linear-algebra

[torn leaf]

It’s rref ref

Right?

[winged prairie]

anybody know how to do this

[wintry steppe]

According to my prof’s lecture videos, g(\alpha) = \beta(f) so I tried it out for arbitrary maps I denoted as \alpha, \beta, f, g, and the compositions don’t match up? Can someone please help clarify this for me?

[vocal prairie]

The usual field structure on reals+the fact that every real uniquely corresponds to a complex number with 0 imaginary part should be used here.

[winged prairie]

ok thanks

[vast iron]

If the zero vector spans a zero space which is 0 dimensional, can we somehow relate it to the field over which the vector space was defined?

[quartz compass]

sure, I guess

a field is a 1 dimensional vector space over itself basically

[vast iron]

The problem I'm trying to solve is to show that intersection of spans of orthonormal basis of R^3 is a point.
What I did is show that it's span(0), can I say that span(0) is just some point?

[quartz compass]

I don't see how that relates to the original question

what is span(0) to you?

[dusky epoch]

span(0) is {0}

not just "some point"

[vast iron]

I see. Thanks. I'll think some more and see if I can figure something else out.

[hard drum]

That phrasing to me seems off as the span of any orthonormal basis is R^3 so the intersection of any set of spans of orthonormal bases is R^3... So i assume its a convoluted way to say orthonormal bases are linearly independent, which is then part of the definition of basis anyway

I'm confused lol

[vast iron]

I misphrased that. Imagine u,v,w are orthonormal basis. Then S is span of u and v, T is span of w, find intersection of S and T.

[teal grotto]

if x is in S cap T, then x = au + bv = cw for some scalars a, b, c. use linear independence to continue

[hard drum]

So yes, show that u and v is linearly independent basically

fair

[vast iron]

This is what I did and got that a = b = c = 0, but I don't know how to interpret that.

[teal grotto]

substitute them in for what x is equal to

x = au + bv = cw = ...

[vast iron]

I don't understand. Is the zero vector a point?

[teal grotto]

you have just shown that if x is in the intersection of S and T, then x must be the zero vector

[nocturne jewel]

It's the 0 vector

In coordinate spaces it's the origin, yes

[vast iron]

Thanks

That makes sense

I thought the answer would be span(0) but that's just 0 anyway

Thank you all.

[little scarab]

why is (AB)^T = B^T A^T?

[teal grotto]

they are the same thing, well, span(0) and {0}

[teal grotto]

work out some matrix multiplication and the answer pops out, unless you're asking for like "a deeper reason"

[runic oar]

How does the inclusion of complex numbers and the function to satisfy being a differential equation change the methods for finding whether it’s closed under addition

[eternal jetty]

what is the condition of a matrix to have real eigenvalues

[hard drum]

a common condition (but NOT necessary) is that a matrix is symmetric, or more generally Hermitian

[wintry steppe]

that it has real eigenvalues

[nocturne jewel]

What complex numbers?

C^2(I) is the space of all twice differentible functions on I

[winter harbor]

I think he's asking whether we can consider such a space as a complex vector space or not. And what does it change wrt viewing it as a real vector space.

[runic oar]

I’m wondering if there’s any difference in how we deal with proving subspaces as opposed to real

[nocturne jewel]

Nope

[hard drum]

No difference rly for any field

[nocturne jewel]

Same requirement regardless of the scalar field V is built on

[runic oar]

Ok thanks

I still don’t understand this stuff especially well but I’ve got enough that I can bs the assignment due tonight lol (got a meeting with the prof later in the week where I can try for the understanding part)

[glad acorn]

I cannot see why it would be a zero matrix

[winter harbor]

Notice that the anti-symmetric matrix $A$ induces an anti-symmetric bilinear map on $V$ given by:
$$
\Omega : V \times V \rightarrow \mathbb{K}
$$
Given by $\Omega(u,v) = u^{t} A v$, where we identify a 1 × 1 matrix with a scalar in the underlying field $\mathbb{K}$.
\
\
But notice that for every anti-symmetric bilinear form $\Omega$ on a field $\mathbb{K}$ with $\text{char}(\mathbb{K}) \neq 2$ we have that $\forall b \in V$
$$
\Omega(b,b) = - \Omega(b,b) \implies 2 \Omega(b,b) = 0
$$
And since $\text{char}(\mathbb{K}) \neq 2$, we have that $\Omega(b,b) = 0$ and via de identification, we have that $b^{t} A b \in \mathcal{O}_{1 ×1}$ is the zero matrix.

[glad acorn]

Your language is abstract to me. I’m a freshman

[stoic python]

MisterSystem

[winter harbor]

Are you familiar with bilinear maps?

[glad acorn]

No. Just year1 student

[winter harbor]

Ok, then we will have to compute things explicitly lol

Notice the following

Since $A = - A^{t}$, we have that $\forall b \in V$ a vector in a finite dimensional vector space (over the reals if you feel more comfortable) we have that:
$$
b^{t} A b = b^{t}(-A)^{t}b = - b^{t} A^{t} b = -(b^{t}Ab)^{t}
$$
But since $b^{t} A b \in \mathcal{O}{1 \times 1}$ is a $1 \times 1$ matrix, it is trivially symmetric, i.e $(b^{t} A b)^{t} = b^{t} A b$.
So
$$
b^{t} A b = - b^{t} A b \implies 2 b^{t} A b = 0 \implies b^{t} A b = 0
$$
So $b^{t} A b \in \mathcal{O}{1 \times 1}$ is the zero matrix ; $\square$

[merry rover]

Is this correct?

[stoic python]

MisterSystem

[winter harbor]

@glad acorn Is this easier to grasp?

This is basically the same argument btw

I just didn't mention anything about fields other than the reals

[glad acorn]

Yeah so why don’t you just answer me in this simple language

[winter harbor]

Because the argument is more general

[glad acorn]

Thx anyway

[merry rover]

@winter harbor Are you familiar w/ matrices?

[winter harbor]

No, I am not

[glad acorn]

I have no idea how to prove this type of question. Do I need to show that all 3 cases would arrive the same conclusion?

That AB-BA is anti-symmetric

[lavish jewel]

should be able to directly substitute

along with the definitions of symmetric and antisymmetric, another handy property is that, for any two matrices C and D, (CD)^T = D^T C^T

so take [A,B]^T and see what happens

[glad acorn]

Case 1: AB are all symmetric case2:AB are all anti-symmetric Case3: one of them is symmetric and another is anti-symmetric

I see 3cases here

I'm very desperate...

[lavish jewel]

yeah try all 3 i guess

the wording is a bit ambiguous

[glad acorn]

Is there an expansion for the transpose (AB-BA)?

[lavish jewel]

i gave you one for the product. for the sum, the transpose of a sum is the sum of the transposes

[glad acorn]

(AB-BA)T=BTAT-ATBT

[lavish jewel]

mhm

now consider all the cases

there's 4 btw

both sym, both antisym, one sym and one antisym, and backwards

[glad acorn]

Very frustrated.

[wispy pewter]

I’m unsure how to solve this can anyone help me?

[midnight trout]

9b: Having trouble thinking about how to express this. Doing row reduction along the first column, new a_i1 would be equal to old a_i1-l_i1*a_11, right?

[stoic python]

CreamyBoy

[midnight trout]

that makes sense if I were expressing this in a computer function, but how do I express this for a linear algebra question?

[formal galleon]

sorry if this is really easy but does anyone know why this system has no basis and a dimension of 0?

i thought you just augment all the equations as rows in a matrix and then see the the dependance relations

[wild fulcrum]

(0, 0, 0) is the only point in the set

maybe check your calculation?

[hot willow]

looks good

[worldly bear]

what does this notation mean

the norm of the negative norm of u * norm of v?

[wild fulcrum]

can you post the whole thing

[worldly bear]

u and v are just vectors

yeah idk wtf is going on lmao

let me check the answer in the back

2 root 966

dont know what they could possibly mean though

[wild fulcrum]

-norm(u) * v, then take the norm of that vector?

[worldly bear]

my professor lol

[wintry steppe]

oh wait lmao that makes sense

[worldly bear]

im just doing the problems he assigned bc we have an exam tomorrow

[wintry steppe]

ignore me

that's one confusing looking expression

[worldly bear]

i agree

how do you know thats what it wants lol

i dont see it

[wild fulcrum]

check dimensions

norm(u) is a constant scalar, so whatever, then times v, now a vector, which you can take norm of

[wintry steppe]

$$|(-|u|)v|$$

[stoic python]

TTerra

[worldly bear]

i see it now, thanks

[worldly bear]

is the 0 vector considered to be orthonormal

it wouldnt be since it doesnt have any length

[nocturne jewel]

0 vector is orthogonal to all vectors, but isnt orthonormal since the norm isnt 1

[worldly bear]

what i figured, thank you

[silver heath]

d

[sullen hazel]

confused on what this question means by "z is closed" does that mean z=0?

yup it was i figured it out nevermind

[midnight trout]

can somebody point me a direction to read about how to do this? elimination by multiplying matrices? this is the first time my book brought it up and I can't find any information about it. googling just has resources on row reduction/manipulation.

it isn't covered by this chapter it seems and I'm suddenly getting questions like this which I don't know how to approach in the slightest...

[pliant blaze]

guys does det(B^n) equivalent to (det(B))^n

[teal grotto]

does det(AB) = det(A) det(B) ?

[pliant blaze]

nvm

yeah

😆

thanks

[torn leaf]

Hi can anyone please help

I got this so then what to do?

[dusky epoch]

what you should do now is remember what the matrix representation of a system of equations really means

and also how to read the rank of a RREF'd matrix

both of which are things which should not be foreign to you

[glad acorn]

BA=AB where A B are two square matrices

Does A exist if and only if B has an inverse?

[dusky epoch]

no

just take B as the zero matrix lmao

[glad acorn]

Is this the only exceptional case?

[dusky epoch]

no

[glad acorn]

A=B-1AB, A=B-1(B-1(B-1...AB)B)B…B)B?

What happens

[dusky epoch]

that's some overcomplication if i ever saw any...

what are you trying to do rn?

there exist non-invertible matrices B which commute with something

[glad acorn]

Ok

[dusky epoch]

in fact, for any matrix B there exists a different matrix that commutes with B
(that every matrix commutes with itself ought to be somewhat obvious)

[torn leaf]

I don’t get it

[dusky epoch]

so what you're saying is, to you these matrices are nothing more but grids of numbers with arcane, ritualistic rules on how to handle them.

[glad acorn]

Can I construct a matrix C-C^t, that the inverse of this matrix exists?

C is an n size square matrix larger than 2

[dusky epoch]

do you mean you want to construct a matrix C such that C - C^t is invertible?

[glad acorn]

Yes

[dusky epoch]

and what do you think?

[glad acorn]

Seems it’s impossible

[dusky epoch]

and what makes you say that?

[glad acorn]

Diagonal entries all become zero

[dusky epoch]

so what?

[lavish jewel]

for even n, can't you always construct a matrix that swaps the rows and shifts the sign of a few of them?

[dusky epoch]

you can have a matrix be invertible despite a zero diagonal.

[lavish jewel]

that's just the easiest example that came to mind, an antidiagonal mat

well, C - C^T antidiagonal

[dusky epoch]

i think we might run into trouble if n is odd.

but phymathdoublemajor didn't specify if we needed an example for every value of n or if just one is enough.

oh, sorry, were we not done with the previous question from your linear algebra homework

but if you insist: you can just take 2B if B isn't 0, and anything you want if B is 0.

[torn leaf]

Yeah?

[dusky epoch]

[torn leaf]

I’m not getting itt

[dusky epoch]

of course you aren't getting it! you will never get anything if you view matrices from such a standpoint!

[torn leaf]

Then elaborate more naaa

[dusky epoch]

do you know how to translate between systems of equations and matrices?

if you do, then you should know that every row corresponds to an equation an every column to an unknown - except the last column, which holds the right-hand sides of your equations.

if you don't know even this basic fact then how can you expect to do anything with systems of equations via matrices?

[torn leaf]

Wait

I think I got it

But

what is rank of a matrix?

@dusky epoch

[dusky epoch]

did your teacher explain nothing to you?

[torn leaf]

Like position?

[lavish jewel]

it's the number of linearly independent cols

[torn leaf]

Well yes she barely taught anything

[dusky epoch]

the rank of a matrix can be defined in many different but equivalent ways. in your case, it's the number of pivots in the (R)REF of your matrix

[torn leaf]

Ok

@dusky epoch

Is this right?

[dusky epoch]

seems ok...

[torn leaf]

Ok thank you

[strange delta]

can someone give me some intuition on how to do this

so far

[lavish jewel]

a janky way is to simply multiply both sides by (I-T)

[strange delta]

yeah thats what i'm doing rn actually

[lavish jewel]

and show that both sides yield the identity mat

[strange delta]

just playing around

[dusky epoch]

that's not janky at all lol

[strange delta]

just got I = 1

[lavish jewel]

i mean, one can associate it to some system x = x_0 + Tx and let y = x - x_0 and y = Tx (like a flow diagram with a loop)

[strange delta]

does not seem alright

[lavish jewel]

and then you get the recursive relationship that directly produces the series

instead of starting already with the series

[strange delta]

OH

actually

i'm an idiot

i got it

[lavish jewel]

but yeah, if you multiply (I-T) from the left and from the right and show you get I both ways, you're done

[strange delta]

is this too good to be true? 😮

no way this question is that easy right?

[dusky epoch]

it isn't because the zero product property isn't true for matrices

[wintry steppe]

What’s the difference between a vector space and a field?

[nocturne jewel]

vector space is kind of like 2 sets merged together and a field is just 1 set

w/ vector spaces you have your set of vectors and your scalar field, and the 2 operations + and *, whereas the field itself would just be.. the field with + and * b/w things in the field

[wintry steppe]

What does “vector field over itself” mean?

[strange delta]

what happened in the 3rd line of the equality

[quasi vale]

@strange delta dot product properties. (x2 + 3x3) * (y1) + (2x1) * (y2)

and then y_1 x_2 + y_1 3x_3 + 2x_1 y_2

[nocturne jewel]

Probably the fact any scalar field K is a K-vector space

[tender valley]

hello, i would like to ask how to solve this? i already found the intersection of the lines, and then what do i do next?

[quartz compass]

3 noncollinear points define a plane

it partially depends on how you want to represent the plane equation too, but I'd just pick one point on either line to be my other 2 points

then make a formula that looks like those lines, but with two parameters like, A+sB+tC

[zinc timber]

find the normal direction of the plane first

[tender valley]

ah i see, thank you

[reef roost]

In every linear algebra resource I can find, we compute eigenvalues/eigenvectors for an operator by first converting it to a matrix, and then getting the characteristic equation by taking the determinant of eigenvalue * identity - matrix. Usually, we then compute the determinant with cofactor, which is horribly inefficient, we’re running in O(n!) time. Is there a better way? Or I’ll start with, is there another way to compute the characteristic equation/eigenvalues without the determinant? I’m looking for any way to speed up computing eigenvalues/eigenvectors of matrices by hand, particularly for 3x3 and 4x4 matrices, but I’d be happy to hear about higher dimensions as well. Further, if anyone has any resources they recommend to consult to learn more about efficient ways to compute eigenvalues/eigenvectors by hand I’d love to hear it.

[winter harbor]

That's a nice question

Indeed there are way more efficient algorithms to compute the determinant

In fact

The most basic one I can think right of the bat

Would be Gaussian Elimination

You can use gaussian elimination to compute determinants and it is about O(n^3) irrc

[reef roost]

Gaussian elimination is very nice for just a matrix of numbers. But unfortunately for the symbolic determinant with the eigenvalues it gets much crazier much quicker with all the multiplication and division by expressions involving lambda, and it seems I can get the answer quicker by just writing down the answer for a 3x3 determinant that I have memorized.

[winter harbor]

And yes, you can indeed define eigenvalues and eigenvalues without making any mention of the determinant at all. Axler "Linear Algebra done Right" famously does this.

But like, you define things intrinsically

[reef roost]

Axler does indeed, I love his book, but he doesn’t provide much in the way of computations sadly

[winter harbor]

Yeah

That's what I was about to say

He defines things intrinsically

So they are theoretically very nice

But computationally unfeasible

[reef roost]

I really like the approach he takes to determinants and trace, but the problem left behind is, what other systematic way can we have to compute the eigenvalues?

all the examples he provides are extremely nice, but it just won’t work for a general case

[winter harbor]

I think in nice cases like hermitian/symmetric or normal operators.

And working over C

We can find eigenvalues by looking at a system of polynomial equations

Namely

[reef roost]

Technically with this method, we only need to 0 out n(n-1)/2 entries, and then multiply together n components, but the problem is these entries get really crazy easily, and for a human like me with poor working memory that slows my processing and computing down, maybe I just need to practice or something

[winter harbor]

At least if you know something about matrix beforehand

Like

[reef roost]

true

[winter harbor]

We have that
$$
\det(A) = \prod\limits_{i=1} \lambda_{i}
$$
and
$$
\text{tr}(A) = \sum\limits_{i=1}^{n} \lambda_{i}
$$
Where $A \mathcal{M}{n}(\mathbb{C})$ is a normal operator and $\lambda{1}, \cdots, \lambda_{n}$ are its eigenvalues.

[stoic python]

MisterSystem

[winter harbor]

In some easy cases we can prolly compute explicitly calculate the determinant and the trace of A

And we can put this information into a system of equations

[reef roost]

yeah for 3x3 that can be done in under a minute

[winter harbor]

And with some maybe other information we can completely specify all the eigenvalues of A by solving this system of polynomial equations on lambda_1, ..., lambda_n

This works nicely for 2×2, but computing 2 × 2 determinants is already so simple that it is kind of boring

Maybe with some more information we can track the 3×3 case too

[reef roost]

2x2 already got the answer memorized in closed form for both eigenvalues lol

[winter harbor]

[reef roost]

it takes a nice form in terms of the determinant and trace

[winter harbor]

Anyways, I think I should know a bit more of these computational aspects.

[reef roost]

iirc it’s (trace +- sqrt(trace^2 - 4 determinant) )/2

and this could further be simplified by bringing in the 2 and viewing trace/2 as the mean of the eigenvalues, and just leave the determinant

I wish we could use that formula to get 2 of the eigenvalues for any matrix and then just guess the rest lol

would make life so much easier

[reef roost]

update: since trace and determinant are easy to compute, I just need to memorize the coefficient on the linear term in 3 dimensions to compute that and I’ll have the characteristic equation quickly

[forest quiver]

Trace is a linear function right?

Like $Tr(A+B)=Tr(A)+Tr(B)$

[stoic python]

Tim O'Brien

[forest quiver]

I wonder if $Tr(kA)=kTr(A)$

[stoic python]

Tim O'Brien

[forest quiver]

Makes sense right?

[zinc timber]

it is

[wintry steppe]

why dont you try proving it

[forest quiver]

Yeah I should do that

Does Trace even have applications?

It seems really useless at least to me

[lavish jewel]

it's equal to the sum of the singular values

you also use it in the definition of the inner product of two matrices

[forest quiver]

Inner product?

I am not there yet

It will come together no doubt

[lavish jewel]

and in multilinear algebra, it's generalized to "contractions"

[forest quiver]

Interesting

Also on a side note

My teach was talking about invertible linear transformatoins

And he said the mapping has to be one-to-one (which makes sense)

and I also asked him if it had to be onto with the range is equal to the codomain

but he said not all onto mappings have [set of range] = [set of codomain]

[nocturne jewel]

Trace is also used in char polynomial

[forest quiver]

That goes against surjectivity no?

[nocturne jewel]

That's wrong yeah, unless I'm smooth braining

[lavish jewel]

maybe they meant image

the wording is kinda weird

[nocturne jewel]

Surj is that the image is the output space identically

[forest quiver]

range = image bruh

[lavish jewel]

oh

then it can be pseudo invertible

[forest quiver]

am I wrong?

[lavish jewel]

idk what the usual terminology is

[forest quiver]

Psuedo invertible

[lavish jewel]

i avoid "range"

[forest quiver]

why? I

[nocturne jewel]

Bruh Penrose... I forget all of that

[forest quiver]

They should make one single convention for function terminology

tbh

[nocturne jewel]

Or whatever the other term for pseudoinversess are

[lavish jewel]

yeah, moore penrose pseudo inverse

[nocturne jewel]

Ye that bitch

[forest quiver]

Functions are cool

I can't wait to learn about non-linear transofmrations

That shit must be mind-beindgin

bending*

[nocturne jewel]

You've already learned a bunch of R to R non linear functions

[strange vault]

I think that some people take the codomain as the "range" because we don't really care about the rest of the y-axis when we're doing functions in high school

it's kinda like giving visual intuition to matrix multiplication for high schoolers

it's great to get an idea of what's going on, but it can be overwhelming

[nocturne jewel]

Yeah ik HS you kinda get the idea of functions map from domain to range, but in reality... no

[strange vault]

😌

[azure bridge]

for this set of vectors is it linear dependent or independent? I'm not sure since theres a dot product which is just a number

[nocturne jewel]

im gonna assume that's meant to be a comma.

[azure bridge]

oh

hmm okay i will ask my prof

[formal galleon]

anyone know how to do that last part?

i have the matrix to change from B to C

i think im overthinking it lol

[quartz compass]

the C to B matrix is what you need for that

and that's the easier matrix to get

[formal galleon]

would these notes be an error then?

they put the c vectors on the left and b vectors on the right side of matrix

then call it B to C

[nocturne jewel]

Nope cause P(B->C) is made of the b vectors written in C

so it takes vectors in B and gives vectors in C (in being wrt that basis)

[formal galleon]

ooo

is that a contradiction with this?

im

it looks like in the first picture its c written in terms of b and thats B-->C where as in the bottom pic i just sent its b written in terms of c and thats B --> C

[nocturne jewel]

Nope

those are the b vectors written wrt C

[formal galleon]

ah ok

in the question i linked, its asking for the B-coordinate vector for -1+2t

it doesnt specify that it starts at C?

[nocturne jewel]

you can easily write -1+2t wrt C

since C is the canonical basis

so you need to "translate" a vector in C to a vector in B

[formal galleon]

😅

oh i think i see waht ur saying

so i found B-->C, i just transpose that matrix and multiply it by the [-1, 2, 0] ?

[nocturne jewel]

$-1+2t=[-1,2,0]_C$

[stoic python]

Mosh

[nocturne jewel]

No, you invert the matrix

[formal galleon]

OH

INVERT NOT TRANSPOSE

thats what i messed up

thanks for the help though i appreciate it a lot

[forest quiver]

Hi :3

This is the prompt for a homework assignment

if I remember correctly an operator is a linear map of the form

$T: \mathbb{R}^n\rightarrow \mathbb{R}^n$

[stoic python]

Tim O'Brien

[forest quiver]

Right?

Like domain = codomain

[nocturne jewel]

yes, linear operators act on the vector space itself

$T:V\to V$

[stoic python]

Mosh

[forest quiver]

I haven't learned about vector spaces yet, but looking forward to it :3

[wintry steppe]

it's asking which equation is true

[silver cradle]

when it says they both diagonalize to the same matrix is that the same as being similar to the diagonal matrix?

[teal grotto]

yes

[silver cradle]

Okay so if we call this matrix X there exists invertible matrices P,Q such that PAP^-1 = X and QBQ^-1 = X?

is that what the question is asking

[teal grotto]

the question is to show that, knowing the above information, show that A and B are similar

so you have PAP^-1 = QBQ^-1

how can you get RAR^-1 = B for some invertible matrix R?

[jagged rock]

I don't understand the generalized form in step k

and this part too in the following slide

[little scarab]

how do you show/prove (AB)^T = B^T A^T ?

[lavish jewel]

you can express the matrix multiplication element by element using sums

just manipulate those sums

[dusky epoch]

@little scarab i remember you asking the same question but 2 days ago in this very channel. you even got a response, but evidently you decided to ignore it, not even asking for clarification.

[lavish jewel]

large oof

[lament vale]

https://tenor.com/view/caught-in-5000k-gif-20880260

[little scarab]

yes, apologies; I think I didn't see the response until much later and it wouldn't make sense to reply to it now

[dusky epoch]

well in any case, the response was repeated by edd just now

proving (AB)^T = B^T A^T really is naught but a matter of writing out the relevant matrix multiplications

[little scarab]

i think i see what ed's referring to but not how to translate it into actual maths. i guess it's better to read over my textbook and see if i can't try to develop the idea myself first before asking for another hint

[dusky epoch]

if you know the definition of matrix product and transpose it really is not hard to write out

[fossil rampart]

For P to be non singular the det(P) must be non zero but how do I prove that it is greater than 0

[dusky epoch]

you don't need to

even if det(P) is negative, det(P'P) will be (det(P))^2 and hence positive

[fossil rampart]

Oh wow okay lol that helps a lot thank you so much

If I had to give a reason for det(P'P) = det(P)^2 ,, would the following be sufficient : P is a non singular matrix meaning P is a square matrix therefore det(P) = det(P') ?

[dusky epoch]

det(P) = det(P') has nothing to do with P being nonsingular

we're talking about square matrices all throughout anyway

[fossil rampart]

Thanks 👍

[midnight trout]

is there a term to describe the identity matrix but where the diagonal of 1s goes from bottom left to top right?

row exchange matrix?

would it be considered the transposed identity matrix? would that make sense since, afaik, the identity matrix is equal to the transpose of itself?

[lavish jewel]

row exchange is right

and as you noted, it's not the transposed identity

[lavish jewel]

oh, i should add

it's column exchange if the mat goes on the right

so permutation mat is better

@midnight trout

[quartz compass]

I think it goes by the specific name 'Exchange matrix' and is denoted by a J

[midnight trout]

yes this is why I was asking because I found it interesting that it changes from row to column depending on where you put it

[midnight trout]

that sounds correct but I have no idea

exchange matrix or permutation matrix sound like completely reasonable names

[lavish jewel]

we were both correct, but mero was more correct

[midnight trout]

I spent 3 hours today trying to understand what 'elimination matrix' is because my book doesn't call them what every other source calls them 'elementary matrix'

[lavish jewel]

exchange mat is more specific

[quartz compass]

lol

well it is probably better to think of it as a permutation matrix unless you have some specific reason not to

[lavish jewel]

i think elementary matrices need not have all 1s as their nonzero elements

[quartz compass]

also nice, you're a mod again

[midnight trout]

when you google 'elimination matrix' all you get is tutorials on how to do row reduction

[lavish jewel]

and you're back to honorable :x

[quartz compass]

haha I left the server for a few months I guess to focus on other things

[midnight trout]

yes the row of an elementary matrix is 1s. then one element is a non zero

which is so simple but my book had to be special and unique

[lavish jewel]

your book sounds more correct

since you don't want the elements to be all 1s when you do elementary operations

[midnight trout]

the whole second half of the chapter was about constructing elementary matrices

constructing and combining them, just they called them something different

[lavish jewel]

the name doesn't really matter as long as you understand what they do, anyway

but yeah

exchange or permutation mat

[midnight trout]

the name matters insofar as I can google it lol

the worst part about getting stuck on something like that is that it makes me sleepy. so I had a second wind starting at like 10 pm. now it's 2 am

[little scarab]

ah, think I got it! thanks guys :)

[winged prairie]

can somebody explain how the x ~ y implies [x] = [y]

i underlined what im referring too in red

like, why is it that if two elements are equivalent, then they belong to the same equivalence class

[surreal otter]

Heya, quick question the Pauli matrices; so I have been trying to derive the Pauli matrices, and have found some information relating to how they behave as Spinors (which I don't fully understand either, as objects, other than the fact that they require two rotations to reach the same place again), and something relating to quaternion rotation? How does one derive these from these bits of information?

[little scarab]

just to confirm, if I have a square matrix which has a left inverse then it must also have the right inverse (and vice versa), and thus is invertible, right?

[lavish jewel]

sounds right off the top of my head

something like

we consider square mat A with right inverse B, so that AB = I

and then the left inverse C, so that CA = I

take the latter and multiply both sides by B, getting CAB = B

we can associate so that C(AB) = B, and AB = I, so C(I) = B

[wintry steppe]

I have a question

so

a complex 2D vector is a 4D object right?

like you need to be able to visualize 4 dimensions in order to visualize a vector space over C^2 right?

[dusky epoch]

from some viewpoints yes

for visualization certainly

[wintry steppe]

got it, thanks

[wintry steppe]

wait wdym by "from some viewpoints"? can you elaborate?

[dusky epoch]

your terminology is a little off

with the "vector space over C^2" thing to be specific

C^2 has dimension 4 as a real vector space, but 2 as a complex vector space

[wintry steppe]

ah so "a complex vector space over C^2"?

thanks

[dusky epoch]

no

"complex vector space" is short for "vector space over C"

C^2 is the vector space

[wintry steppe]

oh right i forgot

[vocal isle]

Hey guys, I don't know how to get the following matrix from this equation

k1 and k2 and n are 3x1 vectors

I is a 3x3 identity matrix

[lavish jewel]

are n and k vectors of real numbers?

[vocal isle]

yes

[lavish jewel]

aight

one thing is that the dot product of vectors of reals can be written as x^T y

and since the vectors are of reals, the order doesn't matter

i.e. x^T y = y^T x

so we can write k_1 dot n as n^T k_1

you'll agree that n^T k_1 is a scalar, and so we can move it to the right

[vocal isle]

I know it has something with this

[lavish jewel]

we get that k_2 = k_1 - 2 n (n^T k_1)

[lavish jewel]

that's literally the same as i wrote, yes

now, the product of matrices is associative

so just associate from the left instead of from the right

[vocal isle]

i understand up to thois point

[lavish jewel]

k_2 = k_1 - 2 n (n^T k_1) -> k_2 = k_1 - 2 (n n^T) k_1

factor out the k_1, and you get the identity you wrote

[vocal isle]

OMG IT GET IT

thanks so much

wow, that was so easy!

thats really beautful actually

thanks 😄

[lavish jewel]

aight coo

the geometric interpretation is nice as well

orthogonal projection onto the orthogonal complement of the normal

... kinda, except for that factor of 2

[vocal isle]

thanks! Out of curiosity, what would happen if k1,k2, or n were not real numbers? Wouldn't the proof be the same?

just wouldn't the matrix be complex?

[lavish jewel]

there would be some extra complex conjugates here and there

you can't just swap the order of the dot product

the dot product in C^n is x^H y, where H is the Hermitian transpose or complex conjugate transpose

you transpose the vector and complex conj all the entries

[vocal isle]

woah

what class did you learn all this in?

(i'm just a dumb engineer)

[lavish jewel]

i'm also an engjneer

must've learned it at some point near the end of undergrad or early masters

[wintry steppe]

Wouldn't defining a vector space over a ring be the exact same as defining it over a field?

[lavish jewel]

a module over a ring is more general

[wintry steppe]

I've heard of "modules" are they just vector spaces over a ring?

[winter harbor]

Aesthetically the definition looks the same, a module over a commutative ring would be just an abelian groups (M,+) endowed with a monoid action (R,*) of the multiplicative monoid of a commutative ring. Which satisfy distributivity properties a*(u+v) = a*u + a* v and (a+b)*u = a*u + b* u.

The definition is very similar

We just change ring to Field

And monoid action to group action

And we get vector spaces

[wintry steppe]

is an abelian group a group where the binary operation of the group is commutative?

I know next to nothing about abstract algebra 😅

and I've never heard of a "monoid action"

[winter harbor]

Yup

That's correct

[winter harbor]

Have you heard about group actions before?

[stable kindle]

monoid actions are like, group actions except monoids

[wintry steppe]

I think so

[winter harbor]

It's the same thing basically

[wintry steppe]

is a monoid action more general?

[stable kindle]

yes

[winter harbor]

I just tried to like, make the definition of a module over a ring as compact as possible, because I am lazy and don't want to write down like 10 axioms.

So it seems really abstract, but it is not.

[wintry steppe]

lol it's okay

[winter harbor]

So like, aesthetically the definition looks the same

But when we are working with rings

Say commutative rings

We lose a lot of properties

For instance, not every module has a basis.

Those that do are called free modules.

[wintry steppe]

oh wow

[winter harbor]

Yeah

[wintry steppe]

damn this is all super interesting, what's a good "intro to abstract algebra" book that talks about modules?

[winter harbor]

Dummit and Foote talks a bit about them

[wintry steppe]

is that a good book for people like me who don't know much? or is it too dense?

[winter harbor]

Nope

In fact

I think most people find Dummit and Foote a bit slow

At least the ones I have talked here in this server

There's also Artin

Which is like

Really good

It's my favorite abstract algebra book

[wintry steppe]

"nope" as in it's not dense?

[wintry steppe]

ah I see

[winter harbor]

Yeah

[wintry steppe]

lol I'll also check that out then

thanks!

[winter harbor]

Np

[torn leaf]

Hii

Can anyone please explain this

[wintry steppe]

I'm pretty sure they mean -2(row1) + row2 by "add -2 times row 1 to row 2"

it's just the basic row operations

[dusky epoch]

there is nothing to explain, just do what you're told.

[torn leaf]

Got it thanks

[torn leaf]

Yes my friend got it now it’s hers not mine

[wintry steppe]

we have linear transformations in linear algebra right? so are "non-linear transformations" just vector functions?

perhaps

well not every vector function I guess but they can be represented using vector functions but never using matricies

[silver cradle]

if we have $PAP^{-1}= QBQ^{-1}$
can we do $Q^{-1}PAP^{-1}Q = Q^{-1}QBQ^{-1}Q = B$?

[stoic python]

clement

[silver cradle]

there was a note talking about how a and b can have different eigenvalues but am i supposed to use that somewhere?

[teal grotto]

yea

[silver cradle]

but can i take my R = Q^-1P? I read somewhere that matrix multiplication is not commutative so we would have R^-1 = QP^-1, but here we had written Q^-1PAP^-1Q = B

[rapid birch]

hello quick question, it asked me is R^2 a vector space

R is all real number but whats the two

[kind cobalt]

R x R

[rapid birch]

and that means...?

[kind cobalt]

all couples (a,b) where a and b are real numbers

[teal grotto]

R^-1 = P^-1Q

[silver cradle]

dont we distribute the ^-1 inside? Since R = Q^-1P then R^-1 = QP^-1

and maintain order of the multiplication

[teal grotto]

no you swap order

[silver cradle]

oh im having trouble understanding as to why that is

Nvm I think I can see

$RR^{-1} = Q^{-1}PP^{-1}Q = Q^{-1}IQ = I*I = I?$

[stoic python]

clement

[silver cradle]

for this are we supposed to solve for det(M - lambdaI) = 0 to show its diagonalizable over C by showing distinct eigenvector

solving using quadratic formulae?

[winter harbor]

Yeah, the characteristic polynomial is:
$$
\lambda^{2} + (-2 \cos \theta)\lambda + 1
$$

[stoic python]

MisterSystem

[winter harbor]

And to show it is diagonalizable we must show this polynomial has distinct roots

[limber kiln]

yo can someone help me with this, "Give an example of a 3x4 matrix A and a vector x when Ax is not defined. Explain."

[stable kindle]

let x = (1)

[thin hazel]

If I have an injective linear map from R^n -> R^m with m >= n, why is it true that there is a suitable choice of basis s.t the matrix is (1_n | 0)?

those are meant to be blocks one on top of the other but I can't format it properly in disc

I know by rank nullity that injectivity means rank = n

so I have n linearly independent rows/cols

so is the idea to choose those as my basis?

[marble lance]

Let T be your map. Let {e_i} be the standard basis on R^n. Then {T(e_i)} is a linearly independent set in R^m. If you extend it to a basis of R^m, that basis works.

@thin hazel

[limber kiln]

If A^2 = I, is A = I?

[marble lance]

No

[thin hazel]

@marble lance perfect, im really rusty on my linear algebra so ill have a think about this. Thank you!

[limber kiln]

@marble lance why

[marble lance]

Consider

-I

[thin hazel]

the pauli matrices are a good example

[limber kiln]

I mean I as in the Identity matrix

[marble lance]

Yes

(-I)^2 = I but -I ≠ I

[limber kiln]

So could you give me a full example

[marble lance]

I just did

The matrix -I

Idk what you mean

A = -I is the counterexample

[limber kiln]

Oh I see, thank you

[midnight trout]

if matrices AB = I, then AB = BA right?

[teal grotto]

correct

[quartz compass]

false

[midnight trout]

uhm

[native ore]

Since y is missing, how do I go about this

I'm assuming I cant just assume y is a static 0

[midnight trout]

y would be any real right?

[native ore]

yea

[limber sierra]

if your matrices are square

[quartz compass]

pick A=[1,0] and B=[1,0]^T, then AB=I but BA is not

[limber sierra]

otherwise this is nonsense

the multiplication doesnt make sense

[native ore]

could I say that this is equal to 4x -z + y -y = 0

[quartz compass]

not true, I just gave an example

[native ore]

meaning to say that the change in y doesnt affect x or z

[limber sierra]

oh yeah

duh

[midnight trout]

I thought I is always square?

[limber sierra]

otherwise its false but not necessarily nonsense

I is always square, but A or B may not be.

[midnight trout]

ah right

[limber sierra]

if A and B are square, then yes, AB = I implies BA = I = AB

left-inverses of square matrices are right-inverses

(this follows from linear injections being surjections)

[quartz compass]

to make a geometrical picture you can think of taking something in 2D space, putting it into 3D space, then projecting it back into 2D space, and you've done nothing

but if you take something in 3D space and project it into 2D, then you've lost information and so putting it back into 3D space you no longer have the identity

[torn leaf]

Is it correct?

[limber sierra]

no.

two of those are filled out incorrectly, check your definitions

[forest quiver]

rref is when each row has a leading one, and the entries above and below each leading one are zero

Row echelon form has each row starts with a number, and has zeros as entries underneath

[torn leaf]

It has be 2

[limber sierra]

the second row is filled out correctly.

[torn leaf]

Rest r wrong?

[limber sierra]

here are your issues.

[torn leaf]

Ihg it’s rref

And neither

[forest quiver]

wait why is 3 not ref?

[torn leaf]

Yes that too

[limber sierra]

nope

check your definitions

[forest quiver]

The leading coefficient (also called the pivot) of a nonzero row is always strictly to the right of the leading coefficient of the row above it.

[torn leaf]

It’s neither

[limber sierra]

that one (#3) is neither, yes

[torn leaf]

Last is ref

Is it correct

[forest quiver]

yea it would be rref if a_14 = 0 though

[torn leaf]

So both or just rref

,?

[forest quiver]

4 is ref as you said

it would be rref if the top right corner was 0 though

[torn leaf]

Ohh

[torn leaf]

@forest quiver

[forest quiver]

is this x_3 ?

[torn leaf]

Yes

[forest quiver]

You are wrong for #1

Convert that to an augmented matrix and put it into rref

and see what the solution is

[torn leaf]

I got no solution

[forest quiver]

Youa re wrong for #2

[forest quiver]

gj

because 0x can never equal 1

[torn leaf]

What’s its

[forest quiver]

Well put it into rref

and you will see what the right answer is

You are correct for #3 gj

That should make sense, two planes that intersect intersect at a line

and there are infinite points on that line that could be a possible solution

you are wrong for #4

[torn leaf]

Infinity ?

[torn leaf]

Unique?

[forest quiver]

and you are correct for #5

[forest quiver]

that's what you said it's wrong\

[torn leaf]

No sotluton

[forest quiver]

yes

[forest quiver]

no bro

[torn leaf]

Wait

[forest quiver]

you are just guessing at this point trying to get the right answer

[torn leaf]

It’s unique

[forest quiver]

No

[torn leaf]

Why not

[forest quiver]

if you are talking about this one

[torn leaf]

I don’t get it

[forest quiver]

Are you talking abt this?

[torn leaf]

Yes this

[forest quiver]

This is the RREF of it

you see that x_4 is a free variable

right?

[torn leaf]

Ohhhh

It’s infilnty

[forest quiver]

I think one sec

wait wait

my bad x_3 is a free variable

x_4= -1

but yeah still infinity

[torn leaf]

Hmmm

Ok

Thank you

[acoustic crescent]

can anyone give me an example of how linear algebra is used in machine learning?

[ocean sequoia]

PCA is legit just LA

[acoustic crescent]

hmm i can see that, but lets say i have a dataset with house sale prices, location and size in square feet, and i want to predict future house prices. I can see why probability and statistics is used but why is linear algebra used? Is is just because python handles data through matrices?

[zealous junco]

multivar probability rely on LA as basis

often times like regression you can derive stuff like max likelihood estimation as a nice linear alg expression

[lavish jewel]

data tends to have a structure that is well captured via linear subspaces

this is related to pca, as someone mentioned above

l9w dimensionality structure

also tge multivar calc you need for optimization inevitably relies 9n vectors

pardon the typos o still have one eye closed

[zealous junco]

n9 w9rries

[boreal shadow]

if i have a set of 4-tuples, if i show it is a basis for R^4, does that also show that its a basis for R^3?

[lavish jewel]

no

you could take a set of 3 linearly independent 4-tuples, which would be a basis for a subspace of R^4 with dimension 3

[boreal shadow]

the question i have to answer gives me four 4-tuples, asks me to show that its a basis for R^3... so I would just take 3 of them, make sure they're linearly independent and that they generate a subspace that spans R^3?

[teal grotto]

they span a subspace of R^4 that is isomorphic to R^3. they do not span R^3 because they are 4-tuples, and consequently not in R^3, since R^3’s vectors are 3-tuples.

[lavish jewel]

exactly as c squared says

[boreal shadow]

yeah that makes sense

The set of ordered 4-tuples {(0, 2, 1, 0), (1, −1, 0, 0), (1, 2, 0, −1), (1, 0, 0, 1)}
presumably forms a basis for R 3 . Show that it is a basis indeed. Then,
starting with this basis use the Gram-Schmidt procedure to constructan orthonormal basis for R 4 . Verify that the resulting vectors are or-
thonormal.

thats the question ive been given

[teal grotto]

that looks like a typo

should be R^4

[lavish jewel]

yeah, should be R^4, otherwise it doesn't make sense

[boreal shadow]

okay cool im not going crazy then

its been confusing me for a few hours now 🙃

ty for the help 🙂

[lavish jewel]

it's basically a "show the vectors are lin indep and then do gram schmidt"

[boreal shadow]

yeah thats what i thought but i didnt realise it was a typo and have been trying to work out wtf it means cause it didnt make any sense to me 😂

emailed my teacher like 48 hours ago about it but he hasnt responded still 🙃

but yeah ty

[lavish jewel]

when it's not diagonalizable

"defective matrix"

these are matrices with repeated eigenvalues, but they don't have the same number of lin indep eigenvectors as the number of the repeated eigvals

yeah

so you look for the eigenvalues, find the eigenvectors, and check whether the eigenvectors are all linearly independent

and if they are, then you can diagonalize the mat using those vectors

[stoic stag]

can someone please explain to me how are these vectors linearly independent in the subspace W. Because from my understanding, the rank of the matrix is lower than the no of unknowns which would give us infinitely many solutions for a,b,c,d. Making the vectors dependent? Unless im missing something?

[lavish jewel]

in this setting, you would've been better served to put the vectors as columns of the matrix, not rows

since you want to generate stuff in R^4, not R^3

but in any case, you found the matrix has 3 pivots

this means rank 3, 3 lin indep rows

so the rows are independent from each other

[stoic stag]

So what would be the condition for dependency here? Or in other words to get infinitely many solutions?😅

[lavish jewel]

infinitely many solutions for what

there is no system of equations here

you aren't solving anything

you were only asked about the dimension of a subspace spanned by a set of vectors

no equation, and no solutions involved

[stoic stag]

okay okay i think i get why i was having this problem in my head, was thinking about these vectors as a set of equations.

anyhow thank you!

[little scarab]

if you have have a transformation S∘T represented by matrices and want to find the matrix representing the transformation, is the algebra ST or TS?

[stable kindle]

S∘T means do T then S

and wrt. matrices, ST(v) = S(T(v))

so it's the same i think

[little scarab]

right, so it's just ST?

[stable kindle]

yes

i think

[ornate fiber]

Can someone explain to me the visualization behind Tensors ? Like, what's the use of these objects in physics ? and why we need to generalize the number of indices to any arbitrary number to work it out in continuum mechanics for example.

like for example, why matrices are not enough ?

[lavish jewel]

technically you can turn them into matrices that represent the same operation

just some elements might have to appear several times

[little scarab]

and if a system Ax=b has one unique solution, what does that say about the solutions of Ax=0

[lavish jewel]

for example, in general if you want to map an array of size m x n x o into another of size q x r x s, you would need a map of some sort that is of size (m x n x o) x (q x r x s)

but you can matricize this

[ornate fiber]

it says that x = 0 but what has that to do with Tensors ?

[lavish jewel]

it says only x = 0 yields Ax = 0

anyway, you can turn it into a matrix

maybe of size mno x qrs, for example

and this can be done in several ways

in general you just save yourself the pain by using einstein notation

since the actual implementation of the map isn't unique

and you might wanna retain some intuition

[little scarab]

so it has at least one solution?

[lavish jewel]

for example some map that takes a scalar field in 3D and maps it into another scalar field in 3D

[little scarab]

probably only one?

[lavish jewel]

you'd need a tensor with 6D to represent this "natively"

but you can just as easily turn the 3D stuff into 1D by "unrolling"

[lavish jewel]

this is completely unrelated to what i said. you were the one that said Ax = b has a single solution, so i told you what that implies regarding Ax = 0

think of the dimension of the null space

[ornate fiber]

As long as you can turn them into matrices, why would you work with them after all ? like why dont you work with matrices only if they are more useful than Tensors in physics ? I just want to know what's the purpose of Tensors in general

[lavish jewel]

it's just a multilinear transformation

how you represent it doesn't matter

you don't need matrices either

[little scarab]

but does what i said follow?

[lavish jewel]

you only care about the underlying transformation and its properties

and anyway, using physics definitions, matrices and vectors are also tensors

[ornate fiber]

For example. scalar product represents the tendency of 2 vectors to align with each others. Now this is like a visualization of scalar product apart from the pure mathematical axioms etc... Can you do the same with Tensors ?

Most often, i see ppl saying Tensors only matter because of how they transform

why is that?

[lavish jewel]

that's a property of vectors

[ornate fiber]

ikr

but why would you talk about Vectors?

[lavish jewel]

scalar products, i mean

[ornate fiber]

okok

[lavish jewel]

because what you said doesn't make sense

unless you make a vector space where the vectors are tensors

[ornate fiber]

oh got it

[lavish jewel]

vectors in a more abstract sense

tensors are one possible representation of a multilinear transformation

[ornate fiber]

so vectors are more general right?

[lavish jewel]

you don't need them

and no

[ornate fiber]

okay btw excuse my dumb ass lol

[lavish jewel]

i think you need to go review again what vectors and vector spaces are

and also linear transformations

[ornate fiber]

okay thanks for that recommendations

but can you please explain the relationship between vectors and Tensors ? which is more general ?

[lavish jewel]

they are different things

generally speaking they are not related

vectors can be tensors though, and tensors can also be vectors

depending on whether they act as linear transformations or members of a vector space

the question doesn't make sense

[ornate fiber]

okok

thanks

[little scarab]

so can I just check again...
I believe: if a system Ax=0 has a solution x0, and a system Ax=b has a solution x1, then that system also has a solution x0+x1

so if Ax=b only has one unique solution, does Ax=0 have one solution or none? i imagine it still has the trivial solution (x = 0)?

[lavish jewel]

only the trivial one (x=0), yes

[ornate fiber]

Dawdle of Doubt. lmao i was in class when i asked those questions and i only looked at half the answers. will check what you said now

[lilac sundial]

i need help

[abstract prawn]

could anyone help me with these?

[lavish jewel]

try the #multivariable-calculus channel

[acoustic crescent]

can anyone tell me the importance of finding the greatest eigenvalue of Matrix A would be important? Like for example using the power iterator method. What applications does it have for data science, machine learning etc?

[lavish jewel]

are you familiar with the gradient descent algorithm?

if you use a fixed step size that depends on the largest eigenvalue of the matrix A = M^TM in a system Mx = b, the gradient descent algorithm is guaranteed to converge

the singular values are related to how well behaved a system is

also you can make iterative algorithms that can find the several eigenvalues and eigenvectors by repeating that method and then projecting on the complement of the eigenvectors you have found

[winter harbor]

Page Rank is also a nice algorithm where knowing the largest eigenvalue of your matrix gives you nice information.

[wild fulcrum]

a bit side note (not related to matrix power) but PCA takes the largest eigenvalue & corresponding eigenvector of the covariance matrix to be the first PC

[wintry steppe]

I have a few general questions about linear algebra and its importance to mathematics. I keep seeing linear algebra pop up in various places as a central topic to study regardless of one's particular interests. Example: University of Arizona Ph.D. in Applied Math --- recommended background reading for incoming Ph.D. Students. "(i) Linear algebra. You can never know too much linear algebra! [snip text] If you only review one thing this summer, it should be linear algebra. This will help you with all three core courses. " https://appliedmath.arizona.edu/background-reading What makes linear algebra so important? Why does it crop up in so many places? Is linear algebra an active field of research?

[lavish jewel]

when you compute stuff on a computer, you have to make discrete calculations

numerical computations tend to involve vectors of samples of a function or random process

and then operations over these vectors are represented as linear operators

[tranquil steeple]

yes, it is an active field (i pretty much only do linear algebra, both pure and numerical)

[lavish jewel]

stuff like approximating differential equations, machine learning, optimization, etc, all realy on this stuff

you make discrete approximations, provide error bounds, etc

what i do is pretty much also linalg

signal processing and machine learning

[wintry steppe]

figure out the matrix A of T in that basis and then compute det(A - tI)

I being the 4x4 identity matrix

[lavish jewel]

that's how all matrices are defined

[wintry steppe]

yeah, the columns of A will be the representations of T's action on the basis elements

[lavish jewel]

some action on the basis vectors

you can also make use of linearity here

two things are happening to the polys, and then they are added

[wintry steppe]

for example, $T(1) = 1 = 1\cdot 1 + 0 \cdot x + 0 \cdot x^2 + 0 \cdot x^3$, so the first column is $$\begin{pmatrix}1 \ 0 \ 0 \ 0 \end{pmatrix}$$

[stoic python]

TTerra

[lavish jewel]

so you could say the matrix M does 2 things, and you could represent that as M = A + B

[stoic python]

Yeetus

[lavish jewel]

yeah the filling in is different

[stoic python]

Yeetus

[lavish jewel]

that's more like it

[winter harbor]

Remember that the minimal polynomial divides the characteristic polynomial and has the same irreducible factors as the characteristic polynomial.

np

[winter harbor]

Well, an endomorphism is diagonalizable iff the minimal polynomial has no repeated roots.

Working over C, ofc.

Yup

[karmic pollen]

Can anyone here go to calculus and tell me if this kinda makes sense

[winter harbor]

Yup.

[winter harbor]

This is easy to see if you think about the Jordan normal form.

[crystal hound]

Hi, I think I’m putting this in the correct section of the server, can anybody provide hints on how to find the matrix of a linear map given only the kernel of the linear map please

[stable kindle]

the kernel isn't sufficient

if M has kernel K, 2M will also have kernel K, for a start

[stable kindle]

right, so you just have to find one such matrix

[crystal hound]

To start this question then, I found a specific form for all vectors within the kernel, and then I think I can proceed to find a basis for the kernel, but I’m unsure how that is going to help with respect to finding one such matrix

[stable kindle]

i'd help but i gtg sorry

consider just the simplest possible projection of the space onto that hyperplane?

ok i have a little more time than i thought

so

take the basis for the kernel

extend it to the whole vector space

then it's just the matrix of the map that maps basis vectors in the kernel to themselves and those not in the kernel to their closest analogue in terms of the kernel basis vectors?

is that coherent? sorry gtg

[teal grotto]

you say that you found a specific form for all vectors within the kernel. do you mind sharing?

[crystal hound]

Sure, gimme 10 mins and I’ll post that

Nearly home just gotta get my soln so far

[teal grotto]

actually

dont waste your time on that

so

lets just "organize" the information you have

we know that

2a - 1b - 0c + 1d = 0
0a + 1b + 1c + 1d = 0
2a - 2b - 1c + 0d = 0
0a + 0b + 0c + 0d = 0

hint : this looks eerily similar to a matrix multiplied by a column vector (a, b, c, d)...

[crystal hound]

Yeah so far so good, that’s how I ended up eventually obtaining a general form for my vectors in the kernel

[teal grotto]

well, im suggesting that you did not need to do that

[crystal hound]

Wait I’m so stupid

Oh crap I think I can use that to literally find the matrix 🤣

[teal grotto]

yea. you just use the equations given to you to form the matrix. it should be

2 -1  0 1
0  1  1 1
2 -2 -1 0
0  0  0 0

[crystal hound]

I can’t believe I actually asked that question 😂you putting that has made it so obvious it’s unreal

[teal grotto]

lol

[crystal hound]

Thank you for the help 😁

[steep stratus]

Hello, I'm asked to check if the following is a vector space. Could I get a hint on how to start this question? I was thinking I could try to write it as a span and then it would be a vector space? But I'm not sure. I got about as far as this: ```
{(x1, x2, (1-2x1-5x2)/9 | x1, x2 for all real numbers}
Then I try to rewrite this as a span but I don't know how to :(
Help appreciated!

[crystal hound]

Think about what it means for something to be a vector space. What must be satisfied to have a vector space?

[teal grotto]

all vector spaces have the zero vector in it....

[steep stratus]

it must have closure under addition and under multiplication right?

[crystal hound]

Correct, think about all the axioms surrounding vector spaces and how you can show if they are or aren’t satisfied

[steep stratus]

oh wait it not be a vector space then? Because I could multiply by two for (1,1,2/3) that is inside the set, and get (2,2,4/3) which isnt inside the set? So it isn't closed under multiplication? Is that right?

ohhh also the 0 vector yea. because 0,0,0 doesn't satisfy this

so it isnt a vector space, and i overcomplicated this

thanks

[teal grotto]

what are E0 and E1

okay. and just to make sure, its f : V --> V, with f^2 = f

i would look at the map
v |--> ( (f(v) - v) / 2, (f(v) + v) / 2)

from V into E0 + E1 (where + in this line is meant to be a direct sum)

you're going to need to show that (f(v) - v)/2 is in E0 and (f(v) + v)/2 is in E1

then show that this map is indeed an isomorphism

which part(s) are you unsure about

cool

[acoustic crescent]

anybody knowledgable on Power method and the inverse power method? For biggest and smallest eigenvalues of A

[tranquil steeple]

what about it?

[acoustic crescent]

so my book says this, i wanted to make the inverse power method in python and an easy way of doing it was just modifying my original power method but converting the matrix A to the inverse first and then run the same algorithm. However in the picture below it says that is not a good idea and they want me to use lu factorization, why ?

[tranquil steeple]

because it is typically frowned upon computing the inverse to solve a system. i do not know how you solve a system in python but most languages has a \ -solve (originally from matlab) maybe you need to solve using a solve function? or use lu explicitly

[acoustic crescent]

yes there is a solve function but idk how to substitute it

idk if this is readable to you but as you can see i set A to the inverse of A in the start, i want to use solve instead but im not sure how

[tranquil steeple]

https://numpy.org/doc/stable/reference/generated/numpy.linalg.solve.html

[tranquil steeple]

it would be readable tomorrow morning, now I am just about to goto sleep. probably somebody else knows

i assume just switch dot(A,x) to linalg.solve(A,x)

and remove where you define A to be A inverse

[acoustic crescent]

bruh, that worked

thx alot man

[crystal hound]

On second glance, I’m not so sure we can do this, those equations are defining the vectors within the kernel(ie (a b c d) transposed such that those equations hold), and since vectors of the kernel are within the domain of the linear map, how does this provide us with information about the matrix for the linear map?

Dont those equations just give us information defining the vectors within the kernel and nothing about the linear map itself

That is, they define the vectors such that $f(v)=0$

[stoic python]

mchambo02

[teal grotto]

the kernel of f (the linear transformation represented by the matrix we found) is precisely U

[crystal hound]

I understand that I think, but don’t the equations therefore define merely what values a, b, c&d can take whilst being contained within the kernel, and therefore what vectors are contained within U? I’m just struggling to see in hindsight how we can use those to obtain the matrix for the map when they’re only defining vectors within U, right?

[teal grotto]

the equations determine the vectors (a, b, c, d) such that f(a, b, c, d) = (0, 0, 0, 0)

[crystal hound]

Ok yeah I’m with you

[teal grotto]

like, idk what else... um. im not sure what the issue is you're trying to bring up

the matrix we get wont be unique in the slightest (if thats what the concern is)

[crystal hound]

Oh wait I get it now I think, instead of trying to find the form of vectors within the kernel, we can use those equations defining vectors within the kernel to obtain the matrix quite trivially. I think I was getting confused because I thought we could only use those equations to find vectors within the kernel, when in fact if we consider general f( a b c d)^T=0 we can also use those equations. It’s hard to explain what I wasn’t getting but it makes sense again now

Thank you again 😅

[teal grotto]

yep, no worries

[crystal hound]

So could the matrix we get be multiplied by any lambda(excluding 0) and still be correct?

[teal grotto]

yes

you could also permute any of the rows of the matrix it would be correct

just not the columns

which is is super effing enlightening because row reduction preserves the kernel and thats the first time ive ever really seen that. like i thought i understood why, but holy

thats so explicit

[crystal hound]

Perfect, thank you 😁

[crystal hound]

Does the rank of a matrix representing the linear transformation = rank of the transformation ? This makes sense in my head but just Want to check

[wintry steppe]

yes

[crystal hound]

Ok thank you

[wintry steppe]

the isomorphism given by the choice of basis on the codomain of your transformation takes the transformation's image to the column space of its matrix representation, and since it's an isomorphism, it preserves dimensions

[gray dust]

🥗

[nocturne jewel]

Linear Salad

[empty hemlock]

Let T : W -> W be linear, where W is an n-dimensional complex vector space. Suppose that T has k distinct nonzero eigenvalues. I want to show that ker T^n = ker T^(n - k). Clearly ker T^(n - k) is a subset of ker T^n. The other direction is giving me a hard time: If T^n v = 0, then T^k(T^(n - k) v) = 0. Assume w := T^(n - k) v /= 0. Then the T-cyclic subspace <w> has some dimension less than k. I'm not sure how to continue here, but it feels like I'm going in the right direction.

[winter harbor]

Notice that we can do this by induction

Notice the following

If $T \in \text{End}(W)$ has $k$ distinct eigenvalues, then let $\lambda_{i} \in \mathbb{C}$ be its $i$-th eigenvalue and $v_{i} \in V_{\lambda_{i}}$ be any eigenvector in the eigenspace $V_{\lambda_{i}$ corresponding to the $i$-th eigenvalue.
\
\
Let $V = \text{span}{v_{1})} \oplus \cdots \oplus \text{span}{v_{k}}$
\
\
Notice that $V$ if we suppose $k < n$. And suppose that the result holds for $k < n$, this implies that $T^{k}$ restricted to $V$ has $k$ distinct eigenvalues. So by hypothesis, $\text{ker} , T^{k} = \text{ker} , T^{k-k} = \text{ker} ,\text{id} = {0}$.
\
\
So if we prove $T^{n-k} (v) \in V$, we see that $T^{n-k}(v) = 0$ follows.

[empty hemlock]

I see how V is equal to that direct sum with T^(k-1). But why T^(k-1)? I don't follow the rest.

[stoic python]

MisterSystem
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[winter harbor]

I am defining V as this direct sum

W is the vector space we started with

[empty hemlock]

Oh, you're right.

[winter harbor]

Yeah, the nice thing is that

We artificially produced a subspace of dimension k

Where T^k has k distinct eigenvalues

And we can apply some induction argument here

I think you can carry out the details

[empty hemlock]

But why T^(k-1)?

[winter harbor]

Oh, yeah I didn't mean that

If $T \in \text{End}(W)$ has $k$ distinct eigenvalues, then let $\lambda_{i} \in \mathbb{C}$ be its $i$-th eigenvalue and $v_{i} \in V_{\lambda_{i}}$ be any eigenvector in the eigenspace $V_{\lambda_{i}$ corresponding to the $i$-th eigenvalue.
\
\
Let $V = \text{span}{v_{1}} \oplus \cdots \oplus \text{span}{v_{k}}$
\
\
Notice that $V$ if we suppose $k < n$. And suppose that the result holds for $k < n$, this implies that $T^{k}$ restricted to $V$ has $k$ distinct eigenvalues. So by hypothesis, $\text{ker} , T^{k} = \text{ker} , T^{k-k} = \text{ker} ,\text{id} = {0}$.
\
\
So if we prove $T^{n-k} (v) \in V$, we see that $T^{n-k}(v) = 0$ follows.

I really meant span(v_i) and so on

Because notice that