linear-algebra

[zinc tapir]

For the second part: since span(S1) is a subset of span(S2), let span(S1)=V which implies that V is a subset of span(S2). S1 is a subspace of V, therefore S2 must also be a subspace of V. And so span(S2) is a subset of V. Therefore span(S2)=V.

[eternal finch]

In particular, if S_1 is a subset of S_2 and the span of S_1 = V, deduce that the span of S_2 = V.
The problem statement.

since span(S1) is a subset of span(S2), let span(S1)=V which implies that V is a subset of span(S2)
I would reorder this for clarity.

Let the span of S_1 = V. The span of S_1 is a subset of the span of S_2, so V is a subset of the span of S_2.

S1 is a subspace of V, therefore S2 must also be a subspace of V.
S_1 is a subset of V, but S_1 is not necessarily a subspace of V. If you replace "subspace" with "subset" here, then you're good.

And so span(S2) is a subset of V.
Yep.

Therefore span(S2)=V.
Yep.

[zinc tapir]

Let the span of S_1 = V. The span of S_1 is a subset of the span of S_2, so V is a subset of the span of S_2. S1 is a subset of V, therefore S2 must also be a subset of V. And so span(S2) is a subset of V. Therefore span(S2)=V.

whoops

dunno how i managed to put subspace in there

[eternal finch]

Yeah, that's it.

dunno how i managed to put subspace in there
I used to make the same mistake. Subset, subspace, subsomething. Similar enough.

[zinc tapir]

arrows mean implies right

[zinc tapir]

ty guys very much appreciated

[teal topaz]

Let me know if I'm interrupting anything in here! Trying to prove a result -- If for an inner product space $\mathbb{V}$ we have $\mathcal{B}={\vec v_1,\ldots,\vec v_n}\subset\mathbb{V}$ as a basis, and we have that $\sum_{j=1}^n c_j\left(\sum_{i=1}^n d_i\langle\vec v_i,\vec v_j\rangle\right)=\sum_{k=1}^n c_kd_k$, can we conclude that we must have $\langle\vec v_i,\vec v_i\rangle=1$ for all $i$ and that $\langle\vec v_i,\vec v_j\rangle=0$ for all $i\ne j$?

[stoic python]

Stract:

[teal topaz]

(where c_i and d_i are real scalars for all i)

I thought about trying something with linear independence of B but I don't know if that's really relevant here

[eternal finch]

Do you suspect it must be true, and why?

[teal topaz]

I suspect it ought to be true because it seems like the only viable lane for the proof I'm trying to do 😆 I'm not sure though, again I thought something with linear independence but I feel that's not relevant, so I'm in the dark

[eternal finch]

Ok, let's explore the linear independence of the basis.

Does the linear independence of B tell you anything about that sum?

Particularly, try to connect linear independence with what you think ought to be true.

You want <v_i, v_i> = 1 and <v_i, v_j> = 0 when i =/= j.

[teal topaz]

Right, I'm not sure if it does though. I'm not sure if there's some weird situation where things might just coincidentally cancel out to yield 0 somewhere, since inner products aren't unique to pairs of vectors

[eternal finch]

I'm not sure if it does though
Ok, I changed "say" to "want" to reflect that.

Ok, so you pointed out that inner products aren't unique to pairs of vectors.

That's certainly true.

Not too useful, tho.

How about thinking about when <v, v> = 1 and when <u, v> = 0.

[teal topaz]

I mean I'm ultimately trying to prove the orthonormality of B here

what I'm doing here is just an attempt at a stepping stone to that

(proving orthonormality of B assuming the inner product equals the dot of coordinate vectors wrt B)

[eternal finch]

Hm, if that's the case, let me attempt to disprove this stepping stone.

Hm, actually, I have a question.

Does the sum need to hold for all choices of c's and d's?

Or is it that we have that sum is satisfied for some choice of c's and d's?

[teal topaz]

I mean yeah, the c's and d's are just coefficients for some arbitrary vectors written in terms of the basis B

i.e. $\vec x=c_1\vec v_1+\cdots+c_n\vec v_n$

[stoic python]

Stract:

[eternal finch]

So, we have that the sum holds for all choices of c's and d's?

[teal topaz]

Yes

[eternal finch]

Ok, got it.

[teal topaz]

i.e. we have $\langle\vec x,\vec y\rangle=[\vec x]\mathcal{B}\cdot[\vec y]\mathcal{B}$ for all $\vec x,\vec y\in\mathbb{V}$

[stoic python]

Stract:

[teal topaz]

that summation I wrote up there was just expanding it out into what I feel is a more useful form

is my approach indeed disprovable then? :P

[eternal finch]

Well, dunno, haven't disproven it yet. Given your original problem statement, I'm more inclined to say it's true, tho.

[teal topaz]

I proved the converse by doing something similar, I'm essentially trying to just reverse what I did there

[eternal finch]

Ok, I can't disprove or prove the statement with the sum.

I'm wondering what would be wrong with proving orthonormality directly from the definition of your inner product.

Let V be an inner product space with a basis B = {v(1), ..., v(n)} and <u, v> be the dot product of the coordinate vectors of u and v, where u and v are in V.

The coordinates of v(1) are [1, 0, ..., 0], the coordinates of v(2) are [0, 1, 0, ..., 0], and so on. If you dot v(i) with itself, you get 1, and if you dot v(i) with v(j), i =/= j, then you get 0. Ergo, B is orthonormal?

[teal topaz]

Huh I think you're absolutely right

The proof of the converse was a bit more involved so I was just overthinking this I think

[soft flicker]

y'''+y''=cscx

[quartz compass]

if you're asking, substitute y'' = u and solve u'+u = cscx first

once you have u, just integrate twice

[soft flicker]

I did that part, but had problems with wronskian

I need to solve with the method of changing parameters

[woeful cedar]

Can you guys help me with this?

[wintry steppe]

ok, try finding A^2

or even

yeah find A^2

https://yutsumura.com/how-to-find-a-formula-of-the-power-of-a-matrix/#:~:text=Diagonalize the Upper Triangular Matrix,b−a0b].

[wintry steppe]

yes

so see how this is the identity multiplied by x^2

[karmic oracle]

@wintry steppe that's certainly the general technique, but probably overkill for this problem

[wintry steppe]

yeah i was gonna tell him how to do that

thats prolly true

but not necessary for this

[woeful cedar]

so see how this is the identity multiplied by x^2
@wintry steppe ye

[wintry steppe]

ok so A^20 is just (A^2)^10

A^2 = x^2 * I

do you see what im doing?

[woeful cedar]

y

so is it just

x^20

[wintry steppe]

multiplied by I

so a diagonal matrix with x^20 as its elements

[woeful cedar]

x^2 * I

is just

x^2 right

so this is the answer?

[wintry steppe]

yes

[woeful cedar]

I got another one

[pallid rampart]

I mean just calculate the determinant

It's a bit tedious, but it's doable

[wintry steppe]

Anyone know how I can draw a function like this in Geogebra ?

Or what should I google to find how ?

The r(t) = (2, t, t^2) I can draw. But I don't know how to tell it that x=2, y=t and z=t^2.

[gritty frigate]

Any expression simplifier like photomath but for PC ?

[wintry steppe]

@gritty frigate https://alternativeto.net/software/photomath/

Personally I type everything in Symbolab

https://www.derivative-calculator.net/

https://www.integral-calculator.com/

and these two are better than symbolab for steps

Dunno of anything that takes pictures, though,.

[verbal bay]

@woeful cedar

i think its a vandermonde matrix

https://en.wikipedia.org/wiki/Vandermonde_matrix

[wintry steppe]

Hey all

I need some help to find the equation of the line tangent to the curve (1+2*sqrt(t), t^3-t, t^3+t) at (3,0,2)

Should my first step be to identify the direction vector of the curve ?

[woeful cedar]

@verbal bay Can you take a bit more high quality picture?

[verbal bay]

better?

[woeful cedar]

yeah

I know how to find the determinant of a 3x3

Is that how you do it

[verbal bay]

YEA

caps*

[woeful cedar]

oh

[verbal bay]

bu t u can just use the formula

[woeful cedar]

thats much better

whats the formula

[verbal bay]

since its a vandermonde matrix

[woeful cedar]

oh

[woeful cedar]

tbh idk what those even mean

on the formula

[verbal bay]

its product rule

[woeful cedar]

ill just find the determinant

what do you do after you find the determinant

[verbal bay]

wdym

if u find the determinant u have the answer

[woeful cedar]

k wait

@verbal bay

[verbal bay]

yea

now

take stuff together

and some elements will dissapear

for example take the z^2 together in the first term

[woeful cedar]

k wait

[verbal bay]

so itll become z*2 (y^2 ( z^2-y^2) - x^2(z^2-x^2))

etc etc

ngl this seems like a longer approach since u'll need 2 spend more time

u should just do the column operations i did and then calculate, saves a lot of time

since 2 of the 1's become a 0

but u can do it how u want ofc

[woeful cedar]

wait

how did u take the z^2 together

and got like an answer with x^2 in it

oh you took it on both things ok wait

I was just trying to do it for this one

ehhh idk

got too complicated

I divided everything and it became a mess

can you write your solution with like explanations

how you turned those 1's to 0's etc

[verbal bay]

K1 = K1-K2

Column 1 = Column 1 - column2

nvm sec

ok so

column 1 = column 1 - column 2

column 2 = column 2 - column 3

thats all u have t odo

[woeful cedar]

wait you solved it like an equation

is that how u do it

like without determinant

you messed with the columns

[verbal bay]

i just

did some column operations

so i could find the determinant easier

[wintry steppe]

I still need some help with this: find the equation of the line tangent to the curve (1+2*sqrt(t), t^3-t, t^3+t) at (3,0,2)

I found vector (2/sqrt(t), -1, 1) but idk if that is useful.

[cold topaz]

is Diagonalizable and ORTHOGNALLY Diagonalizable the same thing?

[eternal finch]

No.

[cold topaz]

oh

[wintry steppe]

Is there something wrong with my question or it's just that no one had time to look into it yet ? Asking 'cause I've had no luck on discord nor reddit so maybe I'm asking something really dumb lol

[slow scroll]

@wintry steppe ur question is more appropriate for #multivariable-calculus but the derivative of the curve at (3,0,2) is the vector tangent to the curve at (3,0,2). Then you can make a line using that information

[wintry steppe]

@slow scroll interesting, I asked here 'cause it's being thought in my linear algebra course, maybe the class covers both courses then

[slow scroll]

yea some linear algebra courses are a sort of hybrid of vector calc, lin alg, and diff eq, so its reasonable to put the question here.

[kindred ingot]

The study of linear differential equations is actually linear algebra, so 🤷‍♀️ There waters of linear algebra run deeply through the study of ODEs.

[distant granite]

is it possible to know what matrix has this vector as solution using gauss algorithm?

[brittle juniper]

idk what you mean by "solution of a matrix", but surely there isn't uniqueness of what you're looking for

[distant granite]

ik

i just need one that has that vecotr as solution

idk how to proceed to find it

:/

[slow scroll]

ur question makes no sense. Take literally any 2x2 matrix A and compute A(3,2) = b. Then the solution to the equation Ax = b is (3,2)

[dusky epoch]

@distant granite if you could post the entire problem exactly as stated it'd help a lot

[distant granite]

it's in german :/

is that ok ?

i tried to translate but i think it did not work

[dusky epoch]

ah! so you want the KERNEL and the IMAGE to be < (3,2) >

[distant granite]

yes exactly

[dusky epoch]

could've said that...

[light brook]

Is there anyone who can help me with this?

i cant think of a way to prove it.

hints are also appreciated

[dusky epoch]

take AB = BA

pre-multiply both sides by A^-1

then post-multiply both sides by A^-1

[light brook]

thank you ❤️

[light brook]

hey one more question

how can i find the inverse of a matrix if i know the the characteristic polynomial of the matrix

[dusky epoch]

do you know the charpoly and nothing else?

[open pivot]

Hey guys can anyone help me with this question? Stumped.

[torn silo]

what does A look like?

you should be able to figure that out form A + 2I = 0

[open pivot]

Yeah okay I think I know whats up cheers

[trim patio]

does transformation matrix T have to have linearly independent columns to be considered a linear transformation

like x -> Tx

[wintry steppe]

no it doesn't

but if it does have linearly independent columns

the dimension of the null space is 0

which means there's an inverse

[trim patio]

can i ask a follow up just to confirm

1 2 0
0 0 1
1 2 0
0 0 1

is a linear transformation

[wintry steppe]

yes it is

the definition of a linear transformation

[trim patio]

thanks

[wintry steppe]

is that T(cu+v) = cT(u) + T(v)

for any two vectors and scalar

and all matrices satisfy this condition

[lunar cape]

guys, how to solve this? Show that U = {A ∈ Mn(R) : At = −A} is a vector subspace of Mn(R). Im having problems proving this u + v ∈ U, for every u,v ∈ U, and αu ∈ U, for any α ∈R e u ∈ U, what is u and what is v in a situation like this?

[dusky epoch]

u and v are both just arbitrary elements of U

you're overthinking it

if you're looking for a choice of notation that's more in line with the problem statement, you could go with the letters A and B instead of the letters u and v

doesn't matter much tho

[lunar cape]

how should i structure my answer then? this is confusing me haha

[dusky epoch]

have you proved "for all" statements before?

[lunar cape]

i have stated that for any case that A may be, the subspace contains 0 because for any matrix 0a, it satisfies At = −A

but i cant wrap my head around doing that for the other properties

[dusky epoch]

you're overthinking it

all you need to prove is that if $A^T = -A$ and $B^T = -B$ then $(A+B)^T = -(A+B)$ and $(cA)^T = -cA$ for any $n \times n$ matrices $A,B$ and scalar $c$

[stoic python]

Ann:

[shy atlas]

oh so thats what At means

[eternal finch]

AT

[shy atlas]

[lunar cape]

do i just pick 2 random matrixes A and B that satisfy the condition and use what you just said? @dusky epoch

never had a question like that, and the teacher didnt really give any examples on this >.>

[dusky epoch]

no??

like

you know

A and B are arbitrary matrices which satisfy A^T = -A and B^T = -B

you know how transposes work right

like literally

[shy atlas]

an example is not a proof

[dusky epoch]

it boils down to acknowledging that $(A+B)^T = A^T + B^T$ and that $(cA)^T = c(A^T)$

[stoic python]

Ann:

[lunar cape]

so i acknowledge this and say that this is valid for any matrixes under the given conditions, right?

thanks a bunch btw, ive been trying to get somewhere with this for days, you are helping me a lot now

[last siren]

How to prove that for: \
Real, positive x,y where x is less than y \
and for real, positive number a \
$\dfrac{x+a}{y+a} + \dfrac{y}{x}>2$

[stoic python]

Niko:

[pallid rampart]

Not here

[wintry steppe]

how would you solve this?

$\begin{cases}a - ab + b = 0\a² + b² = 16\end{cases}$

[stoic python]

Mofumofu:

[wintry steppe]

i had used wolframalpha but the answer was very large when you get the exact form

https://media.discordapp.net/attachments/437044155546337292/718552124275032164/unknown.png?width=589&height=671

Where did the (1/50) come from at the end??

they factored it out of the matrix

[spice storm]

,rotate 90

[spice storm]

Is there a way to reduce the second row? Trying to the the span

[last siren]

@pallid rampart What do you mean?

[wintry steppe]

, rotate

, rccw

[pallid rampart]

@last siren It means the question doesn't belong in linear algebra, I would say

[wintry steppe]

@spice storm multiply the 2nd row by -1/15

so that you get closer to obtaining a leading 1 in the 2nd row

[spice storm]

@wintry steppe then to show linear dependent, how do I do it? I have this

Wait, I think I know what you mean

Hold on

@wintry steppe uhh your method is not working. I can’t get [0 0 1] for the second row

[wintry steppe]

your supposed to get [0 1 0] for the 2nd row

[spice storm]

I got [ 1 0 8/5 ]

[wintry steppe]

how

[wintry steppe]

i nevr said to do that

all i said was multiply the 2nd row by -1/15

[spice storm]

@spice storm multiply the 2nd row by -1/15

But I still end up getting -1/15

[wintry steppe]

to turn the -15 to a 1

[spice storm]

So I do I get rid of the -1/15? Since I don’t know how you get 0

[wintry steppe]

multiply the 3rd row by 1/15(h+11) then add it to the 2nd row to get 0 on the 2nd row

[spice storm]

Okay thanks. I need a break from LA. I’ll do this later. Thank you

[wintry steppe]

?

bruh we are getting thru this

[spice storm]

I’m eating now

[spice storm]

I don’t get zero on the last column. I get a h^2/15+22h/15+8

@wintry steppe

[wintry steppe]

show me what you have

[sick dragon]

running through definitions
does c imply that there has to be a pivot position in each col
if it doesn't, what does that mean? it's not square?

[wintry steppe]

(h+11) * (1/15(h+11) + (-1/15) should equal 0

[woeful cedar]

can someone pls help with this

[spice storm]

Doesn’t tho. @wintry steppe I put it on my calculator and still gives me the same answer

[wintry steppe]

can i see

@woeful cedar put it into matrix form

[wintry steppe]

the (h+11) is supposed to be in the denominaotr

[woeful cedar]

@wintry steppe

[spice storm]

I typed what you gave me and the calculator put it on the top

[wintry steppe]

you didn't type what i said

type this and read it carefully: $$(h+11)/15(h+11)$$

[stoic python]

rent free:

[spice storm]

I did... look

Still got the same answer

[wintry steppe]

omfg

your 2nd (h+11) NEEDS TO BE IN THE DENOMINATOR

do you see how you are multiplying 1/15 with (h+11)?

you are not supposed to do that

[spice storm]

I got zero

[wintry steppe]

exactly

thats good

now you have a row of [0 1 0]

[spice storm]

The way you wrote made it confusing the first time. Then when you did latex it made sense

(h+11) * (1/15(h+11) + (-1/15) should equal 0

[wintry steppe]

ok

yes

[spice storm]

Yea that made it confusing since I wasn’t sure what you were trying to say. So I just wrote what you send

[wintry steppe]

oh ok

makes sense

now do you the same procedure to get rid of the 2 at the top right

[spice storm]

Yea that will be for linearly independent part

Thank you

[wintry steppe]

np

glad to help

@woeful cedar you dont put the x_1, x+2 or x_3 in the matrix

just the coefficents

[woeful cedar]

ye but

I put it into a matrix solver

it said no solutions

bcs determinant was zero

[untold citrus]

can someone help me?

[woeful cedar]

no matter what you put on the right side of the matrix

[wintry steppe]

ok then

[untold citrus]

can anyone help me?

[spice storm]

@untold citrus no one is helping you because it looks like the question is from an exam. Which is not allowed

[untold citrus]

is not exam

is a homework

due monday homework

is first week

of summer class

no test

he just doing final

i can show is homework assignment

not a test

i know the rules

see not a an exam

so can someone help me

my professor not teaching have to learn on my own really so trying to ask for help to understand the material better cause i asked professor and i didnt understand

[wintry steppe]

https://www.khanacademy.org/math/algebra/x2f8bb11595b61c86:solve-equations-inequalities/x2f8bb11595b61c86:num-solutions-linear-equations/v/number-of-solutions-to-linear-equations

[untold citrus]

thank you

[wintry steppe]

np

[untold citrus]

do i have to make a matrix of the 7 equations and 10 variables?

[wintry steppe]

i don't think so

[untold citrus]

cause there no equations

do i make my own?

that what i am confused

[spice storm]

A reflection through the line y= X or (x_1=x_2) @wintry steppe

,rotate 90

[wintry steppe]

your [-4, 8] vector is going in the wrong direction

T rotates vector x, it appears

[spice storm]

Ahh, it should be on the opposite side, but is my reasoning correct?

I see thank you

[sick dragon]

If the determinant of an nxn matrix is 0, does that mean Ax=0 has only the trivial sol?

[wintry steppe]

no

https://www.quora.com/If-A-is-a-square-matrix-with-zero-determinant-then-the-equation-Ax-0-has-a-nontrivial-solution-Why

its only got nontrivial solution

[untold citrus]

Thank you also I just needed to explain in that question and video helped me and now I finish my homework

[sick dragon]

is this correct, if A is nxn matrix: nullspace of A is Ax=0
the subspace would be R^n
the dimension of this subspace is equal to the nullity
Finally, if A is invertible, what is the dimension of that subspace?

also n?

[cursive narwhal]

is this correct, if A is nxn matrix: nullspace of A is Ax=0
the subspace would be R^n
the dimension of this subspace is equal to the nullity
Finally, if A is invertible, what is the dimension of that subspace?

Urm, if A is a matrix, its nullspace is a set, not an equation. It’s certainly is related to that equation you’ve written.

The subspace of what is R^n?

Also, if you have an actual question, post a picture of the problem here. As it stands, what you’ve written isn’t coherent at all.

[sick dragon]

A

its all talking about A dude

[cursive narwhal]

Can you post a picture of the original question?

Which definition are you not sure about?

[sick dragon]

it should be the null space of A right

and the subspace should be R^n right

[half ice]

@sick dragon
So choose some A. Then, for a vector x, if Ax = 0, we say that x is in A's nullspace.

[cursive narwhal]

That first line is not correct.

[half ice]

So the nullspace is a set of vectors

The first line is pretty vague yeah haha

[sick dragon]

the second one elaborates on it

they're allr eferring to the same A

[cursive narwhal]

No, your first line is wrong

[half ice]

Nuu the second line is completely new information

[sick dragon]

how is it wrong

[cursive narwhal]

It is not the solution to Ax = 0. We're not talking about a specific solution to that equation. We are interested in all x for which that holds. In other words, we're looking at the set of solutions.

Saying that something is a solution for an equation is different from saying that it is the solution set of the equation.

[sick dragon]

solution set sorry

i typoed

all x that satisfy ax=0 is the null space

[cursive narwhal]

Yeap, that's okay. So, it's the set of all vectors x such that Ax = 0. This set, along with the same operations as R^n, happens to be a subspace of R^n. Like Kaynex said, that's new information.

[sick dragon]

ok so thatrs correct but just worded poorly?

[half ice]

With new wording, it could be perfect

[cursive narwhal]

I mean, it's okay. It's just not how I would phrase it

[sick dragon]

with some x, so long as the solution is ax=0, then that x is in the nullspace of A

and is one solution of the set

[gray dust]

make an effort in capitalization

[sick dragon]

alright what about the others? im unsure about those for suresy

[cursive narwhal]

Yeap, think of it as the defining condition of the nullspace.

(because that's what it is lul)

Your third statement, urm, is not correct?

The dimension of the image set of the linear map associated with the matrix is the rank. The dimension of the nullspace is the nullity. Two different things but, of course, related by the rank theorem.

[sick dragon]

I see.. yeah, my definition for nullity is non-existent. null you might say

would i write nullity there or null(A)

[cursive narwhal]

And if A is invertible, then the dimension of the nullspace is 0, not n. That's because the nullspace, in that case, consists only of the zero vector. To see why this is true, let f be the linear map associated with the matrix. Since A is invertible, f is invertible too and, therefore, must be bijective.

Let x be a vector in the nullspace of A. Then, f(x) = 0 = f(0). By injectivity, x = 0. So, the nullspace contains only the zero vector.

[sick dragon]

i was going back and forth on 0

[cursive narwhal]

would i write nullity there or null(A)
It's up to you. I don't use either term in my own notes/work. I just refer to it as the dimension of the nullspace.

I see.. yeah, my definition for nullity is non-existent. null you might say
You're null

[sick dragon]

ye

alright its beautiful

[half ice]

nullity is a nice word, if only for rank-nullity theorem

[cursive narwhal]

just like your mom

prank dude prank

[sick dragon]

i dont have a mom i ahve 2 dads

[half ice]

That's nice of you to say

[cursive narwhal]

oh kaynex, i was talking to breakfastbattle

[half ice]

Yaya

[sick dragon]

in b4 hating on the way i wrote it

[half ice]

So dimension = "number of vectors in the basis"

Not "number of ways to represent the basis"

[sick dragon]

sounds the same?

[cursive narwhal]

n ways to represent a basis for H? What does that mean?

The second statement is true.

The third statement is true.

The span of that basis will "cover" all of R^n needs to be clarified. What does it mean for it to "cover" all of R^n?

That last statement is correct too. It's just that typing out vd+1 is somewhat weird because it makes it seem like you're adding 1 to vd, which isn't what you're doing.

[half ice]

There's infinitely many ways to represent a basis for R². Some of them are:
(1,0),(0,1)
(2,2),(2,3)
(5,1),(8,8)
(999999,0),(0,1)
...

But ANY basis I could possibly write always has two vectors.

[sick dragon]

true

ok, you win

idk what an intersection even is

[cursive narwhal]

......

Then why are you doing linear algebra?

[half ice]

"Hint: Do the problem"

Lol that's a pretty useless hint

[sick dragon]

yeah, i know.
by interesection doeds it just mean lines that cross

[cursive narwhal]

Let $A$ and $B$ be sets. Then:

$A \cap B = {x | (x \in A) \text{and} (x \in B) }$

That's the set intersection.

[stoic python]

Abhijeet Vats:

[cursive narwhal]

by interesection doeds it just mean lines that cross
No. Intersections have a more general meaning.

See what I have written above.

[sick dragon]

the staple means intersection?

[half ice]

If breakfast is a STEM like I was, there was no class on set theory

[cursive narwhal]

Yes. That's the intersection of A & B.

[sick dragon]

yeah ive never seen that shit haha

[cursive narwhal]

wot y'all don't have classes on set theory?

[half ice]

As a Canadian engineer, they didn't teach me set theory

Sed

[cursive narwhal]

pranked

Anyways, it's not too hard to understand

[half ice]

Still taught me applied linear algebra though, as if that makes any sense

[sick dragon]

is there supposed to be a symbol for 'and'

[cursive narwhal]

haha my uni has no such thing as "applied classes for physicists or engineers"

they do the same math classes as us

anyways, yes, there is a symbol for 'and'. Typically, you use $\land$

[stoic python]

Abhijeet Vats:

[half ice]

If A and B are sets
We can construct a new set A ∩ B called "A intersection B". This set only has elements common to both sets.

[cursive narwhal]

So, if $A$ and $B$ are sets, then we define:

$A \cap B = {x | (x \in A) \land (x \in B)}$

[stoic python]

Abhijeet Vats:

[cursive narwhal]

But i didn't want to use that because I wasn't sure if you'd be okay with taking that in as an additional bit of notation

[half ice]

So basically the question is saying "take two subspaces of Rⁿ. Show that whatever they have in common is also a subspace"

[sick dragon]

i see

[cursive narwhal]

Hmm, but what confounds me is how you’ve been able to do all of the things we were discussing earlier and not have encountered set intersections at all

That seems very strange to me

[sick dragon]

🤷

online classes

[cursive narwhal]

Like I said, explain what it means to “cover all of R^n”

[eternal finch]

Yo mama so fat, she covers all of R^n.

[sick dragon]

lol

[eternal finch]

Prank bro prank

[cursive narwhal]

And if H = R^n, then it is clear that d = n.

[sick dragon]

by cover, i guess i mean that its everything in that span

like on the plane

[cursive narwhal]

Also, your first bullet point is still not correct.

[sick dragon]

oh, i forgot to fix that

yea below i wrote infinite

[cursive narwhal]

Let B be any basis of H. Then, span(B) = H. Since H = R^n, it follows that span(B) = R^n. That’s what we mean. The span of the basis will be equal to the set of all n-tuples with real entries.

That’s what you mean by “cover”

[sick dragon]

could mean what it's wearing

[cursive narwhal]

No u

[shy atlas]

is it still gonna be 3 if its an augemented matrix

[slow scroll]

im not sure what your picture has to do with augmented matrices, but yes, if you just added a column to make an augmented matrix, it would still have rank 3. In general, rank can't exceed the number of rows (equations) you have.

[shy atlas]

ic

i meant something like this

[slow scroll]

well, its still the same matrix, except that it has a line down separating the last column, so it has the same rank.

If you look at 3x2 coefficient matrix instead you have something of rank 2.

[shy atlas]

mhm makes sense

thnx

[slow scroll]

npnp

[shy atlas]

why is that 1 not a leading 1

i just used the fact that its a leading 1 to say my earlier matrix has rank 3

big thonk

[slow scroll]

it is a leading 1 tho?

its a leading 1 in the augmented matrix, not the coefficient matrix

[shy atlas]

how do i know which one they're talking about

they just refer to it as "system"

oh nvm they say augemented matrix

so in that case 1 is a leading 1

how are they saying theres no leading 1 then

this for example

[slow scroll]

to get the rref(A), you just get rid of the last column, i.e. the column which has that 1 in the last row. so you go from [0 0 ... 0 1] to [0 0 .... 0]

right so if you ignore the last column, you have rref(A)

which has a row of zeros as in your pic there

[shy atlas]

why would i yeet out the last column tho

[slow scroll]

well you want to show that rank(A) < n under the assumption that there is some inconsistent system involving A. U know that if you have an inconsistent system, it looks something like the thing in your pic, and the process that gives you the rref of the augmented matrix also gives you the rref of the coefficient matrix. And then by looking at the rref(A), the coefficient matrix, you can count pivots (leading 1s) to figure out the rank.

the takeaway is that if you have an inconsistent system involving a matrix A, you can conclude that the rref(A) has a row of zeros, and therefore rank(A) < n

[shy atlas]

so rref of a system refers to the rref of the coefficient matrix ?

[slow scroll]

there is the rref of the coefficient matrix and the rref of the augmented matrix. Lets say you have an inconsistent system Ax = b. Let me use the notation A|b to denote the augmented matrix for the system and A for the regular coefficient matrix.

When you do row operations on A|b, you are also just doing row operations on A, and if you take the last column out of rref(A|b), you still have the rref(A).

[shy atlas]

omg

IT LITERALLY SAYS A IS THE COEFFICEINT MATRIX

wtf

im sorry man

i made you write out all of that

where do i learn to read

[slow scroll]

ikr same tbh

[quartz compass]

silly idea, define the binomial coefficient of a matrix as

$\binom{A}{n} := \frac{1}{n!} \prod_{k=0}^{n-1} (A-kI)$

[stoic python]

Merosity:

[quartz compass]

anything interesting we can say about this? when is the determinant 0? lol

[dusky epoch]

when one of {0, 1, ..., n-1} is an eigenvalue of A

[wintry steppe]

if we have vectors <1,2,3>, <5,8,12>, <2,4,6>, <10,16,24>, the span of those vectors equals the span of <1,2,3> and <5,8,12>???

[half ice]

Yes

Two of those vectors are clearly adding nothing to the Span

[wintry steppe]

prove it

[half ice]

No I don't wanna

[gray dust]

it's for YOU to prove

[wintry steppe]

c1 * <1,2,3> + c2 * <5,8,12> + c3 * 2<1,2,3> + c4 * 2<5,8,12>

[slim valve]

you can write 2 of them in terms of the 2 others

[wintry steppe]

thats what i did

how do we continue the proof?

[half ice]

Okay yes I see you did that. Then you can write:
(c1 + 2c3)<1,2,3> + (c2 + 2c4)<5,8,12>

Showing that anything that can be written as a span of the 4, can also be written as a span of the 2

@wintry steppe

[wintry steppe]

sorry for noob question, but im working on a computer science project and forgot what an individual 'tile' in a matrix is called?

the correct parlance slipped out of my mind

[limber sierra]

entry?

element?

im not aware of any other terms

[wintry steppe]

yes entry, tyvm

[wintry steppe]

what do they mean when they say V is a subspace of R^n, does that mean that the vectors in V are of R^n dimension?

[slow scroll]

the vectors in V are vectors from Rn. For example, you could have a plane through the origin in R3. Then the plane is a 2d subspace of R3. The elements/vectors in the plane are still (x,y,z) 3 vectors though

[wintry steppe]

yeah

[half ice]

So a vector space is a set in which you can add/subtract vectors, as well as scalar multiply. An example of a vector space is R².

A subspace is when you find a vector space inside your vector space. The line y = x is a vector space, and a subspace of R²

[wintry steppe]

so whats R^2?

all the vectors that are <a, b>?

[half ice]

Ya ya

[little frigate]

Any good YouTube video for getting started with linear algebra?

I tried reading a translated version (in french) of Gilbert Strang's book but I got stuck at the early beginning at some very simple question so I consider reviewing some online content that could introduce the subject as the book does even if it might not be as good as the book is

[kindred ingot]

Linear algebra done wrong

and linear algebra done right

the latter is available to most uni students via springerlink

[half ice]

3b1b's essence of linear algebra YouTube series is good to go along with a book

[gray dust]

doh sniped

[little frigate]

Alright thanks

[half ice]

Sry ilu

[slender yarrow]

so you're looking for things in french ?

[little frigate]

I'm looking for things in english, but if it's in french then it would be awesome

[slender yarrow]

il y a "algèbre linéaire" de joseph grifone que j'adore les explications sont super

par contre les exos sont pas mal à chier

[little frigate]

D'accord merci pour la référence je vais voir ça 👍

[slender yarrow]

après done wrong et done right sont bien aussi

pour les exos ça peut bien complémenter

[coral dawn]

Hi, as a programmer, I recently have the need to learn maths for game development, and I come across this question in a book I am following(I made the assumption it is linear algebra, but I might be wrong, the book is tailored for programmer and hence didn't specifically mention what exactly it is):

In the attached snippet, I am not sure how to solve the simple matrix equation, I can sort of see the relationship between them(as labelled in the snippet), but I am not sure why is it the case.

Would anyone be able to give me pointers about this topic? I tried looking up matrix equation but wasn't able to pick up anything useful - it can very well be my math background being too limited, I am very bad at maths.

I will really appreciate if people can tag me when answering!

[slow scroll]

are you familiar with reduced row echelon form and gaussian elimination?

@coral dawn

[coral dawn]

I have come across both of the topic in the book, however I won't say I am familiar with them @slow scroll

[slow scroll]

,w rref [[-1, 1, 0],[3, -3, 0]]

[stoic python]

Display results online and refine query

Requested by kxrider#0326

[slow scroll]

if you want to do that manually, you'll need to backtrack a little bit in your book, but once you have the rref, you can convert the augmented matrix here back into a system. So you have
x - y = 0
where y can be anything. therefore x = y so (a, a) = a(1,1) for any real number a would be your solution

[coral dawn]

Thanks, give me a sec I will try to have a look and let you know if I managed to understand it

[slow scroll]

by augmented matrix, i mean if you have a system (matrix equation) Ax = b then the augmented matrix for the system is the matrix formed by joining b to A by making b the last column of A. So if you have Ax = 0, then its just an extra column of 0s

[coral dawn]

,w rref [[3, 1, 0],[3, 1, 0]]

[stoic python]

Display results online and refine query

Requested by Noobnewbier#4143

[slow scroll]

ye so what equation does that give you?

[coral dawn]

I am guessing:

x + (1/3)y = 0
x = -(1/3)y

however, I am just guessing, is it not contrasting with what I should be getting? I would imagine that I should be having something like:

x = -3y

[slow scroll]

x + (1/3)y = 0
x = -(1/3)y

this is correct. Then you can multiply both sides by 3 to get 3x = -y where y can be anything again.
Lets take y = -3b for some real number b because we can. Then 3x = 3b so x = b.
Therefore we have (b, -3b) = b(1,-3) as a solution. You could also just omit the b, and simply know that any scalar multiply of a solution to Av=0 is also a solution

^ thats because A(cv) = c(Av) = c(0) = 0 for any scalar b and any vector v such that Av = 0. That only works for "homogeneous" equations where Ax = b but b = (0, 0, ..., 0)

[coral dawn]

Thanks for clarifying, do you mind gently explaining what is this knowledge called? Like what am I supposed to google to get this piece of knowledge?

I think I am mostly stucked as I hardly know any maths(if I have to estimate it would be pre high school, now I am regretting not taking maths seriously back then)

Because of this I find googling difficult as I don't know what they keyword is

[slow scroll]

its linear algebra. I wouldn't say this is typical high school level knowledge. The Gilbert Strang lectures at MIT (and his textbook) are supposed to be pretty accessible and not overly abstract for programming/engineering purposes. Gaussian elimination, rref, homogeneous equations all appear fairly early on i think

[coral dawn]

Thanks, I will look it up!

Hope you are doing good during lock down

[slow scroll]

tanks, u too

[latent marten]

hi can someone help me with this q

i have found the normal vector to P but i dont know where else to go

[latent marten]

$\frac{x-x_0}{v_i}$=$\frac{y-y_0}{v_j}$=$\frac{z-z_0}{v_k}$

[stoic python]

Elmo_AK47:

[latent marten]

@wintry steppe is that it?

so is $x_0 = 2$, $y_0 = 1$, $z_0 = -7$ and the normal vector at the bottm ?

[stoic python]

Elmo_AK47:

[rustic panther]

Greetings 🙂 need help with part (b)

[elfin ingot]

what did you try

[wintry steppe]

https://services.math.duke.edu/~leili/teaching/uwm/math340s13/extra11.pdf

@rustic panther

[rustic panther]

@wintry steppe omg thanks

[wintry steppe]

np

[rustic panther]

this actually helps me understand it

that*

[wintry steppe]

np

[latent marten]

@wintry steppe ayy thx bro

[latent marten]

could someone help me with this

[wintry steppe]

Multiple B and A

[latent marten]

that it?

$\begin{pmatrix}a&c\ :b&d\end{pmatrix}\cdot \begin{pmatrix}a&b\ :c&d\end{pmatrix}=\begin{pmatrix}a^2+c^2&ab+cd\ ba+dc&b^2+d^2\end{pmatrix}$

[stoic python]

Elmo_AK47:

[wintry steppe]

you can write that as v^T A^T A w = v^Tw, so try making some smart choices of v and w

||smart choices being the usual basis vectors||

if that doesn't give it away, then note that for any matrix M, e_i^T M e_j = M_{ij}. then you can put in the basis vectors to see that (A^TA){ij} = I{ij}, and you know what that means (here B = A^T)

(and that's the solution)

Wait a minute. If (Av) * (Aw) = v * w, then A has to be the identity matrix bc that's the only way that expression can hold true.

If A is the identity matrix, then a and d are 1's while b and c are 0's, which would then make B the identity matrix as well. If you multiply the identity matrix with the identity matrix, then you get the identity matrix.

"then a and d are 1 and b and c are zero" how?

Bc the identity matrix is [1 0; 0 1]

[cursive narwhal]

Let $T$ be the linear map associated with the matrix under some basis ${e_1,e_2}$. Now, $T(e_1) = (a,c)$ and $T(e_2) = (b,d)$. Now, we consider this dot product:

$ab+cd = e_1 \cdot e_2$

We're going to choose particular vectors $e_1 = (1,0)$ and $e_2 = (0,1)$. Now, we know that the columns of $A$ are linearly independent. We know they're orthogonal (their dot product is 0). So, all that remains is to show that they're of unit length. Let $v = w$. Then:

$(Av) \cdot (Av) = (av_1+bv_2)^2+(cv_1+dv_2)^2 = v_1^2+v_2^2$

If you compare coefficients and use the fact that $ab+cd = 0$, then you will find that $a^2+c^2 = 1$ and $b^2+d^2 = 1$. But that is exactly what we wanted to show so that $BA$ is the identity matrix. This proves the desired result.

[stoic python]

Abhijeet Vats:

[cursive narwhal]

Jesus typing on my phone is annoying

@latent marten

[wintry steppe]

If A = [a b; c d] = [ 1 0; 0 1], then a,d =1 and b,c = 0.

ok, but A is not necessarily the identity

also holy crap that's a lot to type on a phone gj

[cursive narwhal]

my phone kept screwing around so i couldn't get it out faster

[wintry steppe]

phone latex is nigh impossible

even just typing the dollar signs is a pain in the ass

[cursive narwhal]

I suppose another way to do this is to recognize that A is associated with an orthogonal map and, therefore, A^TA = I, since B = A^T. But obviously, you'd have to prove that result first and then use it here. Probably also not expected to know what an orthogonal map is too.

and my phone is pretty broken at the moment so whoopee

[wintry steppe]

depending on your definition of orthogonal that could be immediate lol

"proof by definition"

[cursive narwhal]

yesh lol

i've only ever seen one definition of orthogonal for linear maps though

probably just need to read more

[wintry steppe]

i feel like you could use the dual to define it somehow, because if you fix a basis and then look at the dual map wrt the dual basis, the matrix rep is the transpose

but ive never sat down and thought about it

that'd be a strange way to do it

[cursive narwhal]

i've never really done anything with dual maps. klaus janich doesn't really cover them

i'll probably learn about them in my LA class

[wintry steppe]

they are pretty interesting (imo) and they come up in a lot of places

[cursive narwhal]

i'll look forward to learning it, then

[wintry steppe]

they might not seem very useful the first time you learn them but if you take a differential geometry course then you'll see that they appear naturally and give you differential forms

now im rambling

[cursive narwhal]

ooooh interesting, it sounds like fun. I'll probably take differential geometry right after analysis ii (or iii) so i'll get the motivation for them immediately, especially if it doesn't come to me immediately.

[wintry steppe]

you can probably get started with differential geometry (abstract manifolds stuff not just classic low dimensional) after seeing the implicit and inverse function theorems, since those give you very natural examples of manifolds

although you will definitely want topology if it's a manifolds course

[cursive narwhal]

my uni recommends topology before differential geometry

it's actually funny though

they also recommend that the physics students take differential geometry before GR

[wintry steppe]

yeah you should listen to your unis recommendation, my uni doesn't have topology as a prereq and the people who haven't taken it are having a tough time it seems

[cursive narwhal]

ooo

[wintry steppe]

the instructor basically crammed a bunch of topology in an hour of lecture and then put some basic problems on the homework and said "learn it yourself" lol, you don't want to get stuck in that trap

[cursive narwhal]

ohhhh yea that sounds like it sucks

for my uni, we typically have to take a 3 course sequence in analysis and algebra before we're allowed to do anything else. But if there's something you're very interested in, you can do the course

generally, they won't let you go too far with that

[wintry steppe]

that sounds like a decent policy, since algebra and analysis are so commonplace

what does the analysis sequence constitute?

interested to see how it compares to my uni's first and second year """""analysis""""" courses which are just calc and multivariable calc on crack

[cursive narwhal]

uhhh analysis I is everything from logic, ZFC set theory to analytic geometry in R^3. So, like, differentiation and riemann integration. Very standard stuff, i think.

analysis ii is multivariable work leading up to stokes theorem. analysis iii is complex analysis and any additional topics that the lecturer feels like covering. Sometimes, they give a few lectures worth of an introduction to lebesgue integrals but it depends on the individual lecturing

but i'm not in uni right now so i can't really say what the teachers will be covering exactly

[wintry steppe]

that sounds pretty good actually

swap out the logic and set theory with constructing the real numbers to weed out first years to get started with making calculus rigorous and you basically have mine

[cursive narwhal]

oh the set theory stuff includes the construction of the real numbers as well

[wintry steppe]

nice

that's pretty comprehensive it seems

[cursive narwhal]

but like, they typically don't go extremely deep because physicists have to take the analysis courses too

[slow scroll]

y'all don't really have intro to proof type stuff in Europe, right?

[cursive narwhal]

i mean, what constitutes intro to proofs?

[wintry steppe]

^

teaching vellemans book i guess

[cursive narwhal]

i haven't read velleman

well, i read a bit and didn't like it

[slow scroll]

a course on propositional logic, induction, naive set theory, proof techniques. Like something for someone about to take analysis but has never been asked to rigorously "prove" anything

[cursive narwhal]

oh yea we cover that when we begin analysis I but we do axiomatic set theory instead.

[wintry steppe]

my uni has a course like that but i don't think many people actually take it cause the good first year courses assume you already know it

first day of class "ill assume you know how to write an okay proof" or "ill assume you'll pick it up"

[cursive narwhal]

i pestered my course coordinator to let me join into the analysis i lectures being held online. The lecturer was like "Just prove all 300 theorems on your own in Landau's Foundations of Analysis and that'll be an intro to proofs"

[hoary agate]

@cursive narwhal

[cursive narwhal]

wot

[hoary agate]

@cursive narwhal

the t

but wh

er?

[cursive narwhal]

are you drunk

[hoary agate]

noremal

norsde

nor

k

[cursive narwhal]

are you drunk

[hoary agate]

@cursive narwhal

@hoary agate

no nor

nor drunk doed

[cursive narwhal]

your mom

[hoary agate]

whatre

@hoary agate

@cursive narwhal

where have you enrolled?

[cursive narwhal]

ohhh i'm not gonna say lol

[hoary agate]

whyever not

[cursive narwhal]

Just know that i will be studying in Denmark, Sweden, Hungary, Czech Republic and Switzerland

[hoary agate]

oh

are you drunk?

[wind yacht]

Is everyone drunk?

[hoary agate]

i thank so

wei must be rihgt

ritgj

right

[cursive narwhal]

Legit, though, I'll be studying in any two of the countries i mentioned above. That's the only thing I'll say 😄

[hoary agate]

cool

[wind yacht]

👏

[hoary agate]

must be a spy

stinkin spy

[shy atlas]

i think it depends on the domain of T right ?

if the domain isnt R^m we cant have ALL the linear combinations

for example lets say the domain is some subset of R^m. then the x_i 's arent gonna range over all of R and we wont have all the possible linear combinations

[cursive narwhal]

I mean yea, x is implicitly understood to be an element of dom(T). That's why it'll have an image under T.

so pick dom(T) and you get all the vectors that will have an image under T.

[shy atlas]

yeah but

im saying thats not necessarily the span of the columns

unless the dom(T) = R^m

[cursive narwhal]

yeap that makes sense

i guess implictly, the theorem just assumes that dom(T) is R^m, which is very natural i suppose

[shy atlas]

hmm

it would still be a linear transformation if the domain isnt R^m right ?

perhaps a subset of R^m

[cursive narwhal]

Well, linear maps are usually defined with vector spaces being the domain and codomain. It would be a linear map if it was a subspace of R^m

[shy atlas]

[slow scroll]

it sounds like you are conflating linear combinations with what you get in the actual image

[shy atlas]

why does it have to be a subspace ?

whats wrong with like any subset of a vector space

[cursive narwhal]

Subspaces are vector spaces and usually, linear maps are defined on vector spaces.

[slow scroll]

(maybe try to prove that the image forms a subspace)

[shy atlas]

if T: A -- > B obeys T(x + y) = T(x) + T(y) and T(cx) = cT(x) then it should be a linear map right ?

i dont have to refer to vector spaces for this

or do i

[cursive narwhal]

You do, because A might not be closed under addition

[shy atlas]

ooo

so x + y might not be in A

ic what u mean

[cursive narwhal]

Like, when you define a linear map, you say that f:V ----> W is a linear map with V & W being vector spaces.

Now, suppose you wanted to restrict it to some subset of D (not necessarily a subspace). So, you have f: D ----> W. The issue is that x+y might not be in D if it isn't a subspace. Or maybe cx isn't in it. So, you can't talk about the image of x+y or cx if they aren't in D.

Or rather you can but then you're not talking about the function f:D -----> W

[shy atlas]

hmm but my book doesnt refer to vector spaces when they talk about linear functions

[cursive narwhal]

uh show me the definition in your book?

which book are you using?

[shy atlas]

so ig f : A --> B with the added condition that A is closed under addition and scalar multuplication should sufffice ?

[cursive narwhal]

oooh never heard of that

uh okay show me your definition

[shy atlas]

ukkhhhhh

fine

[cursive narwhal]

uh that's definitely a weird definition

but okay, basically, R^m and R^n are vector spaces

[shy atlas]

ye idk why they like to use unpopular definitions

[cursive narwhal]

well, along with the operations they've been imbued with, of course

for example lets say the domain is some subset of R^m. then the x_i 's arent gonna range over all of R and we wont have all the possible linear combinations
That definition also addresses this problem. Remember how I told you that the theorem just assumes that dom(T) is R^m? That definition you just posted confirms it

[shy atlas]

i dont like this defintion

[cursive narwhal]

ye idk why they like to use unpopular definitions
Probably should use a different book haha, this kind of exposition would be very confusing for me

i dont like this defintion

your mom

[shy atlas]

no ur mom

[cursive narwhal]

no urs

[shy atlas]

mmm yes i liked ur mom 😉

[dusky epoch]

@cursive narwhal no urs

prank

[cursive narwhal]

no urs

[wintry steppe]

is there something here that i'm not seeing that'll make this quicker

or do i just have to do row operations

[storm python]

what does 1.7 say?

[lost comet]

hi guys

i have this question

i am unsure on how to go about this one

[cursive narwhal]

Well, it's just showing that two sets are equal

soooo just use mutual containment and other properties of the subspaces involved to show that each set is a subset of the other

[lost comet]

i just need to show that they are a subset of each other

so

oh

ok i see ok ty

what is this notation

just found this

is it the same as <,>

[cursive narwhal]

yea that's probably an alternative notation for the inner product

i think something like that is used in berberian's textbook

Anyways, going to the 6 marks question above, I can give you one direction.

Let $v \in U^{\bot} \cap W^{\bot}$. Then, $v \in U^{\bot}$ and $v \in W^{\bot}$. Now, pick arbitrary but fixed vectors $u \in U$ and $w \in W$. Then, it is clear that:

$$(v,u) = 0$$

$$(v,w) = 0$$

By the bilinearity of the inner product, we have:

$$(v,u)+(v,w) = (v,u+w) = 0$$

But $u+w$ is an element of $U+W$ and the above is simply saying that $v$ is orthogonal to an arbitrary element of $U+W$. It follows that $v \in (U+W)^{\bot}$. This shows that the right set is a subset of the left set.

Now, do the other direction yourself.

[stoic python]

Abhijeet Vats:

[cursive narwhal]

@lost comet

[lost comet]

alright i see that

thanks

[cursive narwhal]

you're welcome

[vestal beacon]

i get x1=-1/2

x2=-2

[cursive narwhal]

?

[vestal beacon]

its wrong and i need help to get it rihgt

righ

t

right

[cursive narwhal]

The second equation is a scalar multiple of the first equation

Let $x_1 = t$, where $t \in \bR$ is a parameter. Then, $x_2 = -2t$. In other words, $(x_1,x_2) = (t,-2t)$. This, loosely speaking, will form the solution set of the given linear equation.

[stoic python]

Abhijeet Vats:

[gray dust]

for context @vestal beacon's computing eigenvectors & only needs 1. you messed up in finding x2

[vestal beacon]

@cursive narwhal can i always set x1=t

and in this case t=1?

[cursive narwhal]

I mean, I don't really see why not

depends on the problem, probably

I've never had much difficulty in just setting something to be equal to some parameter

[vestal beacon]

another problem i have is that this is a bigger problem

because they want me to find sqrtC

i know that C is

jk found it

thx for your help

[wintry steppe]

https://www.youtube.com/watch?time_continue=275&v=lRhjFkcIM_8&feature=emb_logo

[lost comet]

hello again. im stuck on a concept

the question im on is talking about if a 3x3 can be diagonalised

i said yes since the roots of the eigenvalues are distinct

now it wants me to find a matric so that C^-1 AC is diagonal

i found a video showing that C is a 3x3 made up of the eigenvectors

but im sure the order of them should matter right?

[wintry steppe]

https://math.stackexchange.com/questions/1764857/diagonalizing-a-3x3-matrix

[torn silo]

the ordering changes the ordering of the values on the diagonal

[lost comet]

and does a diagonal matrix mean all the values in the non diagonal entries = 0

[torn silo]

yes

[lost comet]

does that mean the diagonals can also be 0

[torn silo]

yes if you have eigenvalue 0

[lost comet]

is the 0 3x3 matrix a diagonal matrix?

[torn silo]

yes

with eigenvalue 0 three times

[lost comet]

ok i was wondering since when i tried to put the eigenvectors into a 3x3 and then did the multiplication the final answer had a 0 in the diagonal entry

so i thought i did something wrong

[torn silo]

thats fine if you had the eigenvalue 0

otherwise not so good

[lost comet]

if one of the eigenvalues = 0 then one of the entries = 0 is fine?

ah

i see

yeah thats the case

alright then

oh and another thing

in my notes

i was looking at the def for

the adjoint operator

T*

for a linear map T

this is what ist says

what is that bar on top

[torn silo]

conjugate

$\overline{a + bi} = a - bi$

😦

[lost comet]

wait i dont understand what lamba is then

a complex number?

[stoic python]

deekaan:

[torn silo]

yes

but the imaginary part might be 0

so it doesn't change anything

really depends on the situation

[lost comet]

ok and how can i use that to determine whether or not a matrix is unitary

they give it like T*T = I

is T* like an eigenvector thing

[torn silo]

take matrix A

then A*

multiply those two and see if it becomes id

[lost comet]

im guessing since my A is

000
1 2 1
-101

i set that = lambda x A

right?

wait no

oh wait is this lambda the same as the eigenvalues?

[torn silo]

maybe read up on adjoint operator again

[lost comet]

yea i think i just confused myself

[torn silo]

just take it one concept at a time, write things down and you're going to get a better grasp of whats happening

[lost comet]

@torn silo sry for the ping but i found that apparently if the abs values of the eigrnvalues are 1 then it is unitary

so since the eigenvalues are 2,0,1 is that enough to say its not unitary

[hazy sonnet]

does anyone know any good video resources for hessian stuff?

[torn silo]

very good

you might want to check out khan, but tbh not sure if im a fan of their mutli var calc stuff

[hazy sonnet]

ill just ask this as i try to review the topics

how would u do a?

[wintry steppe]

How to solve a question like that? I don't know how to "show" it

This question is from the very basics of a linear algebra book, so I don't know if this is the right channel

[limber sierra]

this is more #proofs-and-logic but its fine

anyway uh, there's different methods, and how "accepted" they are depends on your class

perhaps the simplest is just by constructing truth tables and showing that each sentence is always true

there's also formal proofs; you might've seen things like fitch-style proofs

or they might want you to give an informal argument

really depends on how your class introduced this stuff

[wintry steppe]

I see. It was not introduced by a class. I'm learning linear algebra by myself with a book and this is the first chapter

Do you think it's important to do exercises like that or just go straight to matrices and vector space?

[limber sierra]

i'd recommend at least "convincing yourself"

i mean you might think "A or B being the same thing as B or A is obvious"

but the idea is, you want to "convince yourself" that the way you define "or" makes sense

the cleanest way to do this is, as mentioned, with truth tables*

*[in classical logic]

[torn silo]

transformations is where its at

[lost comet]

hello again 😄

i cant seem to understand this one

what is ||Tv||

$||Tv||$

[stoic python]

ChaChaCha:

[lost comet]

the def of normal is when T* T = TT* yea?

[torn hornet]

its just the lenght of the vector

if thats what you mean

[lost comet]

T is a linear map

v is a vector i think

[torn hornet]

yes, and Tv is a vector

[lost comet]

oh

wait

i see

ok so how is the length of this equal to the rhs

[torn hornet]

well you should try playing with it first

[lost comet]

would what i said about the def mean that T = T*?

so the maps are the same

so the length must be the same

[torn hornet]

uh no thats hermitian

normal is what you said

$T T^\star =T^\star T$

[stoic python]

JohnDoeSmith:

[torn hornet]

this does not imply hermitian

[lost comet]

oh i see

but i still dont see how this helps with the questionj

could i use this?

[torn silo]

yes

use this in combinatation with the defintion of normal

[lost comet]

ok thats fine i think i worked it out

it would go like this

$||Tv||^2 =(Tv,Tv)=(v,TT^\star v)=(v,T^\star Tv)=(T^\star v,T^\star v)=||T^\star v||^2$

does that work?

ok no

why are there gaps

ffs

ok

but there

thats what i got

[stoic python]

ChaChaCha:

[torn hornet]

yeah thats right

[torn silo]

3rd and 4th step should be switched no?

[lost comet]

arent they the same

or actually youre right

since its like pre multiplying

idk how to say it

[torn silo]

like chaining functions

[lost comet]

funny because i didnt make that mistake on my paper

😄

oh t he next one is a bit hard

i thought <v1,v2> = 0 iff v1 = v2 = 0

but since the eigenvectors are distinct it means v1 and v2 are distinct yes?

[torn silo]

they're linearly indepedent yes

[lost comet]

so then how can <v1,v2> = 0 ?

unless one of them is 0

[torn silo]

because they're orthogonal to one another

[lost comet]

that doesnt work tho does it

[torn silo]

🤔

[lost comet]

does distinct eigenvalues mean theyre orthogonal

yeah so this is only 0 if v1 is perp to v2 right?

now i gotta show that

[torn silo]

maybe try one and two examples to convice yourself

[lost comet]

i just searched and im guessing a normal map means its symmetric

does that mean t he eigenvalues are orthogonal?

vectors*

[torn silo]

keep in mind A = A^*

[lost comet]

i thought T was normal and not hermitian

[torn silo]

ah right normal sorry

A*A = AA*

[lost comet]

ok so Tv = lamda1 v
Tw = lamda2 w

i got that

and shows that lamda1 != lamda2

so the inner product must be 0

[rustic panther]

Need help with this

[elfin ingot]

what have you tried

@rustic panther

[rustic panther]

@elfin ingot to be completely honest, I don't know where to begin

[elfin ingot]

okay so

just show P^2=P

P^2 here prob means P(P(a))

for all a

[rustic panther]

a not u?

[elfin ingot]

P is a function deifned as p(v) = (u|v)u

for any v in V

v is dummy variable

[rustic panther]

ah a in A is our input/domain

[elfin ingot]

yea

so just show that

by unfolding your definitions ig

[rustic panther]

ig?

[elfin ingot]

i guess*

[rustic panther]

do you have the notes for this?

[elfin ingot]

what notes

[wintry steppe]

The notes