linear-algebra

[spare widget]

What your matrices look like is something like this C = P A

[open locust]

@spare widget Sorry would you mind slowing down a little? I'm a little overwhelmed with all the information

[spare widget]

For rectangular matrices you have to have a different basis, so they have the more general thing there

[open locust]

I'm still back on your previous statement

[fleet sun]

@open locust I don't know if you saw a message I wrote yesterday morning about this, but I came to the conclusion that being invertible implies the matrix is equivalent to the identity, which in turn means there are row operations on it that produce the identity (and at the same time produce the inverse matrix)

[open locust]

I missed that message sorry! (kinda new to Discord)

[spare widget]

This looks like so: [w]_F = A [v]_E

[open locust]

@spare widget Yes indeed.

[spare widget]

an identity map is T(v) = v

So it is necessary that the codomain is the same as the domain

[open locust]

Right, I understand that

[spare widget]

But if you pick two different bases

Then your "identity" is really just the change of basis matrix

Because if you define your matrices using two different bases

Then they have the form C = P A, where P is change if basis, and A is matrix if you were to use the same basis (that's in the case the codomain and domain are the same)

e.g.

[open locust]

@spare widget Well, you could just say it more simply, right? You could just say that [T] is the change of basis matrix

[spare widget]

[w]_F = P [w]_E = P A [v]_E

The above makes sense in the case of V-> V maps

[open locust]

What is A here?

[spare widget]

so linear map

[open locust]

I mean, like, why the distinction between P and A?

[spare widget]

because A maps from E to E

PA maps from E to F

[open locust]

What is the purpose of splitting them up like this? Just to be more clear?

[spare widget]

It's what's usually done in books

Then A = I would always be the identity

[open locust]

Interesting, okay.

[spare widget]

irrespective of the basis

[open locust]

Ah, right.

[spare widget]

But in your case the book tries to be general

And doesn't assume U=V

[open locust]

Interesting, I see

[spare widget]

So it may be even impossible to have the same basis

[open locust]

I'm completely following you

[spare widget]

That's why they have that

But in the trivial case of the identity map you can always pick the same basis

Otherwise you'd just get the change of basis matrix mixed in there too

P * I = P

[open locust]

Well, where is this analysis leading? I agree with your representation

[spare widget]

It's still the identity map

It just also changes basis coord representation

[spare widget]

I mean, that's all

Now I understand what you meant, it's because your book uses different bases on the input and output

[open locust]

I am glad you understand it, but I'm honestly still completely baffled as to what all this means.

[spare widget]

That's valid, just have to be aware of this detail when reading other stuff, especially with V->V maps where people pick the same basis for simplicity

[open locust]

I understand how you split up the matrices there, but I'm not sure what that says about my original question.

[spare widget]

Can you repeat the original question?

[open locust]

(1) Not every invertible matrix represents the identity transform. (2) A "change of basis matrix" is a representation of the identity transform. (3) Every invertible matrix can be seen as a "change of basis matrix".

Given these three facts, if we see an invertible matrix, how can we tell whether or not the original transformation it represents is the identity transformation?

[fleet sun]

I don't think it makes sense to start drawing conclusions about what a matrix represents unless given bases for its domain and codomain

[open locust]

@fleet sun Interesting

[fleet sun]

Many people ordinarily ignore those because for square matrices they tend to just be the standard basis of R^n on the domain and codomain

[open locust]

I think your statement may be correct, and it helps explain my confusion: what I was trying to do is probably impossible in general.

There's a nice reason for it too, which is that you can't really say anything about a linear transformation if you don't start with at least one pair of to and from bases.

Like, we can write A: U -> V, but we can't go much further than that without specifying at least one pair of to and from bases.

Similarly we can write "u in U", but without at least one basis, we can't do anything with u.

So it makes sense that without at least one pair of to and from bases, we can't really take a matrix to its linear transformation form via the isomorphism between matrices and linear transforms, because that isomorphism itself requires a to and from basis to be used.

So let me try to summarize what we can say, given a matrix but no to and from bases. All we can say is that if the matrix is invertible, then it represents a change of basis matrix with respect to at least one pair of to and from bases. We can't say if this matrix represents the identity transform or not (to do that we would need the equivalence thing you mentioned).

Would you agree with all of the above?

[fleet sun]

mostly yeah

[spare widget]

I think the interpretation depends on the context

Just giving me a matrix - it's just an array of numbers

[fleet sun]

You can still write things like 2u or -1/2*u without having any basis specified

[spare widget]

It doesn't say whether it's change of basis or not

The common assumption would be that it is from E in U to E in V

[fleet sun]

or write a transformation like T(u) = 3u. That's basis independent

[spare widget]

where E are the canonical bases

[spare widget]

Yes

[open locust]

@fleet sun Well I do agree with that.

[spare widget]

You can't say how a matrix is to be interpreted otherwise

[open locust]

@fleet sun You could add the phrase "in general" to some of my above statements then 😀

[spare widget]

The common understanding when people don't mention anything is canonical to canonical basis

Then the identity map is always the identity matrix

[open locust]

@spare widget I am not talking about how something should be interpreted really, I'm kind of talking about what we can and can't do

@fleet sun Is there anything else wrong with those statements from your point of view?

@fleet sun If not, I would like to examine the thing you said about equivalence

(the message I missed)

[fleet sun]

I had a short back and forth with @criver, close to two days ago now

I tagged you at the start of it

[open locust]

I see it now 😀 I will read it now, sorry for missing it!

[fleet sun]

no worries

and yeah, I do agree with the rest of what you just said

[open locust]

@fleet sun That is good to hear. Thank you for thinking through this. I just read the back-and-forth.

@fleet sun I think the insight you had about similarity which we later renamed to equivalence is pretty key. I'm going to try to confirm it right now actually

@fleet sun I mean the insight you had towards the end of our discussion a few days ago

[spare widget]

If you want to define a linear map through a matrix you need the bases wrt which this matrix is defined

[open locust]

@fleet sun Here are my conclusions, tell me if these make sense.

Any invertible matrix can be viewed as a change of basis matrix with respect to one pair of two and from bases. Any change of basis matrix represents the identity transform. So if we have a pair of to and from bases, we can say that a particular invertible matrix represents the identity transform. However, if we don't specify a pair of to and from bases, and we look at a square matrix that happens to be invertible, we can't say anything about whether or not it represents the identity transform.

What do you think of that?

[fleet sun]

It feels solid. I plan to double check this stuff carefully when I have a chance later

[open locust]

@fleet sun Cool, thanks for the feedback. By the way, I did see the message you'd tagged me with when I logged back in today, I just didn't know how to jump to it. If you tag me again, I'll be sure to look at your message next time I log in.

[spare widget]

I disagree

Say you have V->V

[open locust]

@spare widget I think you're coming at this from a different perspective

[spare widget]

And bases E and F

And an invertible matrix C wt E->F

Let P be the change of basis matrix

then A = P^{-1} C doesn't have to be the identity

[open locust]

@spare widget I don't think anything I said indicates that it has to be.

I think you're coming at this from a different perspective.

[spare widget]

Ah nvm, I think this agrees with what you say, yes

[open locust]

Oh OK!

[spare widget]

The particular invertible matrix you speak of is then exactly P

[open locust]

I appreciate your help thinking through this today 😀

Thank you both again, I'm going to step away now. Talk to you later 😀

[wintry steppe]

I can do a and c

But I'm unsure of b

Is it the min max eigenvalues

[spare widget]

For b try to find the max eigenvalue

[wintry steppe]

Ok damn I'm right!!!!

[wintry steppe]

Tytyty!!!

[spare widget]

You can then pick the corresponding eigenvector to get:

x^TAx = x^T lambda_max x = lambda_max

The last step is from |x| = 1

[zinc timber]

so many tags

[spare widget]

You're famous

[zinc timber]

no not me

I meant these

[open locust]

@zinc timber Sorry is it bad etiquette? (I am new here)

(Ack, just did it again!)

[zinc timber]

[open locust]

Is it better to avoid doing that?

[zinc timber]

yes

[open locust]

OK, thanks.

(Why, by the way?)

[zinc timber]

you can reply instead (with tag on)

[zinc timber]

or tag off

[open locust]

Ah I see. Not used to having that function.

[zinc timber]

annoying notification and a red symbol showing unread messages, which annoys me, at the least

don't get worked up over it

[open locust]

Understood, thanks!

[zinc timber]

no one's gonna beat you up over this

[open locust]

😀

[wintry steppe]

So I'm not sure what to do with the last step of the eigenvector here

Is it (0,1,0)???

Since x1 =0 and x2 is the free var?

[subtle gust]

it does but i'm making sure

[gray dust]

x1=x3=0 and u get no info on x2 (free), ur form of vector is (0,x2,0)

so u can set x2=1 to get (0,1,0)

[wintry steppe]

Thank you. Quite frustrating to get to the last step then fail :)

[gray dust]

no prob

[compact pike]

Hi how would i use back substitution to invert a n x n matrix?

[torn stag]

@compact pike for each $i$, use backward subsitution to solve $Ax^{(i)} = e_i$ for $x^{(i)}$.

[stoic python]

IlIIllIIIlllIIIIllll

[blissful vault]

How is the boxed in step justified? I understand why the scalar value is there. However, I don't know why we have <vj, ek> - <vj, ek>. Does he use pairwise orthogonality? If so, why is that justified?

[wintry steppe]

denominator is scalar, pull it out. you get one <v_j, e_k> from taking the inner product with the v_j at the start and the e_k. you get another one with a negative sign because all the <v_j, e_i>e_i cancel if i is not equal to k

[blissful vault]

I follow you on the denominator part and how we get <v_j, e_k> and -<v_j, e_k> terms. But why is the inner product between -<v_j, e_m>e_m and e_k equal to 0 if m is not equal to k?

[spare widget]

They are orthogonal?

[blissful vault]

Are you asking a question?

[wintry steppe]

that they're orthogonal is an assumption in the result

[subtle gust]

I've literally been asking the same question for the past 2 days 🥲

.

[blissful vault]

Oh. Is it the inductive hypothesis?

[wintry steppe]

yes

[blissful vault]

Ah, that makes sense. Would there have to be a base case?

[wintry steppe]

as with all induction proofs, yes. the base case is covered here in the first paragraph

[blissful vault]

I mean a base case about orthogonality.

[wintry steppe]

i don't understand what you mean

[subtle gust]

Mind helping me after u finish with mustaf?

[wintry steppe]

i won't

[subtle gust]

Np ...

[wintry steppe]

but someone else might

[subtle gust]

It's just that i've been waiting for sm1 to reply for so long 🌞

So idk

[blissful vault]

Maybe I'm misunderstanding. The first paragraph gives a base case for the proof that span(v_1, ..., v_j) = span(e_1, ... e_j)

Since you say that pairwise orthogonality of the list (e_1, ... e_j-1) is an assumption in the result from the inductive hypothesis, shouldn't there be a base case about pairwise orthogonality as well?

[wintry steppe]

whatever base case there would be concerning pairwise orthogonality is already covered by the base case in the induction proof you're already doing

[lavish jewel]

@subtle gust off the top of my head, can't you always multiply L by a constant c and U by 1/c? at best you have uniqueness up to scaling, i'd say

[zinc timber]

what's the best way to write char poly of a 4x4 matrix

don't wanna brute force it

like is it reasonable for you guys to leave it as a_11-a_22= c then xⁿ+....+c

[silk valley]

'let p be the char poly of a 4x4 matrix'

that'll be $100, please.

[wintry steppe]

in general the coefficients of the characteristic polynomial are the traces of its exterior powers (with some alternating signs)

that might help cut down the necessary computations

[zinc timber]

any ref?

[wintry steppe]

google

[zinc timber]

max I get is 3x3

[wintry steppe]

i don't know of any linear algebra book that covers this

[zinc timber]

tr(A)
tr(A)^2-tr(A^2)

something liek this

https://en.wikipedia.org/wiki/Characteristic_polynomial#Properties

this right?

[wintry steppe]

yes

[zinc timber]

https://en.wikipedia.org/wiki/Faddeev–LeVerrier_algorithm

interesting rabbithole

[subtle gust]

Yeah but that would change the diagonal elements

I'm saying that there is only one matrix L or U such that A=LU and the diagonal of L or U is a certain number n

If we change the diagonal elements to cn the L and U matrices would change but they would still be unique

[lavish jewel]

then yeah

you can look up the uniqueness of the LDU decomp, which is essentially the same as what you describe

[tacit pelican]

could anyone help me understand how the coordinate map works on matrices and linear transformations?

on vectors it seems pretty simple

from my understanding so far, theres a map $\phi_\gamma: V \to \mathbb{R}^n$ that maps a vector $v \in V$ to $[v]_\gamma$

[stoic python]

sean

[tacit pelican]

where $\gamma$ is just a basis in the vector space $V$

[stoic python]

sean

[tacit pelican]

in other words, $\phi_\gamma (v) = [v]_\gamma$, and this is called a coordinate vector

[stoic python]

sean

[tacit pelican]

and this is the column vector containing the scalars $\alpha_k \in \mathbb{R}$, where $v = \sum_{k=1}^n \alpha_k u_k$ where $u_i, i= 1, \ldots, n$ is just the elements of $\gamma$

[stoic python]

sean

[tacit pelican]

so i think thats how that works for vectors

but im having a hard time grasping this idea for matrices

from what i've got so far, the coordinate representation (?) of a matrix $T$ that represents a linear transformation is denoted by $[T]\gamma^\beta$, where this is the collection of columns $\begin{pmatrix} [Tb_1]\gamma \ \cdots \ [Tb_m]_\gamma \end{pmatrix}$

[stoic python]

sean

[tacit pelican]

and $b_j$ denotes hte elements of the basis $\beta$ in $W$ of a the linear transformation $T:V \to W$

[stoic python]

sean

[tacit pelican]

but i'm confused on how this works. For vectors, it is an isomorphsim $\phi_\gamma$, but for matrices there are two indices here so idk how to represent that as one or a composition of operations on a linear transformation $T$. For example, would $[T]^\beta_\gamma=(\phi_\beta \circ \phi_\gamma)(T)$?

[stoic python]

sean

[tacit pelican]

or am i getting this completely wrong?

[fleet sun]

the indices stand for bases

hang on lemme catch up

[tacit pelican]

i guess a tldr is "is there an explicit map that maps a linear transformation to its corresponding matrix wrt bases"

and if so what is it

[fleet sun]

yes

well you just said it

the $i$th column is $[T(b_i)]_\gamma$

[stoic python]

ManifoldCuriosity

[tacit pelican]

oh

well, since $T(b_i)$ is just a vector, then it is true that $[T]^\beta_\gamma= \begin{pmatrix} \phi_\gamma(Tb_1) \ \cdots \ \phi_\gamma(Tb_m) \end{pmatrix}$

[stoic python]

sean

[fleet sun]

yup

[tacit pelican]

ohhh alright

i think i got it now

so the top index $\beta$ is just the thing we take the transformation of, and the bottom index $\gamma$ is just the base we create a coordinate vector wrt

[stoic python]

sean

[tacit pelican]

well, the elements of $\beta$ to be more specific, but you get what i mean

[stoic python]

sean

[fleet sun]

yeah, you feed each domain basis vector into T, and write the results in the codomain basis

[tacit pelican]

finally its making sense

thanks man

[wintry steppe]

My textbook says "all matrices describing the same endomorphism but witg different basis have the same trace, trace characterises a transformation"

Is it correct to conclude that every set matrix of same dimension and with the same trace describes the same endomorphism?

I know this is equivalent to asking whether there don't exist 2 different endomorphisms on the same set whose sum of eigenvalues is the same.

Is this the case?

[fleet sun]

it says representing the same endomorphism with different bases have the same trace

you're asking if the converse is true?

if square matrices having the same trace represent the same endomorphism with different bases

[wintry steppe]

yes thats my question

[fleet sun]

gotcha. I don't know if that's true, but you certainly can't just assume the converse

and I tend to doubt it's true

so, representing the same endomorphism in different bases is called being similar

and it's equivalent to $A = P^{-1}BP$

for some invertible P

[stoic python]

ManifoldCuriosity

[wintry steppe]

okay, I thought thats what they meant by "the trace characterises an endomorpism"

and since they also define the trace of an endomorphism

[fleet sun]

then there's this fact that $\mbox{tr}(MN)=\mbox{tr}(NM)$

[stoic python]

ManifoldCuriosity

[fleet sun]

two matrices can be commuted if they're in a trace

so using that, if A and B are similar, you can go

[wintry steppe]

which I would find weird to define if 2 different endomorphisms can have the same trace

[fleet sun]

$\mbox{tr},A = \mbox{tr}(P^{-1}(BP)) = \mbox{tr}((BP)P^{-1})=\mbox{tr},B$

[stoic python]

ManifoldCuriosity

[fleet sun]

so that's why being similar implies having the same trace

well it's certainly true that having different traces implies NOT being similar

but your question is if having the same trace implies being similar

and that's not clear

or yeah, if two non-similar matrices can have the same trace

I wouldn't rule it out

the trace is just the sum of the diagonal entries

that throws out a lot of information of a matrix

[wintry steppe]

so using that:

say tr(A) = tr(B), then the original statement of "if square matrices having the same trace represent the same endomorphism with different bases" reduces to the question if for all square matrices A and B such that tr(A) = tr(B), there exists a Q such that A = Q^{-1} * B * Q

or to there does not exist A and B such that there doesn't exist such a Q

[wintry steppe]

seemed to me as well, but if the statement is false, then whats the point of defining the trace of a transformation?

[fleet sun]

it has a lot of nice properties

but that may not be one of them

it seems likely the book would have stated it as an if and only if, if it were true

[wintry steppe]

true

thanks tho, I'll check with the professor to be sure

or at least I'll ask her the purpose of defining the trace of a transform if it's not unique to only that transform

[peak plinth]

I’m getting that they r the same

[fleet sun]

consider $\mbox{tr}\begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix} = \mbox{tr}\begin{pmatrix} 1 & 1 \ 1 & 1 \end{pmatrix}$

[stoic python]

ManifoldCuriosity

[fleet sun]

but if you diagonalize the matrix on the right,

it becomes $\begin{pmatrix} 0 & 0 \ 0 & 2 \end{pmatrix}$

[stoic python]

ManifoldCuriosity

[fleet sun]

and that's clearly not similar to the identity matrix

so it's indeed not true that way

[wintry steppe]

ahh okay I see thanks!

I think a general disprove might be to show that tr(A) = tr(B) does not imply det(A)=det(B), and det(A)=det(B) is another property of similar matrices

[wintry steppe]

okay thanks!

[fleet sun]

sure thing

[peak plinth]

Can anyone help with my q above?

[native rampart]

Yea should be same

[fringe burrow]

I need to show that if T is a linear operator on R^n whose matrix A is real symmetric then R^n is direct sum of T’s kernel and image. In order to prove it I distinguished some cases. One of them is where kernel is not empty and I noticed that it is not empty iff there exist some zero eigenvalues of A and then I use the spectral theorem and consider orthonormal basis of R^n where eigenvectors associated to zero eigenvalues span kernel. Is it a good idea ?

[wintry steppe]

Is that an acceptable answer for b??

[zinc timber]

you don't need to do a different case when ker is trivial

[fringe burrow]

Yes so my cases are when ker is trivial and the second when not…

But I asked in particular if my second case looks correct considering zero eigenvalues of A and link between it and ker span

[wintry glen]

Hey all, I'm trying to understand a proof, so I'll write it based on my own understanding and I'd appreciate if someone can tell me if it is right. 😄

Theorem: Let H be a subset of R^n. Then, H = [f : d] if H is a hyperplane.
First, since H is a hyperplane, it is an (n - 1)-dimensional flat, so there exists a vector p so that H_0 = H - p is a subspace of R^n. Now, take any vector outside H_0 and take the component orthogonal to H_0, n, and use it to define a linear functional n • x. Then, [f : 0] is a null space of dimension (n - 1) and since H_0 must lie inside [f : 0], H_0 = [f : 0]. From here, [f : d] = [f : 0] + p = H.

[sharp idol]

Say we want to find the distance between these two points. Although we only know (x,g(x)) and the orientation of the line connecting them.
Can we determine the distance between (x,g(x)) and (x', f(x')) with just this information?

We do also know what the function f(x) is.

[wintry steppe]

I'll wait to ask my question lol

[sharp idol]

Feel free to ask

[wintry steppe]

I'm on part a

So I understand why this works

You got two vectors with the same norm. They're in the same space or sphere

Just in different places and get rotated

Based off the angle between the two vectors

But for b

I'm not quite sure

I gotta turn this into polar cords?

[spare widget]

You need one of the basis vectors to be aligned with a1, a2

e.g. map the basis (1,0) to a / |a|

[wintry steppe]

A1 is cos

[spare widget]

The cos of the angle between these is (1,0) dot a / |a| = a1 / |a|

[wintry steppe]

A2 is sin then right?

[spare widget]

yes

a2 / |a|

but

nah, nvm

I think it's fine

[wintry steppe]

One more question

[spare widget]

final matrix should be something like

1/ |a| [a1 a2; a2 -a1] I think

Try multiplying the matrix I wrote by [a1; a2]

[wintry steppe]

Ight, criver your a boss

[spare widget]

Fixed a sign*

[wintry steppe]

Tyty :)

[rough ember]

Hi everyone!:)

I've been trying to improve my mathematical thinking lately. Today I tried to write my first complete proof. Could you please have a look and tell me what you think of this proof? I would really appreciate it if you could suggest on how I could improve my mathematical reasoning, method of writing proofs, or understanding of linear algebra and mathematics in general.

The proof is based on an explanation from Brilliant https://brilliant.org/wiki/rank/, but I tried to go through the reasoning on my own, trying to be more specific about the procedure.

[spare widget]

It's the intersection of r(t)= (x, g(x)) + t * d and p(s) = (s, f(s))

You have to solve r(t) = p(s)

one of the eq is linear: x + t * dx = s

So you can express s through t

Thus you have to find the roots t of a nonlinear equation

g(x) + t * dy = f(x + t * dx)

And you seem to want the smallest root t > 0

You can solve this through numerical methods

Once you have the intersection the distance is |(s,f(s)) - (x,g(x))|

[hallow edge]

In a 1D vector space (R, scalar field - R), the vectors are basically just scalars or behave like scalars, right?

[spare widget]

It's just R

yes

[wintry steppe]

Criver, I took a look at the solution

Is -a2/|| a |

There's only solutions, no steps

[spare widget]

For sine?

[wintry steppe]

Howd that negative come up? Since -sin relates to a2 in g?

Yes

[spare widget]

Yeah, I told you I fixed a sign

[spare widget]

This

note that -sin = a2/|a| from it

The above matrix is [cos -sin; sin cos]

[spare widget]

Because both a1^2 + (-a2)^2 = |a|^2 and a1^2 + a2^2 = |a|^2

So you cam get it up to sign just from pythagoras

So I additionally used that they want |a| in the first component

Basically I had the choice between

1/|a| [a1 -a2; a2 a1] and 1/|a| [a1 a2; -a2 a1]

But only the second is a rotation

The first is a rotation + reflection

[wintry steppe]

How can you tell?

[spare widget]

det = -|a|?

Wait did I kess up the scaling

nah nvm it's det = -1

You can also try multiplying both by [a1; a2]

Only the second produces the desired result

[wintry steppe]

Allright, thank you very much man

[open locust]

Is there a way to show that a particular definition of a linear transformation (\forall j \in [n] \quad A(e_j) = \sum_{i=1}^n a_{ij} u_j) is an isomorphism without resorting to matrix stuff?

[stoic python]

joesmith1042

[spare widget]

should be a bijection

injection + surjection

[open locust]

Ah, good point.

[spare widget]

That is if you take any 2 elements u,v

T(u) != T(v) if u!= v

also for every w there exists a v such that T(v) = w

[open locust]

Right, thanks.

[wintry steppe]

$\begin{pmatrix} A_{11} & A_{12}\\ A_{21} & A_{22} \end{pmatrix}$

Whats the characteristic polynomial of that block matrix?

[stoic python]

mate

[wintry glen]

Pretty sure that's impossible to answer without further info.

[wintry steppe]

can we express it in terms of char polynomials of A11 A12 A21 and A22?

[spare widget]

If you can bring it to block triangular then det[M] = det[M11]det[M22] afaik

[wintry glen]

Sorry for the ping. I misunderstood the proof. Lemme read it again.

[zinc timber]

you can also use schur complement

[spare widget]

See this: https://en.m.wikipedia.org/wiki/Block_matrix#Block_matrix_determinant

[zinc timber]

@spare widget do you have any good ref for showing sign of eigen values are +ve / -ve (real part) without calculating char poly / RH criterion?

[wintry steppe]

thank you very much!!!!n thats EXACTLY what i needed!! in my case one block is 0!!

thank you so much!

[spare widget]

Not really? What do you need - to figure whether a matrix is indefinite?

Are you trying to figure out whether there are different signs?

[zinc timber]

it's for local stability of a system

I only need to show the signs are -ve

all I have is some identities

[spare widget]

All eigenvalues being negative?

[zinc timber]

their real parts, yes

difficulty is, my matrix is 4x4 and entries are parameters

[spare widget]

I am assuming you know of sylvester's criterion, though idk whether your matrix is hermitian

[zinc timber]

it's not symm

been stuck with this for days

[spare widget]

well this is clearly solvable through ferrari though a pain in the ass

[zinc timber]

I tried

[rough ember]

I was trying to find the mistake for some time, in any case it was a good exercise 😄

[zinc timber]

2 of the 4 terms are bruteforcable

[spare widget]

if you know the eigenvalues are complex then you know they come in conjugate pairs

maybe that can help

[zinc timber]

I have 0 knowledge of the eigen values

that's the issue

also what is this ferrari you were saying?

[spare widget]

But you can find those using ferrari no?

4th degree poly roots

[zinc timber]

I thought you are talking abt faddev-adlfadfja algo

[spare widget]

no, I mean just write the char poly

And then find the roots analytically

[zinc timber]

[spare widget]

It's very painful for 4th degree

[zinc timber]

wish I could

[spare widget]

but there's a formula

You can force maple to do it for you prob

[zinc timber]

unless yr matrix looks like this

already tried sage

[spare widget]

That's alright, just rename the terms

[zinc timber]

finding evs is not the problem, finding their sings is the issue

[spare widget]

idk whether the 0s help

[zinc timber]

in case of det, yes but not the other terms

it's still huge ass long

[spare widget]

I mean, if you have the eigenvalues then you should be able tp figure out the sign

https://encyclopediaofmath.org/wiki/Ferrari_method

[wintry glen]

Nah, the proof seems fine. Looks logical. Though you don't really need to introduce linear transformations at all. You can simply show that Ax_1, ..., Ax_r are linearly independent. Also, I think your R(A)^T and C(A)^T should be R(A^T) and C(A^T), but eh, just minor gripes about presentation. Good work otherwise.

[spare widget]

Sounds like a lit of work, but at least feasible

You should get some inequalities for the terms at the end if you want them negative

[zinc timber]

let me see

though I don't even have my q yet

not in an explicit form

[spare widget]

enjoy: https://en.m.wikipedia.org/wiki/Quartic_function#Solution
😛

https://upload.wikimedia.org/wikipedia/commons/9/99/Quartic_Formula.svg

😂

[zinc timber]

if only I could do that in a reasonable time

I'll ask my supervisor to do this part

[spare widget]

Isn't it usually the reverse

The supervisor outsources tedious work lower down the chain

[zinc timber]

I'll ask him to let me skip that part

It's not for a paper or anything, it's for grading

[open locust]

@spare widget I have another question related to those I've been asking recently

[spare widget]

@zinc timber can this help in your case? https://en.m.wikipedia.org/wiki/Gershgorin_circle_theorem

Or are the bounds too loose?

If Re(aii) + Ri < 0 (for all i) then you can argue that Re(lambda) < 0

[spare widget]

yes?

[rough ember]

You mean that instead of using the concept of linear transformation T that transforms the basis vectors I could just use the matrix A that multiplies that basis vectors, which implies the linear transformation?

As regards R(A)^T, I deliberately used it to emphasize that when you transpose R(A) you get the column space of A. But R(A^T) would actually convey the same meaning, so maybe it would make more sense in terms of style.

[open locust]

@spare widget Never mind actually, I think I "get it."

Oh, well, let me try to ask this anyway

It seems that if we have a square matrix, we can choose any pair of to and from bases for its interpretation.

And in particular, if that matrix is invertible, we can still choose any to and from pair of bases for its interpretation.

[spare widget]

You can, but usually people choose the same

[open locust]

(OK, but I'm trying to be general here)

I'm assuming (1) the linear transformation is from a space into itself, and (2) that we have to and from bases which can be different.

And, if we choose that pair of to and from bases, this invertible matrix represents the identity transformation (I think you may disagree with this)

Before I continue to my question can you tell me if you disagree with that last sentence?

[spare widget]

I would say that people will generally be confused if you say that, but I understand what you mean since I am familiar with your problem.

[open locust]

Okay thanks. 😀

[spare widget]

My suggestion for same space stuff is to define it wrt same bases

Then you can compose it with a change of basis if you so wish

[open locust]

My question is: does this really work with every single possible pair of to and from bases we could come up with? The reason I'm asking is, say we choose the to basis in advance, say it's (\mathcal{B}1 = {u_1, \ldots, u_n}). Then it turns out that if the matrix is columns of numbers (S = [s_1, \ldots, s_n]), the from basis works out to be the vectors (\widetilde{s_1}, \ldots, \widetilde{s_i}) where ([\widetilde{s_i}]{\mathcal{B}_1} = s_1).

[spare widget]

Which matrix

[stoic python]

joesmith1042

[open locust]

The matrix we started with, (S), the square matrix of numbers.

[stoic python]

joesmith1042

[spare widget]

If you have bases E, F with a map T:V -> V and a matrix S corresponding to T with basis E from and basis F to

Then you can always compute the change of basis matrix P from E to F

And compute A = P^{-1} S

which corresponds to T from E to E

[open locust]

Well, you don't have to compute anything right? Isn't the matrix just already the change of basis matrix?

(If it's invertible that is.)

[spare widget]

If T is the identity then yes

If S is just invertible then no

[open locust]

So, I meant to say, we start with an invertible matrix.

So S is invertible, and we just declare to and from bases of vectors.

[spare widget]

That's not enough for it to be the change of basis matrix

Ah

You have S

[open locust]

Then S is the representation of the identity transform in those to and from bases

[spare widget]

And you claim it is the change of basis

You have to then specify at least one of the bases

[open locust]

Yes, I'm saying, start with a square matrix of numbers S that is invertible, and declare that the interpretation of it is with respect to a particular to and from basis

[open locust]

Yes, this is my question. Why is that?

Like, why can't it be arbitrary for both bases?

[spare widget]

Because imagine S - rotation

you have to fix the starting point from which this rotation starts to give you the second basis

Or the end point

You can of course take by default E = canonical basis

Then it's fine

[open locust]

Well, I guess my point is, say I have S. We know that the columns of S are the representations of the from basis vectors in the to basis, right?

Like, Su_1 represented in terms of v1, ..., vn is the first column, and so on.

[spare widget]

ok, but then you have to define the to basis

[open locust]

Why? I mean, why can't it just be "v1, ..., vn". Why must it be more specific than that?

[spare widget]

Because the change of basis matrix is relative

Again say I pick a rotation

[open locust]

I mean, mathematically speaking

[spare widget]

I can now pick any basis

Then rotate it

And end up with another

And the change of basis matrix is R

it's like saying: I give you the distance between two points - tell me what the points are

You cannot

[open locust]

So you're saying that S, an invertible matrix, may not be the change of basis for some pairs of to and from bases.

[spare widget]

No, it will be for infinitely many

There are infinitely many bases that are related through it

I can take any basis, then apply S, get another basis, and then these two are related by S

[open locust]

Ah, but we do have to sort of "derive" one of the bases.

from the other (and the matrix).

[spare widget]

yes

But you need at least one

[open locust]

I still don't really understand why that's the case, is there a way to prove that with the formulas?

[spare widget]

basis + change of basis matrix -> other basis

[open locust]

I mean, the formulas for linear transformations, matrix multiplication, etc.

[spare widget]

$\vec{f}i = \sum_j s{ji} \vec{e_j}$

[open locust]

Ah yes.

[spare widget]

You need either the es or fs except for s

In order to find the other

[stoic python]

criver

[open locust]

Oh, so you're saying that just putting any e's and any f's doesn't mean that this equation works out

[spare widget]

it does

[open locust]

But aren't the e's and f's just symbols? Or you're saying, they have to be like, specific numbers.

[spare widget]

No es and fs are vectors

[open locust]

My bad, I meant to say vectors of numbers.

[spare widget]

Not necessarily of numbers

[open locust]

But doesn't that equation say that the equation works for any symbols e's and f's?

[spare widget]

they are vectors

Functions can be vectors

[open locust]

OK. But in general aren't those just letters?

I guess this is the source of my confusion

[spare widget]

they are elements from a vector space

we denote them with some letters

[open locust]

So, the act of "choosing" what the e_i's are specifically is going beyond just labeling them with letters, and getting more specific?

And that's why it doesn't really work for any choice of e_i's and f_i's?

[spare widget]

idk what you mean by doesn't really work

if you are given the es and S, you can find the corresponding fs

[open locust]

Well you said that if you chose say the f_i's, then that implies the e_i's be something specific. They can't be just anything anymore.

[spare widget]

If you are given the fs and S you can find the correspondimg es

[open locust]

But it looks in the equation like you could have anything for the f_i's and e_i's since they're all just abstract symbols

[spare widget]

If you have the es and fs you can find S

[open locust]

Do you see what my confusion is?

[spare widget]

no, I do not

[open locust]

Like, if we look at that equation, e_i and f_i are just there, which makes me think they can be anything

[spare widget]

If it's easier for you think of es and fs as elements from R^n

After all finite dim real space are isomorphic to R^n

[open locust]

Yeah, so I think the problem is the difference between 1) labelling something abstractly and 2) keeping that label but assuming a more specific representation

Like, if I have e_i there, it seems e_i can be anything, and same with f_i.

[spare widget]

not anything, vectors

[open locust]

OK, any vector.

[spare widget]

Linearly independent ones too

And spanning the vector space

And S has to be invertible in that case

[open locust]

See if I look at the equation x = y + z, it seems like this works for "any" y, z, and x, except of course we know it doesn't, because we're trained of think of them as numbers

So we naturally think "well of course this wouldn't work for any x, y, and z, obviously given y and z there's only one x that would work for it"

[spare widget]

If it's easier for you think if es and fs as vectors from R^n

[open locust]

Yeah, that might help, do you see how it's confusing me now?

Like, in all the proofs I've been doing and everything, we always "set" the basis to be "e1, ..., en"

[spare widget]

expressed wrt the canonical basis (1,0,...0,0), ..., (0,0,...,0,1)

[open locust]

But in my head, that's like just saying "x" which can be anything at all

So in a way it's like we have fixed absolutely nothing.

[spare widget]

Setting the basis in practice

[open locust]

So I think I need to think more like, if I ever write e1, ..., en, that those are specific vectors

Does that make sense?

[spare widget]

would involve you specifying some number repr

For the es say in R^n

[open locust]

Right. Or, functions, like you said before. Like x^2 or something.

(the usual polynomial basis)

[spare widget]

well let's keep to R^n for simplicity

[open locust]

Well I get it now, if this is indeed the issue I'm having

Like, it seems obvious now

[spare widget]

Then e1,...,en are just arrays of numbers

[open locust]

Yeah.

[spare widget]

That are lin indep

And span R^n

Then the system arising is something like

F = S E

[open locust]

The last thing I have to do (I think) on this topic is to prove the statement I quoted to you at the beginning of my question, I just copied it out of a matrix algebra book

[spare widget]

Some of those may need to be transposed, gotta expand the sum to figure it out

[open locust]

Right.

[open locust]

Yes, and doing that would "prove" the statement I was just mentioning.

That book doesn't have a proof of it, it just quotes it.

Maybe I should just take it as a given and move on from this topic, or do you think I should try to prove it actually?

[spare widget]

Try to, it will probably help with understanding

E.g.

Assume e1,...,en are given

And S is given

Rewrite

[open locust]

Here it is: it's the sentence beginning "Conversely"

[spare widget]

$\vec{f}i = \sum_j S{ij} \vec{e}_j$

[stoic python]

criver

[spare widget]

If you stack those I think you get something like F^T = S E

[open locust]

Yes.

[spare widget]

$\begin{bmatrix} \vec{f}_1 & \ldots & \vec{f}_n\end{bmatrix} = \begin{bmatrix} \vec{e}_1 & \ldots & \vec{e}_n\end{bmatrix} S$

I think this is better

[stoic python]

criver

[open locust]

I don't actually fully follow that.

[spare widget]

let me think whether this is correct

[open locust]

How did you go from the summations to that?

[zinc timber]

skill

[spare widget]

$F_{ij} = \sum_k E_{ik}S_{kj} \implies \ \vec{f}j = \sum_k S{kj} \vec{e}_k$

Did I mess this up?

[open locust]

What is F? The matrix of f_i's?

[stoic python]

criver

[open locust]

And E is the matrix of e_i's?

[spare widget]

F is made of the columns f, E of columns e

[open locust]

The second line is correct

[spare widget]

if you take i = 1,...,n then you get the vectors

[spare widget]

Did gershgorin help?

[zinc timber]

tried

thing is parameters are really arbitrary

[spare widget]

But you should get some inequalities from it

[open locust]

The first formula is correct as well criver.

(just finished checking it)

[spare widget]

max_i Re(a_ii) + Ri < 0 -> Re(lambda) < 0

[spare widget]

Well that's your change of basis if you have e and f as numbers

The implies should read iff btw

It's just rewriting the vector lin comb as a matrix equation

[open locust]

Yes I was going to mention the iff

[spare widget]

I think they pick the canonical basis

e.g. E = I

[open locust]

Oh they did!?

No wonder I couldn't figure the stupid thing out

This is one of those "matrix algebra" books so it's super computations focused...

[spare widget]

F = ES = IS = S

[open locust]

Ugh.

Thank you. That's the missing part I couldn't get

Alright, well that pretty much buttons this topic up. Again I really appreciate your help on all this.

[spare widget]

That or F=I, I am not sure from their notation

[open locust]

It's OK, you helped me figure out the key piece, which is how one basis sort of implies the other.

Thank you again!

[spare widget]

I think I am wrong reading what you posted

They assume B1 is defined

Yes, they assume B1 to be defined and S to be given

Then they argue they can find E

[open locust]

Eek, okay.

[spare widget]

which is clear

[open locust]

I'm not quite sure how they got those specific numbers, but I think I could use the formulas you just derived above to find that, right?

[spare widget]

What they call \tilde{s} are really the basis vectors of B afaik

[spare widget]

For R^n you could yes

[open locust]

Well they're the representations of those basis vectors

[spare widget]

For abstract stuff

[open locust]

Er, no, you're right, the tilde s's are actually the basis vectors.

[spare widget]

It just means that you are given f and S, and you can find e from that

[rough ember]

Thank you very much for your answer:)

[open locust]

Why does it make a difference whether or not it's R^n or not?

These things are just coefficients of vectors, right?

[spare widget]

because for R^n I can write es and fs as array of numbers

[open locust]

Like, s_i is coefficients of \tilde{s_i} in a particular basis

[spare widget]

And then form F = ES

If f and e are not arrays of numbers

Then I have to work with

$\vec{f}_i = \sum_j \vec{e}j S{ji}$

[stoic python]

criver

[spare widget]

Alternatively:

[open locust]

But how does that prevent you from getting their result there?

[spare widget]

$\vec{e}j = \sum_i (S^{-1}){ij}\vec{f}_i$

[stoic python]

criver

[open locust]

Right.

[spare widget]

It doesn't

But for instance if f is a polynomial

I can't just make a matrix F and E

I can only do this by mapping to R^n through an isomorphism

Ley's try an example

f1 = 1, f2 = x

[open locust]

OK.

[spare widget]

s11 = 1, s12 = 2, s21 = -2, s22 = 1

Then you can find polynomials e1, e2

Or vice versa so you won't have to invert s

[spare widget]

e.g. using this

[open locust]

I see.

OK well, that all makes sense.

I think you just proved their result, right?

[spare widget]

If you construct a map between polys and R^n

Then you can use 1 -> (1,0,...,0), x -> (0,1,0,..,0) etc

[open locust]

Like, all we needed was s_ij and f_j's, and we get e_i's

[spare widget]

Then you can work in R^n instead

[open locust]

OK. Is that what they implicitly did there?

[spare widget]

no

[open locust]

Or are they just stating the final, put-together result?

[spare widget]

I am just saying how you can go to R^n

Do your computations with arrays

Then return to your other fin dim real space

[open locust]

Ah OK. I'll keep that in mind. I actually like sticking with the original spaces better personally but this is a cool thing to know about.

[spare widget]

They are stating this afaik

that you can find e, give f and S

[open locust]

I think so too, I think they're just sort of finalizing it a bit too much for my tastes. Really appreciate you walking me through the details on that.

[spare widget]

\tilde{s} being the es

[open locust]

Yes.

Alright, I'm going to take a break from math now, thank you again criver!

[spare widget]

you're welcome

[spare widget]

Can you prove your matrix is diagonally dominant?

And then show that all the diagonal elements are negative

well not quite I guess

since it's complex

you need $-Re(a_{ii}) > \sum_{j\ne i} |a_{ij}|$ after all

[stoic python]

criver

[spare widget]

which looks a bit like diagonal dominance

At least it gives you inequalities that are sufficient conditions for your matrix having negative real parts of the eigenvalues.

The bounds are fairly loose though probably.

[wintry glen]

Sorry for late reply, but essentially, yeah. Since you are working with the underlying matrix A anyways, you don't need to talk about the transformation, since the vectors Ax_1, ..., Ax_r are already linearly independent vectors in the column space of A by the definition of column space.

[zinc timber]

nope

one condition for RH is to show det >0

I can use schur complement to break it up

even still no help

[spare widget]

rh?)

What I wrote are sufficient conditions, so you can get inequalities for your variables

[zinc timber]

Routh Hurwitz

[spare widget]

Beyond that ferrari will give you the exact bounds for the variable ranges for which this holds

or are you trying to prove that the eigenvalues have negative real parts regardless of what params you pick?

[zinc timber]

there's one restriction

R0>1 where R0 is

[spare widget]

You can use this to further intersect it with thr gershgorin disks

may yield a tighter bound

[wintry steppe]

A is a singular transformation. is Ker A an eigenspace of 0?

[leaden tide]

Yes

[wintry steppe]

thanks :)

[open locust]

@spare widget By the way, the thing we discussed before about given a matrix and one basis and knowing we want the matrix to be a change of basis matrix, the other basis must be derived from the matrix and the known basis. Does that apply to matrix representations of isomorphisms in general? (i.e. not just to matrix representations of the identity transform)

My guess is no.

[wintry steppe]

Prove $\alpha$ is the only eigenvalue of $\alpha I$.

Let $\beta$ be an eigenvalue of $\alpha I$. Then there exists a vector $x \neq 0$ such that $(\alpha I)x = \beta x$, ie. $\alpha x = \beta x$, which implies $\alpha = \beta$.

Is this proof ok? I guess no, doesnt sound good to me.

[stoic python]

mate

[leaden tide]

Yeah it's fine

You also have that (αI-β) is non-invertible (zero, in fact) iff α=β

[wintry steppe]

thanks :)

[spare widget]

I do not understand the question

[open locust]

Well the determination of the other basis thing was when we were talking about a change of basis matrix

But now I'm wondering if the same requirement applies to invertible matrices in general.

[spare widget]

what requirement

[open locust]

I.e. for a given invertible matrix to be representative of an isomorphism, and we choose one of the bases, is the other basis determined by those two choices? Or can we choose it to be anything we want.

[spare widget]

Yes it is determined

[open locust]

Really. Even though we're no longer being as strict on what the matrix must represent?

[spare widget]

$\vec{f}i = \sum_j S{ji} \vec{e}_j$

[stoic python]

criver

[spare widget]

It's just linear combinations

[open locust]

Hm, I see what you're saying.

[spare widget]

It's clear that given the es and S you can compute the fs

And vice versa using S^{-1}

[open locust]

It's just interesting that this requirement looks like, almost exactly the same as what we were looking at before.

Before, we wanted S to represent the identity transform. Now, we want S to be any isomorphism

I thought the requirement would be a little looser.

Do you see what I mean?

[spare widget]

What requirement

Requirement for what too

[open locust]

Well, we start with a matrix S. We demand that this matrix represent the identity transformation. We fix one basis. Then, that demand, the numbers in the matrix, and the basis give us the other basis.

Now, we start with a matrix S. We demand that this matrix be some isomorphism. We fix one basis.

How much freedom do we have in choosing the other basis? Is it fully determined, just like in the first case?

[spare widget]

it's fully determined, I don't see the difference between the two cases

[open locust]

Then there probably isn't a difference 😀 I just wanted to make sure. I'm writing up a summary of a lot of this and this question popped into my mind.

Thanks for checking it.

[spare widget]

You start with an invertible matrix, pick a basis, you can get another basis

whether you interpret this as an identity, isomorphism or w/e doesn't matter

[open locust]

I see.

[spare widget]

Note that there are two ways to get another basis

Using S or using S^{-1}

depending on the choice the from/to flips

[open locust]

Right. 🙂

[spare widget]

and nite that fi = sumj ej Sji is not the same as fi = sumj ej (S^{-1})_ji

So it matters whether you are given the to or the from basis

[open locust]

Right.

Thank you again criver.

[spare widget]

Also something to keep in mind

If you have f_i = sum_j e_j S_ji

This expresses f in terms of the e

So S actually transform vectors from F to E and nit vice versa

[open locust]

Yes.

[spare widget]

[v]_E = S [v]_F

that's why these vectors are typically termed contravariant

[open locust]

I see, hadn't heard that term before.

[spare widget]

since they transform in the opposite way of the basis

e.g
F = E S vs v_E = S v_F

covectors transform in the opposite way (like the basis)

Your covectors will be from V^* (the dual)

[open locust]

Appreciate you mentioning that, I am familiar with the dual

Back to writing up this summary 😀

[peak plinth]

Did I do this right

Cause I know dim < 6 if there is one L.D. vector

But with span, I think you can have both?

My assumption was that if basis spans R^6 then dim has to be R^6 since it’s a compilation of only the L.I. vectors, since well, the redundant L.D. vectors would have to be removed

But idk

I’m not sure if I’m going down the right direction or if I’m overly complicating this

[spare widget]

span(v1,...,v6) = S -> dim(S) <= 6

To be precise dim S = max # of linearly indepent vectors from v1,...,v6

[fleet sun]

you can take spans of linearly dependent sets

like span{ (1,0), (0,1), (1,1), (2,3), (-5, 1/2) } = R^2

so yeah, dim(S) <= 6

[peak plinth]

It’s not asking if it’s less than or equal. It’s asking if it’s less than 6

Less than or equal would be true, but if it’s just less than, then it’s false given my explanation

Or so I think?

Am I going down the right direction

[fleet sun]

why couldn't it be equal to 6?

you're not given {v1, ..., v6} is linearly dependent are you?

yes, your proof / answer is fine

[spare widget]

What I wrote is the correct thing, what's written in b) is wrong unless v1,...,v6 are linearly dependent

[fleet sun]

no

he's saying the claim that dim(S) < 6 is false

the writeup could be more polished but I think the right idea is there

not saying what you said was wrong, their answer agrees with it

unless the inequality is really meant to be a <=, in which case the answer should have been True

looks like it was printed as < and then lines drawn under it

[peak plinth]

Yeah

I underlined it lol

My bad

[fleet sun]

Looks like a "just do it" proof

[mossy coral]

Prove that if T : V →W is an isomorphism then T^-1 is also an isomorphism.

can someone help me with this. I know I have to prove inverse T is bijective, but I am not sure how I would do that for a generic linear transformation

[silk valley]

It's literally telling you what to do

[fleet sun]

the inverse of a bijection is always a bijection - that's set theory

you have to show the inverse is linear

[mossy coral]

oh ok, ty

[mossy coral]

for this, dont I write the basis a as a linear combination of B?

oh wait im bad at arithmetic

[wintry glen]

So, my textbook says that this is "clear", which I didn't think it was, so I tried to prove it. Is this right?

[sudden ruin]

Anyone know what's the theorem that let's you know a column of a matrix is a generating set?

[zinc timber]

generating set of what?

col space? that's the definition of col space

[languid sphinx]

outer product of two vectors is always rank 1 matrix right

Wait no

nvm let me reread up on outer products

[languid sphinx]

Oh wait yeah it is max rank 1

[lavish jewel]

yeah rank one

you can write it as u v^T. then it becomes clear that the columns are all scaled copies of u

[rugged wren]

I send it first one wrong but my question is second one. If you locked at I would be so glad

[lavish jewel]

try playing around with a few special cases

[opal aurora]

trying, i think it's C

would that be correct

[lavish jewel]

that sounds about right, what was your reasoning though?

[opal aurora]

subspace if null, not otherwise

[lavish jewel]

yeah, that seems right

[opal aurora]

coolsies ty

[wintry steppe]

could somebody help prove that linear transformations $A$ and $\alpha A (\alpha \in \bR)$ have the same eigenvalues?

[stoic python]

mate

[dusky epoch]

good luck proving a false statement!

unless α = 1, A and αA will in general have different eigenvalues.

@wintry steppe

[wintry steppe]

oh thank you, i wanted to say that the set of all eigenvalues of alphaA is alpha*(the set of all eigenvalues of A). i apologize

[dusky epoch]

okay then it becomes very simple

for α=0 it's obvious (what are the eigenvalues of the zero map?) and for α!=0 it suffices to show one direction (λ is an eigenvalue of A => αλ is an eigenvalue of αA), and then convince yourself that what you just proved can be applied to αA scaled by 1/α

[wintry steppe]

thank you, Ann.

if $\beta$ is an eigenvalue of $A$, then
$(\alpha A)x =\alpha A(x) = \alpha (\beta x) = (\alpha \beta) x$. thats one direction, right?

basically we just use $(\alpha A)x = \alpha A(x)$ to prove both directions?

[stoic python]

mate

[dusky epoch]

if you wish to word it that way sure

[warped flame]

hi

if I have for example

$e^{\begin{bmatrix}
t & 0 & 0\
0 & t & 0\
0 & 0 & t
\end{bmatrix}}$

[stoic python]

dervaa_

[warped flame]

is this equal to

[spare widget]

expand in a series

[warped flame]

$\begin{bmatrix}
e^{t} & 0 & 0\
0 & e^{t} & 0\
0 & 0 & e^{t}
\end{bmatrix}$

[stoic python]

dervaa_

[spare widget]

$e^x = \sum_{k=0}^{\infty} \frac{x^k}{k!}$

[stoic python]

criver

[warped flame]

Yea, I know it can be done like that

[spare widget]

Now substitute x with a matrix

[warped flame]

but in textbook its like this and I am asking and wondering if this some mistake

or is it possible to just "input" e into matrix like tomorrow doesn't exist hahah

[spare widget]

It's correct

expand it and see

you get a matrix of the form

$\begin{bmatrix} \sum_{k=0}^{\infty}\frac{t^k}{k!} & 0 & 0 \ 0 & \sum_{k=0}^{\infty}\frac{t^k}{k!} & 0 \ 0 & 0 & \sum_{k=0}^{\infty}\frac{t^k}{k!} \end{bmatrix}$

[warped flame]

aha, okey thanks

[stoic python]

criver

[spare widget]

You turn the sums into e^t and done

Here's a more interesting variant of this - compute e^A, where A = P^{-1} D P

[wintry steppe]

thank you very much, Ann! have a great day

[crude star]

can anyone tell me what a' and b' are?

does that mean subsets?

i have to show that these statements are equivalent

[haughty berry]

They’re anything

That’s the point of the question….

[fleet sun]

a, a', b, and b' are elements

you're showing the definition of ordered pair in (1) makes sense

[hallow plover]

can anyone help me prove this

[stuck tendon]

Show that adj(A^-1)adjA = I

[hallow plover]

i got it thanks

[zinc timber]

I'm guessing they are defining ordered tuple here?

[crude star]

yes

i got it thank you

[queen pagoda]

can anyone help me solving b)? i did part a), but tbh have no clue solving b)

[wintry steppe]

If map A from V to V is bijective, then V is the direct sum of its kernel and image?

[wintry steppe]

yes, but in a rather trivial way

the kernel is 0 and the image is V

thank you

[gray pond]

Does anybody understand this argument? I’m totally confused

All the a’s, lambdas, and l’s are positive. Lambda* is always less than corresponding lambda and for commodity i the numerator will have l_i - δ for some positive δ less than l_i

I don’t get how we are getting that the average is greater than the smallest ratio

In case it helps this is coming from a matrix equation $$\Lambda = \Lambda A + L $$ where $\Lambda$ and $L$ are vectors and A is a matrix. Imagine we decrease the ith entry of L. He’s trying to show that the ith lambda decreases proportionately the most $\frac{\lambda_i^*}{\lambda_i}$

[stoic python]

casualpolemic

[gray pond]

Is this proof just wrong?

[dull rose]

I have a question and i can provide more context to the question if needed

Does it matter which basis we take for v1?

Like can I instead use [-1, 0, 1] (matrix with blue underline) as my v1?

[static fjord]

Does it matter if the highlighted integers are equal to 1? Or can they equal another number

[sullen escarp]

They can be different so long as your solution set is still the span of [-2, 1, 0] and [-1, 0, 1]

So it'd be fine if you wrote your solution set as s[-4, 2, 0] + t[-3, 0, 3]

[safe schooner]

heyy i have a doubt can someone helpp

[dusky epoch]

what is your doubt? @safe schooner

[safe schooner]

disss

[dusky epoch]

i can see the image just fine

in it there is a problem (Example 12.15) with a partially worked-out solution

can you please be more specific as to where your doubt is?

[rancid fiber]

A bit confused. When solving for Eigenvalues of a square matrix, why is it that sometimes the formula is A - lambda*identity and sometimes it's lambda*identity - A?

[dusky epoch]

doesn't matter

$\det(\lambda I - A)$ and $\det(A - \lambda I)$ only differ by a constant multiplier of $(-1)^n$, so finding the zeros of either one will give you the same result

[stoic python]

Ann

[rancid fiber]

that makes it a lot clearer. Thank you!

[zinc timber]

though when talking about characteristic polynomial, prefer |xI-A| over |A-xI|

[stuck tendon]

why?

[quartz compass]

|xI-A| if you want it monic, |A-xI| if you don't want to have to write out a bunch of negative signs

[stuck tendon]

oh, I see. thank you

[gray pond]

I think I figured this out

[wintry steppe]

hello

[runic musk]

i know you can pre-multiply a matrix by its inverse to make it identity, can you post-multiply by its inverse to also make it the identity?

[restive raft]

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

[stoic python]

holazach

[runic musk]

thankyou 🙂

[sharp idol]

Here it is @peak marsh

[pure path]

I need some help. Regarding Gaussian Elimination. I need to find the quadratic polynomial that is passing through three points using Gaussian Elimination. (1,4), (2,0), (3,12)

That must mean that I have
x1=1, y1=4
x2=2, y2=0
x3=3, y3=12

But how do I move on from here?

[quartz compass]

you make a system of equations with y=ax^2+bx+c by plugging in the 3 points to this to get 3 equations

then you solve for the 3 variables a,b,c kind of backwards of what you might expect

[lavish jewel]

building up on what mero says, note that the variables are a,b,c. with that in mind, inspection of the RHS of y=ax^2+bx+c should let you see this looks the same as a dot product

namely, [x^2, x, 1] dot [a b c]

then consider that matrix vector products Mv can be written in terms of dot products between the rows of the matrix M and the vector v

[pure path]

Thanks guys!

[quartz compass]

I guess as a bit of handy trivia, the general matrix for doing it this way with a polynomial is called the Vandermonde matrix

[subtle skiff]

Sup guys. Going over some review for a test tomorrow. Is this question even possible? I feel like I am missing a vector here

Because for linear combinations in R2, we are given 3 vectors

for R3 shouldnt we be given 4?

[lavish jewel]

hmm?

to have a basis for all of R2, you need 2 linearly independent vectors

for all of R3, you'd need 3

but part of the question is whether you can express the given vectors in the basis S at all, i.e. whether they are in the span of the given set

[jaunty thorn]

notice the "if possible" bit

[subtle skiff]

Ya, thats making me second guess myself. Rarely is it so easy as to just say "nope"

Thank you

[lavish jewel]

it's not "so easy" though, since you have to show why

[jaunty thorn]

np, I get you lol, I'm always super paranoid when it says if possible and just think I'm being dumb

[subtle skiff]

So just say that there isnt enough vectors to span R3, so a linear combo isnt possible

maybe?

[lavish jewel]

no

that's not right

[quartz compass]

one way to check is take the cross product of the vectors in S, then dot it with z and v and see if you get 0 or not

[lavish jewel]

the vector could still be in the span of the set

mero's approach is very clever if you understand it

if not, gauss jordan will never let you down

[subtle skiff]

it has a determinant that != 0

[subtle skiff]

is that the same as this

[quartz compass]

yeah

[jaunty thorn]

you need n vectors in Rn to be able to write EVERY vector in Rn as a sum of those vectors, you can have n-k vectors which can write a subset of Rn

[lavish jewel]

checking the determinant of the augmented matrix is equivalent to what mero said, yes

since it means adding the vector to the set increases its rank

i.e. the system is inconsistent

[quartz compass]

only real reason why I didn't suggest determinant is cause there are 2 vectors, z and v to check, which means two 3x3 determinants. I thought it'd be easier to do just the cross product, so that way you're only doing 2 dot products instead

[jaunty thorn]

finding the determinant is equal to zero is equivalent to showing an inconsistency though^^

just for clarification

[quartz compass]

basically just reusing part of the determinant being evaluated rather than do it from scratch twice

[subtle skiff]

That makes a lot of sense

I appreciate you guys help with this 🙂

[twilit minnow]

doing this practice final

and im unsure about the rank of B

would it be 2?

[restive raft]

well it inputs vectors of dimension 4 so it has 4 columns, since its null space has dimension 2 by rank-nullity its rank is 2, so yeah.

[twilit minnow]

how about the possible m x n dimensions

[spare widget]

wdym possible mxn dim?

Ah, I got it

V is 4x2

Since vectors in the nullspace are vectors from {v : Bv = 0} and V is a basis for those, then they are in R^4

This means that n = 4

Since the product Bv is not defined otherwise

As far as m goes, the rank is 2, so m>=2

Note R(B^T) + N(B) = 4

And B needs to have 2 linearly independent rows and cols

[twilit minnow]

makes sense

[twilit minnow]

in that case

could i make

[spare widget]

No, this is rank 3

If you want to make an explicit matrix you must make sure that BV = 0

[peak plinth]

I need an example of this

Idk seems impossible

How do you make rank 2 with no 0 in your rref matrix lol

[hard drum]

Well consider matrices not in rref

[peak plinth]

Still impossible

[hard drum]

Note it's equivalent to finding a set of 4 vectors in R^4 whose span has dimension 2

And that columns may repeat, for example

[peak plinth]

Yeah but no zero entry

Can you just give me an example

Of such a matrix

[spare widget]

Matrix with 2 linearly dependent columns (with no zero components)

Just pick two lin indep vectors, and then build 2 linear combinations of those (resulting in no zero components, e.g. you can just take copies of the two vectors)

[hard drum]

I'm a bit confused why it seems impossible though. Would you say the same if the question said 1 or 3?

[twilit minnow]

yep, my bad

didnt see that

i get it now

[spare widget]

It's obviously not, they're just trying vectors that have 0 components

[hard drum]

I meant why it seems impossible to them

Since otherwise it seems they think all matrices are 0 or rank 4 or smth but yes

[spare widget]

Because they are picking vectors with 0 components, I don't think there's another option that yields these 0 cases

It's easy to pick orthogonal vectors if one plugs 0 components

[twilit minnow]

i found P = 3+2x-x^2

dont know how to do a b c d though

[spare widget]

Construct an isomorphism to R^3

then represent B2 and B3 as linear combinations of the vectors of B1

(You can do the latter even without the isomorphism)

[twilit minnow]

how would you do it without isomorphism

[spare widget]

1 = 1 * 1 + 0 * x + 0 * x^2

x-1 = -1 * 1 + 1 * x + 0 * x^2

(x-1)(x-2) = 2 * 1 + -3 * x + 1 * x^2

[hidden berry]

hello guys please someone can help me with this problem