseudoinverse?

#

Like if i have

#

$$X X^\dagger y$$

stoic pythonBOT Sep 27, 2019, 2:35 AM

#

BamBam:

📎 481839404869222400.png

cedar solar Sep 27, 2019, 2:36 AM

#

Assuming X is not invertible because the case when it evaluates to I is obvious

#

So $X X^\dagger$ is a projection but projection onto what

stoic pythonBOT Sep 27, 2019, 2:37 AM

#

BamBam:

📎 481839404869222400.png

native lodge Sep 27, 2019, 2:41 AM

#

thinking it's row space

#

but looks like I'm wrong lol

#

📎 unknown.png

cedar solar Sep 27, 2019, 2:59 AM

#

Oh

#

Thanks

#

I just realized it is explained in wiki but i mustve scrolled past it earlier

junior shard Sep 27, 2019, 4:53 AM

#

📎 3aLinAlg.JPG

#

📎 3aaa.png

#

📎 3aaaaaa.png

#

is this correct?

#

im not sure if i can multiply c_1,c_2,c_3 by the different derivatives

pallid swallow Sep 27, 2019, 10:50 AM

#

you can because it's true for all x

hardy blaze Sep 27, 2019, 2:01 PM

#

yall this is a dumb question

#

if im asked to find 3(AB)

#

i find AB then scale it by 3 right ...

dusky epoch Sep 27, 2019, 2:09 PM

#

you could do that, sure

short lava Sep 27, 2019, 5:19 PM

#

I’m not asking for homework help but I came across this problem that I need help with

#

“Prove L(x)=(u•v)v is a linear transformation”

#

Any insight?

dusky epoch Sep 27, 2019, 5:20 PM

#

are you sure you didn't mistype anything

#

as written it either doesn't make any sense or is underspecified

#

post the whole question exactly as it is stated

short lava Sep 27, 2019, 5:26 PM

#

@dusky epoch hey sorry I don’t have access to the question anymore

#

R^3–>R^3 mapping

scenic acorn Sep 27, 2019, 5:35 PM

#

Are you sure it's not something like "Show that L(x)=(x \cdot v)v is a linear transformation for any fixed v in R^3."?

lunar sable Sep 27, 2019, 11:10 PM

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Can someone guide me through to find the special solution to

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1

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📎 Screen_Shot_2019-09-27_at_7.12.32_PM.png

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Sorry about the pic

pallid swallow Sep 28, 2019, 12:04 AM

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you mean, a particular solution?

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Just gaussian elimination it

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WAIT

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erm

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It has no solution lol

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because you can't multiply a 3x4 matrix with a 3x1 matrix

charred stirrup Sep 28, 2019, 12:20 AM

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Why cant vector n be [1, 1] and vector x be [2x, y]?

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📎 unknown.png

steady fiber Sep 28, 2019, 2:07 AM

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does it say anywhere that it can't be [1, 1] and [2x, y]

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but it's better to split the coefficients and variables

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and I believe it's necessary for normal form

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it gives some useful properties (as outlined in the description you linked)

charred stirrup Sep 28, 2019, 3:40 AM

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oh so it convention and good practice

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@steady fiber thank you

urban bough Sep 28, 2019, 6:18 AM

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📎 image0.jpg

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What specifically is meant by this?

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Are ui and vit supposed to be full matrixes in of themselves?

dusky epoch Sep 28, 2019, 6:25 AM

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they're vectors

urban bough Sep 28, 2019, 8:04 AM

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Never mind, I got it

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I thought multiplying them was supposed to produce a scalar

undone garnet Sep 29, 2019, 12:16 AM

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$(I+AB)$ is invertible, prove that $(I+BA)$ is invertible too

stoic pythonBOT Sep 29, 2019, 12:16 AM

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Nguyễn Thành Trung:

📎 461183575627137034.png

undone garnet Sep 29, 2019, 12:17 AM

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hm...any idea

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my proof is

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if lambda is an eigenvalue of AB then it is also an eigenvalue of BA

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so det(I+AB) = (1+lambda_1)(1+lambda_2)...(1+lambda_n) = det(I+BA)

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I hope to see another idea for this problem :p

quartz compass Sep 29, 2019, 12:30 AM

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my gut instinct when I see something of this form is to use a geometric series if possible

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$(I-X)^{-1} = I+X+X^2+X^3+\cdots$

stoic pythonBOT Sep 29, 2019, 12:30 AM

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Merosity:

📎 622029249053982720.png

quartz compass Sep 29, 2019, 12:30 AM

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just make sure things converge

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well, idk if that's the best approach here, I think making use of this is a good idea:

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$B(I+AB) = B+BAB = (I+BA)B$

stoic pythonBOT Sep 29, 2019, 12:31 AM

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Merosity:

📎 622029249053982720.png

quartz compass Sep 29, 2019, 12:32 AM

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there should be the same kind of relation for A as well, just some general ideas to think about that might help

pallid swallow Sep 29, 2019, 12:34 AM

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A and B are square matrices?

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or not necessarily?

undone garnet Sep 29, 2019, 12:34 AM

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hm..don't remember exactly

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maybe square

steady fiber Sep 29, 2019, 12:48 AM

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BA and AB have the same eigenvalues except for a bunch of zeros (in this case, they're square matrices of equal dimensions, so they have the exact same eigenvalues with the same algebraic multiplicities). Since (I+AB) is invertible, -1 isn't an eigenvalue of AB, so it's not an eigenvalue of BA either. If -1 isn't an eigenvalue of BA, (I+BA) must be maximal rank, and therefore must be invertible

undone garnet Sep 29, 2019, 12:49 AM

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well

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I mean

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is there another proof for this problem

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without using eigenvalue

steady fiber Sep 29, 2019, 12:50 AM

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Probably

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Is there a restriction on using eigenvalues

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This is probably the simplest way to prove it

undone garnet Sep 29, 2019, 12:51 AM

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yeah

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but this problem was on a midterm exam

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eigenvalue hadn't have introduced

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so I'm looking for another method

steady fiber Sep 29, 2019, 12:56 AM

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Oh there is another way

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Find an explicit expression for the inverse of I+AB in terms of I, A, B and BA

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It will automatically prove the existence of the inverse of I+BA

lone cove Sep 29, 2019, 12:57 AM

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assume the inverse of (I + BA) is (I + C)
for some C

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then find what C can be

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this is important for that

📎 622029249053982720.png

undone garnet Sep 29, 2019, 1:00 AM

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hm...

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$(I+AB)(I+C) = I \Leftrightarrow AB+C+ABC=0$

stoic pythonBOT Sep 29, 2019, 1:01 AM

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Nguyễn Thành Trung:

📎 461183575627137034.png

lone cove Sep 29, 2019, 1:01 AM

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you mixed up the order of A and B

undone garnet Sep 29, 2019, 1:02 AM

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huh?

lone cove Sep 29, 2019, 1:03 AM

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we want to find the inverse of I + BA not I + AB

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we already have the inverse of I + AB

undone garnet Sep 29, 2019, 1:04 AM

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already have?

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where?

lone cove Sep 29, 2019, 1:05 AM

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from the question

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we know it exists

undone garnet Sep 29, 2019, 1:05 AM

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ah ah we just know it exists

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so

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$(I+BA)(I+C) = I \Leftrightarrow BA+C+BAC=0$

stoic pythonBOT Sep 29, 2019, 1:05 AM

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Nguyễn Thành Trung:

📎 461183575627137034.png

lone cove Sep 29, 2019, 1:07 AM

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$(I + BA) C = -BA$

stoic pythonBOT Sep 29, 2019, 1:07 AM

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Sigma:

📎 187332490958995457.png

undone garnet Sep 29, 2019, 1:08 AM

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hm...

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$C=-(I+BA)^{-1}BA$

stoic pythonBOT Sep 29, 2019, 1:11 AM

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Nguyễn Thành Trung:

📎 461183575627137034.png

undone garnet Sep 29, 2019, 1:11 AM

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$C = -(I+C)BA = -BA - CBA$

stoic pythonBOT Sep 29, 2019, 1:11 AM

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Nguyễn Thành Trung:

📎 461183575627137034.png

lone cove Sep 29, 2019, 1:13 AM

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📎 622029249053982720.png

undone garnet Sep 29, 2019, 1:15 AM

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@@

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let $C = (I+AB)^{-1}$\
$(I+BA)BCA = BCA + BABCA = B(C+ABC)A = B((I+AB)C)A=BIA=BA$\
$(I+BA)(BCA-I)=(I+BA)BCA - (I+BA) = BA - I - BA = -I$\

stoic pythonBOT Sep 29, 2019, 1:28 AM

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Nguyễn Thành Trung:

📎 461183575627137034.png

undone garnet Sep 29, 2019, 1:28 AM

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done

scarlet hamlet Sep 29, 2019, 2:11 AM

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hey quick nub question here

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to compute the norm of a vector v in an inner product space

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📎 unknown.png

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does that work for all of 2x2 matrices in R?

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ok ignore i asked that

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i wrote out a general one and figured it out myself :p

scarlet ermine Sep 29, 2019, 3:08 PM

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i'm pretty confused by this question and what its asking

📎 unknown.png

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what does it mean when it says 'set of prices when the price for output is 100 units'?

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i've gotten to this stage but when i row reduced i got the 1 1 1 matrix which isn't useful

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📎 0oypIXIbLjqySWWC6f308U9B4z9rtwfJmjzEj5Zk5Yt53d6kmJn2-jk4uyeFnR5IPogjWRZuPD8yy4Uba_kh1qGQvuCJrmTaN3m5.png

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so i think i made a mistake somewhere

prisma prawn Sep 29, 2019, 3:35 PM

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guys why a linear transformation is injective if and only if the kernel is 0 ??

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I need some help to understand it conceptually, geometrically

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this is how i think

dusky epoch Sep 29, 2019, 3:43 PM

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do you understand injective => zero kernel?

rancid sonnet Sep 29, 2019, 4:19 PM

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derivatives can be represented as matrix operators right?

half ice Sep 29, 2019, 4:20 PM

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Kind of. They'd have to be infinite if your vector space is all polynomials

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And we don't like to call them matrices anymore if they're infinite in size

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But if you're limiting the degree, which is still a vector space, then yes definitely

dusky tartan Sep 29, 2019, 4:23 PM

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this is a good explanation for the linear transformation 0 kernel thing https://math.stackexchange.com/a/2193375

Mathematics Stack Exchange

Showing a linear map is injective if and only if kernel is {$ {0} $}

So my prof gave me this proof:

$f(x) = f(y) ⇐⇒ f(y − x) = 0 ⇐⇒ y − x ∈ Ker f.$

I dont see why this proof is enough, this only says $y-x \in Ker f$

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the main thing is understanding what linearity means and then it's clear

steady fiber Sep 29, 2019, 4:26 PM

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@rancid sonnet

📎 unknown.png

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Calculus on Manifolds, by Michael Spivak (Chapter 2) might give you more information on what you want

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about derivatives being linear operators (and being represented by matrices)

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there's a lot more to it, so check out the book if you'd like (and don't skip Chapter 1)

rancid sonnet Sep 29, 2019, 4:28 PM

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ok I will

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But if I represent polynomials with column matrices, then can I represent complex polynomials with a matrix too, by adding a second column for the imaginary components?

dusky tartan Sep 29, 2019, 4:33 PM

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you don't need to add an extra column

rancid sonnet Sep 29, 2019, 4:33 PM

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you mean just make the elements complex?

dusky tartan Sep 29, 2019, 4:33 PM

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yeah, complex numbers are still a field so they're not really any different in terms of polynomials

rancid sonnet Sep 29, 2019, 4:35 PM

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I want to solve some eigenvalue problems by trying to represent the hamiltonian operator of quantum mechanics as a matrix by writing some python code, so since I can write derivatives and polynomials as matrices, I'm gonna give it a try

steady fiber Sep 29, 2019, 4:36 PM

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I mean complex numbers are different, and you can't assume complex valued polynomials are differential everywhere (or even, anywhere)

dusky tartan Sep 29, 2019, 4:37 PM

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yeah but in terms of just defining a polynomial by coefficients it makes more sense to just keep it as one column vector

rancid sonnet Sep 29, 2019, 5:14 PM

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wait how do I represent polynomial multipllication in matrices?

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I got my second derivative operator in the hamiltonian

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but if I want to express the potential as an operator

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I should express it as an nxn matrix instead of a nx1 polynomial matrix right?

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Should I just

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enter coefficients of x^0 to x^(n-1) in the first row and

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cycle them

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in the next rows?

quaint heart Sep 29, 2019, 7:18 PM

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"yeah, complex numbers are still a field so they're not really any different in terms of polynomials"

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laughs in Algebraically closed

steady fiber Sep 29, 2019, 7:36 PM

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laughs in ordered field

hardy blaze Sep 29, 2019, 8:51 PM

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is the only way to get an identity matrix

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by multiply a matrix & the inverse of that matrix

gray dust Sep 29, 2019, 8:52 PM

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you can also have this because I is its own inverse

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$II = I$

stoic pythonBOT Sep 29, 2019, 8:52 PM

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RokettoJanpu:

📎 150414892225134592.png

hardy blaze Sep 29, 2019, 8:52 PM

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what if AB = Identity

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wuld that mean B is the inverse of A

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or A is the inv of B

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like would it mean both of those statements*

stoic pythonBOT Sep 29, 2019, 8:55 PM

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RokettoJanpu:

$A^{-1}A = AA^{-1}=I$

📎 150414892225134592.png

hardy blaze Sep 29, 2019, 8:59 PM

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so if AB = I, then A=IB^-1 correct?

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couldnt you make it A=B^-1 since I is like 1

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wait

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AB = I, so B is the inverse of A

wintry steppe Sep 29, 2019, 9:03 PM

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that's assuming both B is a square matrix

hardy blaze Sep 29, 2019, 9:03 PM

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yes

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A,B,C are all nxn

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so is it possible for AC to = 0 without C = 0? 🤔

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i dnt think so bc if A is invertible, then AC=0 -> AA^-1C=0 -> C = 0

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but maybe bc u dont have to put A^-1

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but then if u solve for C u would just get 0

wintry steppe Sep 29, 2019, 9:06 PM

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try A = [[1, 0], [0, 0]] and C = [[0, 0], [0, 1]]

hardy blaze Sep 29, 2019, 9:07 PM

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see

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how yall come up w matrices like that that fast 😭 😭

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so it is possible for AC to = 0 if C = that matrix u put

wintry steppe Sep 29, 2019, 9:07 PM

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for AC = 0, the image of C has to be in the null space of A

hardy blaze Sep 29, 2019, 9:07 PM

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well

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i havent learned that 😳

wintry steppe Sep 29, 2019, 9:08 PM

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obviously, if C = 0, then the image of C is just the 0 vector and that condition is trivially satisfied

wintry steppe Sep 29, 2019, 9:31 PM

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I find that most of the time when people say that they haven't learned it, they really have, but just didn't understand it.

formal moss Sep 29, 2019, 10:11 PM

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question

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if y = mx+b and m (slope) is undefined does that mean it's equal to 0?

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or is a 0 slope undefined??

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I'm given two points with an undefined slope so how do i figure the slope intercept form

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when y-y1 = m(x-x1)

gray dust Sep 29, 2019, 10:14 PM

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0 slope is not undefined slope

formal moss Sep 29, 2019, 10:15 PM

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so how can i solve the point slope formula if i have a undefined slope

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slope intercept*

dense sonnet Sep 29, 2019, 10:40 PM

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@hardy blaze if AB=I then that means B is the inverse of A if and only if A is the inverse of B. Also yes you can have two matricies not equal to the zero matrix that multiply to be zero. You can show this for a 2X2 case easily as someone else has shown you. Just multpily any arbitrary matrices 2X2 and then find what conditions will satisfy it to be the zero matrix. Linear is abstract but you'll get it eventually

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@formal moss 0 slope is not the same as undefined slope. Think of it as rise over run. 0 slope can be written as (0/n) for any number n so we rise 0 and move right any amount. undefined slope is kinda like (n/0) so we rise any amount and move right zero. however this is kinda over simplifying it since you cannot divide by zero that is why it is undefined

formal moss Sep 29, 2019, 10:44 PM

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so the points are (2/3 , 1/5) and m is undefined and to find the answer in slope int form the answer is 2/3 and idk why

dense sonnet Sep 29, 2019, 10:47 PM

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if you have an undefined slope which coordinate will stay the same on your line?

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@formal moss

formal moss Sep 29, 2019, 10:48 PM

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b?

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oh

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the x coordinate will

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hmm i dont think thats right either

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its the y

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coordinate

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that stays the same

dense sonnet Sep 29, 2019, 10:49 PM

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yes! so usually we have the equation y=mx+b but what you just said was our x will always be 2/3. so instead of y= youll have an equation x=

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do you know what the equation x= will be?

formal moss Sep 29, 2019, 10:50 PM

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umm

dense sonnet Sep 29, 2019, 10:50 PM

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and you were right first with the x stays the same for an undefined slope

formal moss Sep 29, 2019, 10:51 PM

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x = 2/3

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?

dense sonnet Sep 29, 2019, 10:51 PM

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perfect!

formal moss Sep 29, 2019, 10:51 PM

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OOH

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i see

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2/3rds is our constant and with an undefined slope(rise over run) it's just x = 2/3?

dense sonnet Sep 29, 2019, 10:52 PM

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you got it

formal moss Sep 29, 2019, 10:52 PM

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oddly enough that sounds like the same as if it was 0 slope though

dense sonnet Sep 29, 2019, 10:53 PM

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draw the line x=1 and y=1. what do you notice? they are similar but have a big difference

scarlet hamlet Sep 29, 2019, 10:53 PM

#

o

formal moss Sep 29, 2019, 10:53 PM

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one is horizontal and the other vertical?

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waiit

dense sonnet Sep 29, 2019, 10:53 PM

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yes. x=1 is vertical and y=1 is horizontal

formal moss Sep 29, 2019, 10:53 PM

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oh

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ok

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so it's an undefined function where the constant is 2/3rds

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if a slope is undefined?

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its not even a function actually

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since x=2/3 fails the VLT

dense sonnet Sep 29, 2019, 10:55 PM

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yes. x=2/3 is not a function

formal moss Sep 29, 2019, 10:55 PM

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i see

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thank you!

dense sonnet Sep 29, 2019, 10:55 PM

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no problem

sand oak Sep 30, 2019, 1:05 AM

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Can anyone help me with a problem

#

Let V = the x1x3 plane (also called xz plane) in R3, and v1 = (1,0,2), v2 = (0,0,1). Is V contained in Space {v1,v2}?

half ice Sep 30, 2019, 1:25 AM

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@sand oak
You mean Span{v1, v2}? Yes, it is. In fact, they're the same space

toxic pendant Sep 30, 2019, 2:36 AM

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What are complex numbers used for? Searching online gets me stuff like "used for electricity" but no examples.

gray dust Sep 30, 2019, 2:37 AM

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lots of things. circuit analysis is certainly one application

toxic pendant Sep 30, 2019, 2:38 AM

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Can you give an example of how complex numbers are used to solve problems

gray dust Sep 30, 2019, 2:38 AM

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~~quantum mechanics too but that's for the physics discord server~~

toxic pendant Sep 30, 2019, 2:39 AM

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wait I'm dumb, complex variables is literally right under this channel

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also physics discord? gib

gray dust Sep 30, 2019, 2:39 AM

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#old-network

half ice Sep 30, 2019, 2:39 AM

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Circuit analysis isn't a small one lol.

You can't get into any real integral transformations without including complex numbers

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Differential equations will include complex at some point, and differential equations are a lot of engineering

gray dust Sep 30, 2019, 2:40 AM

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did i say it was a small one?

half ice Sep 30, 2019, 2:40 AM

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No, you didn't

toxic pendant Sep 30, 2019, 2:43 AM

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I'm still ignorant of how to apply them so I can't really "imagine" a scenario using them. Can you give a problem which would require them

half ice Sep 30, 2019, 2:43 AM

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Know what a derivative is?

toxic pendant Sep 30, 2019, 2:44 AM

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I think so

gray dust Sep 30, 2019, 2:45 AM

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you have a circuit which consists of a resistor, inductor, and capacitor strung together in series with an alternating voltage source (it drives AC current). you can use complex numbers to model the current through the circuit over time (look up phasors too)

half ice Sep 30, 2019, 2:46 AM

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A differential equation is an equation that relates an equation to it's derivatives. Lots of natural phenomena follow differential equations. Getting information out of them isn't always easy, but integral transforms are a common way to do it.

The Fourier transform needs complex numbers, even if the system you're looking at is real

toxic pendant Sep 30, 2019, 2:48 AM

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Huh, interesting

steady fiber Sep 30, 2019, 2:48 AM

#

I mean you can do the "easy" fourier transform with R->R^2

gray dust Sep 30, 2019, 2:48 AM

#

discussion belongs in #real-complex-analysis tbh

steady fiber Sep 30, 2019, 2:48 AM

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which is how it's often first taught

#

sadly

#

where you do cosine and sine seperately

half ice Sep 30, 2019, 2:49 AM

#

What Roketto and I are describing are related, a transform allows one to find phasors

#

I think anyway? I know you can use the Laplace to get circuit stuff

toxic pendant Sep 30, 2019, 2:50 AM

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Well, this was helpful, I have a lot of wikipedia to read though now

#

thanks

half ice Sep 30, 2019, 2:51 AM

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Np, feel free to ask if you have any questions about it

dense sonnet Sep 30, 2019, 3:19 AM

#

@toxic pendant so one quick thing you can notice for complex numbers is the formula for the roots of a cubic. now for specific cubics youll naturally run into imaginary numbers using the formula. but after computing it youll get a real solutions (all the imaginary parts cancel each other) and you end up with a real solution. however you go through a "world" of a number system that real numbers cant explain in order to get your solution. so imaginary numbers can arise to get real solutions. in the lift of a plane to model some stuff imaginary numbers will naturally arise having a real meaning in how the system behaves

toxic pendant Sep 30, 2019, 3:22 AM

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Thank you for the explanation

#

~~you may or may not have saved me from the wikipedia rabbit hole~~

quartz compass Sep 30, 2019, 3:32 AM

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I can show you some stuff if you have some time

#

I find it nice to take a geometric perspective and see the complex numbers as extending the number line into a 2D number plane

#

gives a very clear interpretation of exponents and multiplication, just ping me if you're around

toxic pendant Sep 30, 2019, 3:34 AM

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will do

upbeat niche Sep 30, 2019, 4:55 AM

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Hey yall I got a quick question

#

📎 20190929_215459.jpg

#

I'm sure this is actually super simple but idk where to start

#

Or what to get from this

dusky epoch Sep 30, 2019, 4:56 AM

#

,rotate -90

stoic pythonBOT Sep 30, 2019, 4:56 AM

#

📎 628092545020198912.png

dusky epoch Sep 30, 2019, 4:56 AM

#

do you have access to determinants

upbeat niche Sep 30, 2019, 4:59 AM

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Nope that's all I was given, or were you asking if I can find them from this system of eq?

quartz compass Sep 30, 2019, 5:02 AM

#

I guess you have to get your hands dirty and do gaussian elimination then

dusky epoch Sep 30, 2019, 5:07 AM

#

@upbeat niche i'm not asking whether or not you were given anything else, i'm asking whether your class has ever talked about determinants in any capacity

upbeat niche Sep 30, 2019, 5:09 AM

#

Right now I'm just trying to solve the augmented matix

#

Into rref

cobalt tartan Sep 30, 2019, 5:50 AM

#

So in my lin alg class I have this,

📎 unknown.png

#

When they say that "If and only if v = 0", say for example that we have some p in P^2

#

p(x) = ax^2 + bx + c

#

Right?

#

When they say that the vector has to be the 0 vector do they mean that we have to have p(x) = 0x^2 + 0x + 0?

#

Or does p(x) = 0, i.e, the values of its roots...?

quartz compass Sep 30, 2019, 5:51 AM

#

the first statement

#

the actual "value" of x doesn't matter when talking about the vector space P^2

cobalt tartan Sep 30, 2019, 5:51 AM

#

hrm

quartz compass Sep 30, 2019, 5:52 AM

#

think of them as just letter placeholders, they're basis vectors taking on no value

cobalt tartan Sep 30, 2019, 5:52 AM

#

I'm a bit confused about this question then,

📎 unknown.png

#

Oke

#

As I can like... find p, q, and values of a, b, c such taht it's not positive definite...?

quartz compass Sep 30, 2019, 5:53 AM

#

here a,b,c are like separate x values, they're not the coefficients on the polynomials

cobalt tartan Sep 30, 2019, 5:54 AM

#

Hrm

quartz compass Sep 30, 2019, 5:54 AM

#

it's like they've taken the original version but they've repeated it three times in a row

cobalt tartan Sep 30, 2019, 5:54 AM

#

Ok so like is the question basically saying

#

Let p(x) = p_1x^2 + p_2x + p_3, q(x) = q_1x^2 + q_2x + q_3, and then go and show that for any a, b, c, the inner product defined in the question will equal 0 if and only if p_1 = p_2 = p_3 = 0 or q_1 = q_2 = q_3 = 0?

quartz compass Sep 30, 2019, 5:55 AM

#

very close, it's not wrong exactly, but

#

the last part I think should be said as when p=q and then p_1 = p_2 = p_3 = 0

#

since they're not separate

#

I think I might be nitpicking in a confusing way

cobalt tartan Sep 30, 2019, 5:56 AM

#

Oh right

#

Because positive definite requires them to be the same right?

quartz compass Sep 30, 2019, 5:56 AM

#

yeah

cobalt tartan Sep 30, 2019, 5:57 AM

#

Err because positive definite is'nt on different p and q, it's on the same p and q

#

Like

#

Do I just expand it out...?

#

Like I can do that but it's a lot of work lol

quartz compass Sep 30, 2019, 5:58 AM

#

well, I'd just rewrite it with p=q and then

cobalt tartan Sep 30, 2019, 5:58 AM

#

err

#

I'd jsut write it with <p, p>

quartz compass Sep 30, 2019, 5:58 AM

#

all the terms are really just (p(a))^2 + (p(b))^2+ (p(c))^2

cobalt tartan Sep 30, 2019, 5:58 AM

#

Ya

quartz compass Sep 30, 2019, 5:58 AM

#

yes exactly

#

is it clear how this is gonna work out, you don't have to expand anything

cobalt tartan Sep 30, 2019, 5:58 AM

#

Oh

#

Hrm

quartz compass Sep 30, 2019, 5:59 AM

#

p(a) is just a number

#

it could be positive or negative

#

p(a)^2 however is always positive

cobalt tartan Sep 30, 2019, 5:59 AM

#

Yea, <p, p> = 0 iff p(a)^2 = p(b)^2 = p(c)^2

#

But because it's in p2

#

You can only have 2 roots

#

So it's impossible for all 3 to be 0

quartz compass Sep 30, 2019, 5:59 AM

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not necessarily you also have to have p(a)^2 = p(b)^2 = p(c)^2=0

cobalt tartan Sep 30, 2019, 5:59 AM

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Err yea

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Only in the case where it's all 0

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Oh wait

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Hrm

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Well, showing positive/negative is easy enough, p(a), p(b), p(c) are all real numbers, when you square them , they will always be positive, so their sum will also be positive

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Right?

quartz compass Sep 30, 2019, 6:00 AM

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yeah good

cobalt tartan Sep 30, 2019, 6:01 AM

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Hrmmm yea my roots idea would have like 3 cases which is too much work

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Would need to show the case where 3 distinct, 2 distinct, and no distinct

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And my idea only holds for 3 distinct/no distinct

quartz compass Sep 30, 2019, 6:05 AM

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it's good cause you have been given all 3 are distinct

cobalt tartan Sep 30, 2019, 6:05 AM

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Oh wait

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Hrm

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Then does my idea work then if I just show the 2 cases?

quartz compass Sep 30, 2019, 6:07 AM

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I don't think I follow your reasoning, I'm imagining 3 quadratics each having at most 2 roots

cobalt tartan Sep 30, 2019, 6:07 AM

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Well because it's <p, p> we only have one quadratic

quartz compass Sep 30, 2019, 6:07 AM

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err, 1 quadratic

cobalt tartan Sep 30, 2019, 6:07 AM

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One quadratic would have at most 2 roots

quartz compass Sep 30, 2019, 6:08 AM

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3 results of plugging in 3 roots to 1 quadratic which can have at most 2 roots

cobalt tartan Sep 30, 2019, 6:08 AM

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Ah wait no

quartz compass Sep 30, 2019, 6:08 AM

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so one of them can't be 0