linear-algebra

[boreal wadi]

But this is also an inner product:

$\forall \bm{x} = [x_1,x_2] \in \mathbb{R}^2 , \bm{y} = [y_1, y_2] \in \mathbb{R}^2
\\
\langle \bm{x}, \bm{y} \rangle = x_1y_1 - ( x_1y_2 + x_2y_1) + 2(x_2y_2)$

[stoic python]

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

[boreal wadi]

Exercise 3.1 from this book asks me to prove it: https://mml-book.github.io

Any suggestions on something that I can read that will explain why the second formula is an inner product? Do i need to prove that it’s bilinear?

[zinc timber]

try to prove all the axioms of an inner product

or you can show the bilinear form is symmetric and positive definite

@boreal wadi

[boreal wadi]

Thanks

[plain robin]

Can there be a Linear Transformation that goes from, for example R^6 to R^9?

[tribal willow]

yes i believe so

e.g. something like (x1,x2,x3,x4,x5,x6) |-> (x1, x2, x3, x4, x5, x6, 0, 0, 0)

you would just need to define a function which does so

[plain robin]

I thougjt Linear Transformations dont increase dimensions

[tribal willow]

alternatively, something like f(x) = (x,x) will map from V to V^2

a linear functional does not increase dimension

a linear functional is a linear transformation which maps onto itself

e.g. T: V -> V

essentially the dimension of the codomain can be larger than the dim of the domain

but the dim of the image cannot be larger

[plain robin]

In that case is the dim(Im) langer tho

Or is it because its not a Linear functional

[tribal willow]

in this case, dim(codomain) is larger

image =/= codomain

but dim(im) is not larger

[plain robin]

Ah okay

[winter harbor]

Yes, in fact, the whole mathematical formulation of the standard model of particles is based on the representation theory of lie algebras. There's also a lot of lie theory machinery used in the study of quantization of classical systems. A good book to learn a bit about these topics after learning some linear algebra would be "Quantum Theory for Mathematicians" by Brian C. Hall.

[tribal willow]

this is so sexy. thank you so much

what timing too

i was just looking for a replacement for mcintyre

[winter harbor]

Wdym by timing?

[winter harbor]

Hmmm idk mcintyre

[tribal willow]

mcintyre is a qm textbook

which one of the professors at my uni described as

[winter harbor]

Yeah, idk I personally like Brian C. Hall tho

I do not like reading physics textbooks written with the physicist in mind

[tribal willow]

agreed

well

taylor classical mechanics was an enjoyable read

mcintyre is a painful read

so i stopped

[winter harbor]

So Hall doss a nice thing by presenting a sort of exposition to QM is more friendly to the mathematician.

It in fact introduces a lot of the relevant notion from classical mechanics through the textbook.

Which is good

[tribal willow]

music to my ears

[fleet sun]

I've been trying to learn from that book. It seems good but be warned, if you're not up on your functional analysis, it will be hard

as soon as they start talking about domains of self adjoint bounded operators, which is early, it's rough going (for me anyway)

[zinc timber]

which book?

[tribal willow]

this one

@zinc timber

why are cyclic subspaces called cyclic

[wintry steppe]

you're cycling through the powers of an operator acting on a vector

[tribal willow]

ah ok

[fresh obsidian]

How I tell if these have inverses? (And find them if they do)

[zinc timber]

not a place to ask but it has something to do with Bezout

[fresh obsidian]

where shall I ask it then? fwiw my linear algebra prof. is going into this stuff bcz he seems to think crypto falls under the same umbrella :P

[zinc timber]

#groups-rings-fields

He's not wrong

(idk if 1001 is prime of not)

[tribal willow]

,w is 1001 prime

[stoic python]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by anamono#1010

[tribal willow]

damn

1001 seems like the kinda number to be prime

[zinc timber]

wait it's divisible by 11

should have seen that

[fresh obsidian]

yeah lol

[fresh obsidian]

thx! that's just the hint I needed ;)

[errant mist]

Can someone tell me which types of matrices are corresponding to the case where the geometric multiplicities are not equal to the algebraic multiplicities for the matrix? I believe I heard something about Jordan normal form matrices displaying this behavior for example

Thats where you have eigenvalues on the diagonal and at least one 1 right above the diagonal I believe if I am not mistaken?

[wintry steppe]

the geometric multiplicity of an eigenvalue is the number of jordan blocks corresponding to an eigenvalue, and so the geometric multiplicity equals the algebraic multiplicity for that eigenvalue if and only if the jordan blocks corresponding to the eigenvalue are all trivial (i.e. 1x1 jordan blocks consisting of just the eigenvalue). you should compare this with the statement "a matrix is diagonalizable if and only if its characteristic polynomial splits and all the geometric and algebraic multiplicities coincide")

so non-trivial jordan form matrices are exactly the matrices for which the geometric multiplicities do not agree with the algebraic multiplicities (at least for matrices with splitting characteristic polynomials - those which actually have jordan forms)

[bright shale]

Hello everyone, i know the picture has german language in it but i will try to translate it, maybe someone can help our group

we need to find the matrix M of the projection on the line with and angle of 1.9696 [Rad] to the first axis (x-axis) using the mirror matrix for the x axis

[errant mist]

@wintry steppe would the following matrix be such a case I presume $$\begin{equation*}
\begin{pmatrix}
2 & 1 & 0 & 0 \
0 & 2 & 0 & 0 \
0 & 0 & 3 & 0 \
0 & 0 & 0 & 3 \
\end{pmatrix}
\end{equation*}$$

[stoic python]

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

[wintry steppe]

yeah

[errant mist]

Here the eigenvalues give a algebrac multiplicity of 4

[wintry steppe]

the geometric multiplicity of 2 is 1, but the algebraic multiplicity of 2 is 2

[errant mist]

while the geometric multiplicity is not equal

[wintry steppe]

be careful to say geometric/algebraic multiplicity of what eigenvalue

[errant mist]

oh, right. What do you mean by splitting characteristic polynomials exactly?

[wintry steppe]

you may write them as a product of linear factors

anyone know LA reading groups or sth like that

for example, x^2 + 1 does not split over R, but it does split over C as (x - i)(x + i)

every matrix whose characteristic polynomial splits has a jordan form

think of it as the next best thing after diagonalization (indeed, diagonalization is a special case of jcf)

[errant mist]

@wintry steppe ok , I think I got that part thanks. Could you explain more what you mean by trivial Jordan blocks?

[wintry steppe]

1 x 1 jordan blocks

[true musk]

I am stuck on proving that the algebraic multiplicity of an Markov matrix's eigenvalue 1 is one

[errant mist]

Isnt that just the diagonal though if the dimension is 1?

[wintry steppe]

if the dimension of what?

[errant mist]

the Jordan blocks being 1x1

[wintry steppe]

if you have a jordan block that's 1x1 it's a diagonal entry of the matrix (with no other entries in its row or column)

#chill

ah yes very cool matrix manipulation

[errant mist]

yeah, thats what I thought. Like a 1x1 matrix if you will

[wintry steppe]

"jordan form" means a block diagonal matrix consisting of jordan blocks

all matrices whose characteristic polynomials split are similar to jordan form matrices (which are unique up to permuting the blocks)

[errant mist]

hmm, interesting

[wintry steppe]

so although you may not be able to diagonalize a matrix whose characteristic polynomial splits, you can always put it into jordan form

[errant mist]

And then the trivial form is just when you have the "regular" case if you will of diagonal eigenvalues on the diagonal meaning the alg. mul. = geo. mul

[wintry steppe]

<@&268886789983436800>

[frosty vapor]

banned

[wintry steppe]

thanks metal

[wintry steppe]

pretty much

but note that you could have a non-trivial jordan block and a trivial jordan block for the same eigenvalue, which still wouldn't be diagonalizable

[frosty vapor]

also banned

[wintry steppe]

for example, $$\begin{pmatrix} 2 & 1 & 0 & 0 \ 0 & 2 & 0 & 0 \ 0 & 0 & 2 & 0 \ 0 & 0 & 0 & 3 \end{pmatrix}$$ has two jordan blocks corresponding to the eigenvalue $2$, namely $$\begin{pmatrix} 2 & 1 \ 0 & 2 \end{pmatrix}, \begin{pmatrix} 2 \end{pmatrix}$$

[stoic python]

TTerra

[wintry steppe]

so although you have a 2 on the diagonal (being the only one in its row and column), you still aren't diagonalizable

[errant mist]

So, its good to think almost diagonalizable for these Jordan forms

[wintry steppe]

yes!

that's exactly how you should think of them

[errant mist]

good , Im learning

[wintry steppe]

the idea for constructing jordan forms follows that

a matrix fails to be diagonalizable because the eigenspaces are "too small" (captured by "diagonalizable if and only if algebraic, geometric multiplicities coincide"), so to get the jordan form we "enlarge" the eigenspaces to what are called "generalized eigenspaces"

[errant mist]

What about this matrix where I have $$
\begin{equation*}
\begin{pmatrix}
0 & 1 & 0 & 0 \
0 & 0 & 1 & 0 \
0 & 0 & 0 & 1 \
0 & 0 & 0 & 0\
\end{pmatrix}
\end{equation*} $$

[stoic python]

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

[wintry steppe]

and those happen to be large enough, in the sense that our underlying vector space is a direct sum of them (and so we can get block diagonal matrix forms where the blocks come from bases for the generalized eigenspaces)

then one proves that generalized eigenspaces have nice bases in which you get the jordan blocks you see

that's the general story, at a basic level

[errant mist]

I will pause and read slowly what you wrote thanks)

[wintry steppe]

the more high-brow view is that it follows from the decomposition theorem for finitely-generated modules over principal ideal domain (if you have an operator T on a finite-dimensional vector space over a field k, then you get a f.g. k[x]-module to which the big theorem applies), but you don't need to think about this if it's your first time

[wintry steppe]

this is a 4 x 4 jordan block with eigenvalue 0

[errant mist]

we get a singular eigenvalue of 0 and if we calculate the null space we get a multiplicity of 4. How do you count the Jordan blocks?

[wintry steppe]

it has one jordan block

the 1's tell you what the jordan blocks are

[little crater]

Does someone have a nice set of theormes that discusses pivot rows and pivot columns regarding what it means about a linear system Ax=b, Ax=0, and maybe stuff regarding span and linear (in)dependennce of a set of vectors made up of the columns of A?

[wintry steppe]

in general, a jordan block is a matrix of this form:

that's how to count it off of a matrix in jordan form

if you aren't in jordan form, then counting the jordan blocks is a little trickier

generally, you get the jordan blocks for a specific eigenvalue by constructing a certain kind of basis for the corresponding generalized eigenspace

[errant mist]

So if there is a corresponding 1 above the diagonal we count it as a jordan block if it is in this form?

[wintry steppe]

i should probably say now: if $T\colon V \to V$ is a linear operator on a finite dimensional vector space $V$ over a field $F$, then the generalized eigenspace corresponding to a $\lambda \in F$ is the subspace $$G_\lambda = {v \in V : (T - \lambda I)^k v = 0 \text{ for some } k \in \bZ_{\geq 0}}$$

[stoic python]

TTerra

[errant mist]

yes, we compute and evaluate the multplicity by finding the null space

[wintry steppe]

this generalized eigenspace has dimension precisely the algebraic multiplicity of \lambda, and you can find a basis of it consisting of (geo. mult.) many distinct cycles {v, Tv, ..., T^{k-1}v}, where v is an eigenvector with eigenvalue \lambda. if you compute how the operator T looks in such a basis, you'll see that you get a block diagonal consisting of jordan blocks with eigenvalue \lambda

that's exactly how jordan form works

(proving you can actually do all this is the hard part)

[wintry steppe]

it's not quite so simple, at least at first glance. G_\lambda looks instead like the union of a bunch of null spaces, right? and the union of subspaces need not be a subspace

but it does turn out that, if the algebraic multiplicity of \lambda is m, say, then G_\lambda is the kernel of (T - \lambda I)^m

which tells you how to find the generalized eigenspaces

[errant mist]

Well I computed the kernel for the eigenvalues of 2 and 3 for the first matrix I posted.

$$\begin{equation*}
ker(\mathbf{A}- 2 \mathbf{I}) = \begin{pmatrix}
-1 \
0 \
0\
0
\end{pmatrix},
ker(\mathbf{A}- 3 \mathbf{I}) = \begin{pmatrix}
0 \
0 \
0 \
1 \
\end{pmatrix} , \begin{pmatrix}
0 \
0 \
1 \
0 \
\end{pmatrix}
\end{equation*}$$

[stoic python]

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

[errant mist]

so the first eigenvalue 2 has geometric multiplicity equal to 1 and eigenvalue 3 has geometric multiplicity equal to 2

[wintry steppe]

yes

and note that there's one jordan block corresponding to 2, and two jordan blocks corresponding to 3

[errant mist]

So, we get these three eigenvectors returned for the 4 eigenvalues.

yeah, got it!

[slim geyser]

I have this exercise on bilinear/quadratic form and the Sylvester theorem. Q(u,v)=uˆ2-vˆ2 and g((u1,u2),(v1,v2)) = u1v1-u2v2. So that means that we have (p,q)=(1,1) but then my prof says that two vectors, a and b form a Sylvester basis for Q iff a1ˆ2-a2ˆ2=1 , b1ˆ2-b2ˆ2=-1 and (and this is the part i don't understand why) a1b1-a2b2=0.

So my question is why does a1b1 - a2b2 have to be equal to 0?

[wintry steppe]

Someone help me pleaseee

[little crater]

#calculus or #precalculus would probably be the channel you are looking for

more than likely #calculus

[wintry steppe]

Thank you

[plain robin]

,tex Why cant i replace here rk $(f \circ g) \le min(rk (f), rk (g))$ the less than equal with a = ?

[stoic python]

Herakles

[plain robin]

why does it have to be ≤

[wet stratus]

I assume rk is the rank of a linear map and f, g are linear maps?

let f, g be two linear maps such that f "eliminates half the space V and leaves the other half fixed" and g does the "opposite, ie leaves the stuff f eliminates and eliminates the rest". then fg is the 0 map but each has rank n/2

[zealous onyx]

Let's say A and B are quadratic matrices such that AB=BA. Also there exist natural numbers m and k such that A^m=0 and B^k=0. prove that det(A+B)=0

Hello

how would you prove this?

[slim geyser]

Hint: calculate the determinant of (A+B)^s where s:=max{m,k}

Using the binomial expansion

[zealous onyx]

ooof

[slim geyser]

also important to note is that det(Aˆk)=det(A)ˆk

[slim geyser]

yep, thats a big one

but everything cancels out

[zealous onyx]

but how do I find max m k?

[slim geyser]

not important, just assume m>=k or vis versa

in either case the proof is the same

[zealous onyx]

I will try

can't promise anything

but I will

[zealous onyx]

@slim geyser

Does that look convincing?

[frozen trellis]

I generally get to multiply by 5 and divide by 8

but how do the r and a switch

[tribal willow]

this does not look like linear algebra

i think you're looking for #prealg-and-algebra

[wet stratus]

You cannot divide by A^i. First A is a matrix so you can't divide. And second A is not even invertible

[zealous onyx]

A^m is zero

it's given in the problem

should I still care about the division?

[wet stratus]

You cannot divide by a matrix

It just doesn't make sense

[slim geyser]

Yes but A is a matrix, not a scalar

[wet stratus]

You could multiply by A^-1 if A was invertible. But it isn't

[slim geyser]

Also the whole proof is in that binomial expansion. In the expansion use the AB=BA to always get something in the form of A^mB^something to make that always equal to zero since A^m =0

[zealous onyx]

yeah I get that

[slim geyser]

After that you will have that tou have a series of 0+0+...

[wet stratus]

But not every term will be of that form A^mB^something

(A+B)^2=A^2+2AB+B^2=2AB for m=k=2. But now?

[vestal magnet]

Hey guys, any idea regarding this?

[slim geyser]

Can you explain the notation?

[zealous onyx]

also why max?

[zealous onyx]

here they are equal

k and m

[wet stratus]

So?

[vestal magnet]

Need to prove that if S is none negative then Phi is also none Negative

[wet stratus]

I assume non negative means something like xSx>=0 ?

[slim geyser]

What is T*?

[tribal willow]

i assume the dual space of T

[slim geyser]

Ill try to find my proof, wait a min

[tribal willow]

sorry not dual space

dual transform

[vestal magnet]

Yes indeed

[wintry steppe]

what does "non-negative" mean?

[slim geyser]

all eigenvalues are non negative i think

[wet stratus]

<TST^*w, w>=<S(T^*w),T^*w>=<Sv,v> >= 0 with the substitution T^*w=v. I assume probably like that

And with the inner product <,>. Or rather inner products as there are two different ones used here

First in W, the other ones in V

[slim geyser]

so instead of taking the maximum, we actually take the sum of the two orders

[zealous onyx]

Thank you so much!

[slim geyser]

np, lmk if u dont understand sm

[wintry steppe]

isnt $E_{rs} = I + I_{rs} +I_{sr}-I_{rr}-I_{ss}$

[stoic python]

Jester

[wintry steppe]

cant see shi

[fickle mica]

https://gyazo.com/69037342f24a1aefae0a11e8a0f061ef

can u see this one

[wintry steppe]

what is the q?

[fickle mica]

how did he get the basis

for that

cuz i dont undestand where the basis comes from

[fickle mica]

Nvm i got it

[wintry steppe]

^

[slim geyser]

Its a seperate condition from what i've read, not something you dedjce from the fiest 2 because the first two can both be equal to + or - 1

[distant schooner]

from spivak's calculus on manifolds, 1-7... the solution i saw was much more complicated

am i missing something

[wintry steppe]

are you proving (b)?

your argument looks fine. you might want to point out after the last line that every element of R^n is of the form Tv, which gives that T^{-1} is inner product preserving

[little crater]

Can anyone comment on my solution for the following question?

My attempted Solution:
B => A:
because if the columns of A only have the trivial solution for Ax=b (definition of linearly independent), then applying theorem 1.6 ("The structure of solution sets") would tell us that Ax=b has at most 1 solution because Ax=b is made up of a particular solution vector, p, plus any vector solution of the homogenous equation Ax=0, v_h. Since Ax = 0 only has the trivial solution and p + v_h = p+ 0 = p, then b only has at most one solution.

A =>B:
because if we let b = 0 and know that for any matrix A that A(0)= 0, then there can be no more solutions to Ax=0 given (a) is true that any b in R^m has at most one solution. Since Ax=0 has one solution, then by definition the columns of A are linearly independent.

C=>A:
because by theorem 1.2 (“Existence and Uniqueness Theorem”) if you have no free variables which is equivalent to saying there is a pivot in every column of A, then you have 1 unique solution if the system is consistent or no solution at all.

A => C:
because have having at most 1 of the equation Ax=b for each b in R^m, this means you would have no free variables in your columns of A and that means you have a pivot in each column of A.

Since B=>A, A=>B, C=>A, A=>C, then A <=> B <=> C

Question

2 theorems mentioned

definition from book

[dusky epoch]

whats your definition of an affine subset?

uh huh

which direction are you struggling to prove in char 2?

alright, so take a nonempty set A in a space V over a field of characteristic 2

take an arbitrary point u ∈ A

we wish to show A-u is a subspace

that is, we wish to show that for every v, w ∈ A-u and λ ∈ F we have v + λw ∈ A - u

so we have v + u in A and w + u in A and our goal is v + λw + u in A

ok wait too many symbols lets prove closure under addition and closure under scaling separately

we want v+w+u in A

...hm

[vestal magnet]

Can you assume that you could use the inner product?
I mean, I do understand why this - <TST^*w, w>=<S(T^*w),T^*w>=<Sv,v> >= 0 . is true considered you have a given inner product, but can you just define one?

[vestal magnet]

or am I missing something else?

[wet stratus]

I assumed there was an inner product already defined on W. Not sure how you would define T^* in the first place otherwise

[vestal magnet]

Sorry but I’m confused with this part, I know S(v) is non-negative but why the inner product is non-negative

[wet stratus]

I thought per assumption we have <Sv, v> >= 0 for all v

otherwise you have to say what S being non-negative means

[vestal magnet]

Our definition for that was,
S=S^* , and S eigenvalues are real non-negative. also theres exist another transformation T where S = T*T

[wet stratus]

hmm

well I guess then we have to prove something completely different

[vestal magnet]

Yeah cuz from here, S is the 0 transformation

Sorry for didnt mentioning that earlier

[wet stratus]

well first $(TST^)^ = T^{**}S^* T^= T ST^$.

[stoic python]

Denascite

[wet stratus]

let $S=V^V$. Then $TST^=TV^VT^=(VT^*)^VT^=Z^Z$ with $Z=VT^$

[stoic python]

Denascite

[wet stratus]

not sure about the eigenvalues right now. the problem is that speaking about matrices, all eigenvalues positive does not mean that the matrix is positive definite in general

but well your definition for some reason includes S=S^* which in matrix terms would be called symmetric (at least over R) and which is enough for matrices

so you can work with that. not sure yet how tho

[wet stratus]

\begin{align*} \lambda ||v|| = \lambda \langle v, v\rangle = \langle v, \lambda v\rangle = \langle v, TST^* v\rangle = \langle v, Z^* Z v\rangle \= \langle Zv, Zv\rangle = ||Zv|| \geq 0 \end{align*} for an eigenvalue $\lambda$ with eigenvector $v$ of $TST^*$.

[stoic python]

Denascite

[wet stratus]

you might need to worry about conjugation if you work over C. so first show that lambda is actually real

[vestal magnet]

I'll dive deeper into that soon, sorry had a class

Thanks a lot 🙂

[tribal moss]

is a linear map from V to V always surjective? (so linear maps from a vectorspace to the same vectorspace)?

[dusky epoch]

no

[golden pelican]

Hi I'll be in my freshman year (high school) next year and i was trying a couple of questions to get my basics sorted
There are around 45 questions and I doubt in most of em
Could anyone help me out?
I sorted my way out but still have doubt in 30 questions
So, if you could help me, I'd really be grateful 🙏
It's really important for me🙏

They're basic HS questions
I just wanna verify

[tribal moss]

Then I'm wondering, how the one before last equivalence holds. I know that if it's not injective then it's not an iso but I don't know why it doesn't hold to the other direction (this is a proof to show that you can compute eigenvalue calculating the determinant)

[dusky epoch]

a map from a vector space to itself is either both inj and surj or neither

[tribal moss]

Ah yes now I get it, I found the theorem, thanks

[molten pilot]

ok, first thing I'm noticing is that you're proving more then you need: you only need to prove A=>B, B=>C, C=>A (can you see why?)
second, A=>B and B=>A and C=>A look fine to me but A=>C it seems to me you're just restating the problem

hmm

[molten pilot]

Thm 2 is written a bit vaguely: does "when there are no free variables" means "if and only if there are no free variables"?

[wintry steppe]

Why is this true?

How can I show this?

[tacit pelican]

why does it say "isomorfism" tho

is it not "isomorphism"

or is that also used

[tribal moss]

Nooo I'm sure you're right it's with ph, I just translated it quickly from my language where it is with f my bad

[tacit pelican]

oh right i see

[short eagle]

how do i find the base change matrix, when i am given B and C.

B is {u1 , u2, u3} and C is {v1, v2, v3}

from C to B or even the other way around

[wintry steppe]

Follow the definition of change of basis

Write each vector in one basis as a linear combination of the others

[short eagle]

So by this do you mean i should make a Matrix consisting of [u1|u2|u3] and then find the basis for that matrix

haha of course you meant, finding the basis of the space spanned by the set of vectors

so i found the basis {[1,0,1,1]', [0,1,1,2]', [0,0,-1,0]'}, what do you mean

by 'as a linear combination of the others'?

[grave kettle]

how do row operations preserve pivot columns

[grave kettle]

<@&286206848099549185>

[tranquil steeple]

google tridiagonal symmetrization

[wintry steppe]

Thanks

[molten pilot]

what do you mean by that?

you mean they remain linearly independent?

[grave kettle]

for any matrix, the pivot columns remain unchanged when row operations are applied

[molten pilot]

each matrix has a unique reduced row echelon form

[grave kettle]

i was trying to understand why row equivalent row echelon forms have the same pivot columns

[grave kettle]

ig we should prove this first?

[molten pilot]

pivot columns for matrices that're not in echelon form are defined to be the pivot columns of their respective reduced row echelon form

[molten pilot]

there's nothing to prove, it's by definition

[grave kettle]

definition of what

[molten pilot]

pivot columns for matrices that're not in echelon form

[grave kettle]

i thought pivot columns are defined to be columns which contain a pivot (in ref)

[molten pilot]

oh, I think I get what you're asking now

[molten pilot]

row operations are reversible

[grave kettle]

yea

so there doesn’t exist a row operation that changes the pivot columns

idk what to do

contradiction?

[molten pilot]

so you can be sure that applying a row operation to a matrix doesn't change its rref

[grave kettle]

now that you put it that way, i can’t immediately see how ;-;

row equivalent matrices have the same rref

[molten pilot]

since the original matrix and the matrix with row operations on it have the same rref they then have the same pivotal columns

does that make sense now?

[molten pilot]

ye

[grave kettle]

this doesn’t really make sense still

[molten pilot]

I don't really know what you don't get

[grave kettle]

so (row equivalent matrices have the same rref) -> ??? -> same pivot columns

[molten pilot]

suppose $M$ is a matrix, the columns which have a pivotal 1 in $rref(M)$ are the pivotal columns of $M$

[stoic python]

DarQ

[molten pilot]

that's how we define pivotal columns for matrices that are not already in rref

[grave kettle]

i don’t understand how they remain unchanged through row operations then

[molten pilot]

let $M$ be a matrix and $M'$ be the $M$ matrix with some row operations applied to it. We know that $rref(M)=rref(M')$

[stoic python]

DarQ

[molten pilot]

we know that because rref(M) is unique

no matter what row operations you apply to M the final result will always be rref(M)

are you asking why that is?

[grave kettle]

i’m still confused about why M, M’ have the same pivot columns 😦

[molten pilot]

please be more specific on which step you don't understand

[grave kettle]

(same rref) -> ??? -> (same pivot cols)

i don’t see how the uniqueness of rref is related

and by pivot cols, i admit to being a bit unclear
according to the definition you’ve given, the pivot columns of a non-ref matrix is the same as its rref form

[molten pilot]

.

[grave kettle]

i think i’m looking for justification in your definition

[molten pilot]

you're looking for justification for why they're defined like that?

[grave kettle]

yes

it’s more of a theorem than a definition imo

it’s kinda both

[molten pilot]

[molten pilot]

this definition is mathematician's attempt for generalizing the idea of pivot columns

pivot columns don't really make sense for matrices that aren't in rref

[grave kettle]

yea i get it

i’m just looking for proof

i think it can be justified

[molten pilot]

it's justified because rref is unique

no matter what row operations you use you'll always arrive at the same pivotal columns

[grave kettle]

hm so let me rephrase

suppose M_1, …, M_n are row equivalent matrices in rref

according to your definition, they should have the same pivot columns

but pivot columns can already be seen from any ref matrix

[molten pilot]

yeah

[grave kettle]

like if M_1 follows from your definition, the columns that we can see are pivotal should match the columns that follow from the definition

i’m just looking to verify this for every ref matrix i guess

[molten pilot]

is M_1 in rref?

[grave kettle]

no

just ref

[molten pilot]

what's ref?

[grave kettle]

row echelon form

[molten pilot]

what's the difference between row reduced echelon form and row echelon form?

[grave kettle]

rref is reduced row echelon..?

rref is apparently stricter than ref

[molten pilot]

I'm sorry but I'm not really sure how to help you

[grave kettle]

;-;

do you know row echelon form?

[wintry steppe]

lol darq

[molten pilot]

I tried, I swear

[grave kettle]

i am dumb

question: how to verify all row-equivalent row echelon form matrices have the same pivot columns

[wintry steppe]

just prove it lol

verify that row operations don't change the pivots

[grave kettle]

‘just prove it lol’

👍

[grave kettle]

yea this is what i’m stuck on lol

[halcyon spindle]

hint: row operation.

[grave kettle]

???

0 idea what you’re insinuating

<@&286206848099549185>

[wintry steppe]

okay

you have three row operations

they all have extremely explicit forms

check that they don't change where the pivots are in a ref matrix

what part are you stuck on?

[grave kettle]

here

i don’t even know how to begin

[wintry steppe]

why not try a few examples

[grave kettle]

mainly i don’t know how to represent a general ref matrix

[wintry steppe]

fair

that's probably the hardest part

if you just want to convince yourself, try a few different examples

[grave kettle]

i’m fairly convinced

[wintry steppe]

maybe a really careful proof would do something like induction on the size of the matrix or something

[grave kettle]

suppose it has n columns and k (k ≤ n) pivot columns?

or something

not much idea

[gray dust]

@uncut dust #prealg-and-algebra

also why the rush?

[grave kettle]

how to do ;-;

[wintry steppe]

honestly

for something like this, just convincing yourself is enough. writing an actual, rigorous, full proof of this probably isn't gonna be very insightful

if you've convinced yourself with a few examples that you could do so, that's probably good enough

... unless this is homework

[dusky epoch]

maybe begin by writing out the defn of a pivot column

and of a pivot

[grave kettle]

pivot of a matrix in ref form is the first non zero entry of a row

pivot column is a column containing a pivot

[wintry steppe]

I have a generic question about orthogonality

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

"In mathematics, an orthogonal polynomial sequence is a family of polynomials such that any two different polynomials in the sequence are orthogonal to each other under some inner product."

At the moment I'm trying to understand legendre polynomials

so if I'm interpreting this correctly i could take the polynomial associated with n=2 and relate it to the polynomial associated with n = anything else by an inner product and it would be orthogonal?

https://www.youtube.com/watch?v=k7RM-ot2NWY&list=PL0-GT3co4r2y2YErbmuJw2L5tW4Ew2O5B&index=3

[rustic salmon]

oh thanks

[wintry steppe]

np

[wet stratus]

so if I'm interpreting this correctly i could take the polynomial associated with n=2 and relate it to the polynomial associated with n = anything else by an inner product and it would be orthogonal?
yes. but you have to use the correct inner product. this is not true for some arbitrary inner product, it's just true for the one used to construct these polynomials. or a multiple of that if you don't care if the polynomials have norm 1

[grave kettle]

i’m also concerned about the motivation of defining free/basic variables as those corresponding to non-pivot/pivot columns

like is there some sort of ‘proof’ behind this as well

in fact, it’s this that led to what i was asking

perhaps ‘for a system Ax = b, the i-th column of A is a non-pivot column iff there exists a solution with x_i = r for all r ∈ R’

(x = [x_1 … x_n]^T)

[wintry steppe]

Gotcha, well at least my rough intution is pointed in the right direction. Thanks @wet stratus !

[grave kettle]

<@&286206848099549185>

[wintry steppe]

👀

no idea good luck

[grave kettle]

[vestal magnet]

Hey guys, any help regarding this?

[wintry steppe]

is there a nxn matrix where AB=I but BA!=I?

[wintry steppe]

no

it's a good exercise in basic linear algebra to figure out why

the only technical fact you need is rank-nullity

[vestal magnet]

Any idea regarding this guy? 🙏🏻

[wintry steppe]

do you know any way to show that operators have roots? i think there's a long list of equivalent statements in axler you could use

but i dont know what rank nullity is

[fringe fjord]

Google it and see if the statement sounds like something you know with different words.

[wintry steppe]

do you know at least that "injective" and "surjective" are the same for square matrices?

i know what injective/surjective functions are.

i asked if you knew this specific fact about them

if you know it, you can prove the thing you asked about

matricies can be injective?

i mean the matrices, thinking of them as linear maps R^n -> R^n (or replace R with whatever field)

[wintry steppe]

you prove this fact (appropriately stated) by using the rank-nullity formula. if you're uncomfortable with that, see tropo's message

is there no easy way to prove it?

rank nullity is easy

but let me think

[vestal magnet]

I need to find another operator with the sqrt of the same eigenvalue?

[wintry steppe]

maybe a row reduction argument works

AB = I implies A has full rank, so you can do a bunch of row operations to A to get the identity I (why?). that is, there's a matrix E which is a product of elementary row matrices that gives EA = I. now show that AB = I and EA = I imply E = B

and then that proves what you want

[vestal magnet]

not speaking to me, right?

[wintry steppe]

i am not.

[wintry steppe]

@wintry steppe is this okay? i'm willing to clarify anything written here

[grave kettle]

out of curiosity, any necessary/sufficient conditions for matrices A and B to commute?

[wintry steppe]

why would AB=I and EA=I imply E=B

[wintry steppe]

at least for diagonalizable matrices, commuting is equivalent to the matrices sharing a basis of eigenvectors. if your matrices have splitting characteristic polynomials, then you might be able to drop "diagonalizable" and put "generalized" before the word eigenvectors. only thing off the top of my head

[wintry steppe]

E = EI = E(AB) = (EA)B = IB = B

ah

I see thank you

I guess I will fully understand when I learn ranks and stuff

[grave kettle]

any idea?

[wintry steppe]

rank-nullity is usually how this is proved for linear operators on vector spaces, but it's nice that it wasn't really needed here

it's probably hiding behind the scenes somehow lol

[little crater]

Is this a valid solution

[grave kettle]

looks good

just have to show au - v ≠ 0 with u, v being independent

shouldn’t be hard

[little crater]

True I was just thinking about that

[wintry steppe]

the classic rubber duck moment

[grave kettle]

also, where this problem from

curious

[little crater]

my problem set

umm so either from

or the prof made it up

or found it somewhere

[grave kettle]

everyone uses that book lmao

[grave kettle]

wait, you have the fifth edition pdf?

[little crater]

yes?

have

[grave kettle]

i only have the fourth edition

is it possible you could share it to me

[little crater]

just googled [book name].pdf

[little crater]

||https://github.com/vzardlloo/trans/blob/master/origin/Linear Algebra and Its Applications 5th Edition.pdf||

not originally where I found it but seems someone posted it on github

seems to be complete

[wintry steppe]

Can someome help me, can someone explain to me how spatial covariance matrix is made for 2D arrays

Like a gridded dataset plotted on world map

[fair wren]

let f be an endomorphism from a vector space E of finite dimension such that f^(2) = -id; the second composition of f, assuming f has no real eigenvalues.
prove that the dimension of E is even

https://cdn.discordapp.com/attachments/903481109420048385/978819204276961360/324736074750623748.png

is this valid for an answer to this question ?

[slow scroll]

no, that is not the char poly

[fair wren]

i picked for example dimE = 3

[slow scroll]

cubics always have a real root, so if lambda^3 + 1 is the char poly, then f has a real eigenvalue. Besides, since f^2 + id = 0, lambda^2 + 1 divides the char poly, but lambda^2 + 1 does not divide lambda^3 + 1

[fair wren]

i don't understand

whats the charpoly then

also how does f² + id = 0 relate to the charpoly

[slow scroll]

its a 2n degree polynomial for some positive integer n

[fair wren]

what?

[slow scroll]

that's basically what you have to prove. the degree of the char poly is equal to the dimension of E

[slow scroll]

do you know about the minimal polynomial?

[fair wren]

yes i need to prove that

[fair wren]

no

[slow scroll]

probably no need to worry about it actually. My reasoning there is slightly wrong

[wintry steppe]

one way to do this is to show that such an f defines a complex vector space structure on E

i think you can guess how a complex scalar a + bi should act on a vector v in E

[fair wren]

dont overcomplicate it

[wintry steppe]

this is not overcomplicating it

[fair wren]

it is

• i haven't went thru complex vector space structure

[wintry steppe]

it just means a vector space over C

[fair wren]

ye still idk

[wintry steppe]

ah

take determinants

[fair wren]

on what?

[wintry steppe]

f^2 = -id

[fair wren]

how do u even do a determinant on that?

[wintry steppe]

they're linear operators on finite dimensional vector spaces, so the determinants are defined

defined by picking a matrix representation and proving that determinant is preserved up to similarity, if you like

the specifics don't matter for this problem

all you need to know is that all of the properties of the determinant you know for matrices carry over to operators on finite dimensional vector spaces

[fair wren]

how do u pick a matrix representing the operator?

also i'd like if u do the charpoly method

[wintry steppe]

by choosing a basis. you don't need to for this problem

[fair wren]

much better

[fair wren]

we have that in the later question

they asked to prove that for any x in E, (x, f(x)) is stable by f

[wintry steppe]

that has nothing to do with choosing a basis and taking matrix representations of operators on E

[fair wren]

it does

[wintry steppe]

it's not like it's the reason you can pick matrix representations of operators

[fair wren]

since there exists (e_1, .., e_n) such that (e_1, f(e_1), ..., e_n, f(e_n)) forms a basis for E

[wintry steppe]

that's not what i'm talking about

i'm talking about the fact that if you have a linear map T: V -> W of finite-dimensional vector spaces, and if you choose bases of V and W, then you can represent T by a matrix in terms of these bases

[fair wren]

idk what u talkin about either tho

[wintry steppe]

if you take V = W and you take the bases to be the same, then you get a square matrix. you can take the determinant of this, and this is how the determinant of an operator on a finite dimensional vector space is defined

allowing you to take determinants of both sides of f^2 = -id

you must have seen this if you've defined the characteristic polynomial of operators on vector spaces

[fair wren]

no i haven't seen it

[wintry steppe]

then how did you define the characteristic polynomial

[fair wren]

det(A - lamda I_n)

but when i try to extract the A, the operator is already very explicitly expressed

so i can extract basis and stuff

but here its confusing for me

[wintry steppe]

so if you want to use the characteristic polynomial of f (which is defined on an abstract vector space E, hence f is not necessarily a matrix), you need to know how the determinant of operators E -> E is defined

and it is defined how i wrote above

the determinant of an operator T: V -> V is defined, in general, by choosing a basis of V and taking the determinant of the corresponding matrix representation of T

you do not need to worry about finding a specific matrix representation. you only need to know that facts such as "det(AB) = det(A)det(B)" and "det (cA) = c^n det A" and "det I = 1" carry over

[fair wren]

what basis do we take for E then

[wintry steppe]

any

but you do not have to. you only need to use the properties of the determinant that i wrote

[fair wren]

which

[wintry steppe]

the ones that i wrote

[fair wren]

on what f² = -id?

[wintry steppe]

yes

for example, det(f^2) = det(f)^2

[fair wren]

but f is an operator not a matrix

[wintry steppe]

i've told you a few times now how to define the determinant of an operator

for all intents and purposes you may pretend that everything is a matrix

[fair wren]

ok

but det(f^2) = det(f(f(v)) != det (f * f)

[wintry steppe]

f^2 is the composition f o f, and composition corresponds to matrix multiplication

[fair wren]

i see

det(f)^2 = - det(id) = -1 hm

[wintry steppe]

det(-id) is not necessarily -det(id)

you need to be careful with pulling scalars out of the determinant

[fair wren]

det(cA)

[wintry steppe]

equals?

[fair wren]

= c^n detA?

[wintry steppe]

c^n det A

[fair wren]

ye

what is n ?

[wintry steppe]

the dimension of the vector space

i should have included that

[fair wren]

lol

how do we know the dimension when we are already looking for it

[wintry steppe]

well

the vector space has a dimension

[fair wren]

ofc

[wintry steppe]

it's not like we know it, all we know is that it's an integer n that must satisfy det(cA) = c^n det(A)

[fair wren]

yes

[wintry steppe]

it is using this property that we can figure out that the dimension is even

back to the problem

[fair wren]

hm

[wintry steppe]

so det(f)^2 = det(-id), which equals...?

[fair wren]

det(f)² = (-1)^n det(id) = (-1)^n

[wintry steppe]

correct

now why should this imply that n is even?

hint: compare the signs of the two side. what if n is odd?

[fair wren]

but det(f) can be complex?

[wintry steppe]

can you post a screenshot of the original problem?

[fair wren]

No i don't have the paper on me

i only remember the question

why

[fair wren]

the determinants is always a real number?

@wintry steppe r u still here?

[wintry steppe]

i am thinking

what field is E a vector space over?

[fair wren]

the field

[wintry steppe]

which

[fair wren]

yes

[wintry steppe]

is it the real numbers or something else?

[fair wren]

i mean since f has no real eigenvalues

then it can't be over the real

[wintry steppe]

every single version of this problem i've seen very explicitly states that E is a real vector space

and note that an operator f with f^2 = -id can't have real eigenvalues anyways, since if f(v) = cv for a non-zero v, then (c^2 + 1)v = 0, which implies that c^2 + 1 = 0, not true for any real number

so that assumption is redundant

[fair wren]

yes i did this in the question before

[fair wren]

which?

[wintry steppe]

"f has no real eigenvalues"

[fair wren]

yes but can be used ig

[wintry steppe]

sure

i'm just wondering about the specific statement of the problem as it appeared on your question sheet

[fair wren]

wym?

btw i have a question

[wintry steppe]

i would like to compare

[fair wren]

the determinant is always a real number?

i thought this is a yes/no question

[wintry steppe]

would you give me more time to type?

[fair wren]

ye sure

[wintry steppe]

the problem is false if you assume E is a complex vector space. take E = C^3 and T: E -> E given by T(x, y, z) = (ix, iy, iz). this satisfies T^2 = -id, but the dimension of E (as a complex vector space) is 3, which is odd. therefore, you need to ask that E is a real vector space.

i had assumed in the solution that i was guiding you through earlier with determinants that E was real. the determinant of an operator on real vector spaces is always real, but the determinant of an operator on a complex vector space may be a complex number (not real)

so now, if you assume E is a real vector space, you have det(f)^2 = (-1)^n, and you can actually talk about the sign of the left-hand side

[fair wren]

dim_C (E) = 3 but also dim_R(E) = 6

[wintry steppe]

sure, but if you assume E is a complex vector space, it having even real dimension is trivial

there is no problem to solve there

the interesting problem statement (and the one that's more likely to have been assigned to you) is the one that if E is a real vector space with an operator f with f^2 = -id, then E has even dimension

[fair wren]

ye i see

[fair wren]

E being attributed "real" wasn't mentioned

[wintry steppe]

it should have been

[wintry steppe]

it should have been

[fair wren]

yes but u can't deduce that from f having no real eigenvalues?

[wintry steppe]

did you read anything i wrote above?

[fair wren]

yes

[wintry steppe]

i deduced that from the problem being false (or trivial, if you ask for its real dimension) if E is complex

[fair wren]

no im saying can't u deduce that E is a real vector space from the fact the f has no real eigenvalues?

[wintry steppe]

from only that fact? no

[fair wren]

Is there something else?

[wintry steppe]

what do you mean?

[fair wren]

nvm

so E being a real vector space is a mandatory

[wintry steppe]

yes

you must assume this at the start

and once you do this, you can extend the scalar multiplication action of R on E to one of C on E (i.e. giving E a complex vector space structure) by making i act by f. that's how you should think of this problem

so now that we have that E is a real vector space, we know that det(f) is a real number, and det(f)^2 = (-1)^n, where n is the dimension of E. what does this say about n?

this finishes the problme

[fair wren]

yes i see so n must be even

[wintry steppe]

exactly

[fair wren]

aight i see good thing

btw

[fair wren]

determinant can be applied to vectors?

[wintry steppe]

nope

unless you want to call endomorphisms vectors (since they're elements of a vector space)

[fair wren]

assuming dimE = 2n , and there exist vectors (e_1, .., e_n) such that (e_1, f(e_1), ..., e_n, f(e_n)) forms a basis for E, what is the matrix representation of f in that basis

[wintry steppe]

you tell me

[fair wren]

we have M(f^2) = M(-id) = (-1 0 .. 0) / ( 0 -1 0 .. 0) / ... / (0 .. -1)

@wintry steppe is this correct tho ^

[wintry steppe]

well sure, but this is the matrix representation of f^2, not of f

[fair wren]

Yes sec

M(f)² = (-1 0 .. 0) / ( 0 -1 0 .. 0) / ... / (0 .. -1)

[wintry steppe]

you should compute M(f) by seeing how f acts on (e_1, f(e_1), ..., e_n, f(e_n))

finding M(f) from a matrix equation sounds a little hard

[fair wren]

f((e_1, f(e_1), ..., e_n, f(e_n))) = (f(e_1), -e_1, ..., f(e_n), -e_n)

right?

[wintry steppe]

maybe i should have said "how f acts on the elements of (...)"

you have computed that correctly

[fair wren]

(x, f(x)) is stable by f

then M(f)² = -M(f) ig

[wintry steppe]

i don't see how that first fact helps us and i don't see how the second follows

[fair wren]

first fact : (x, f(x)) stable by f?

[wintry steppe]

so you have the basis (e_1, f(e_1), ..., e_n, f(e_n)), and you know how f acts on its elements, which tells you what the columns of the matrix representation will be

[wintry steppe]

that is what i meant. it's true, but i don't see how in particular it helps us find the matrix representation

i guess it gives you a hint as to what the representation will be, but you can just write it down

[fair wren]

well M(f) = (1 0 .. 0) / (0 -1 0 .. 0) / .. / (0 .. 1 0) / (0 .. -1)

[wintry steppe]

that doesn't even make sense, how are the entries of your matrix representation elements of E?

they need to be scalars

[fair wren]

ye i was dazed

check again

[wintry steppe]

it's not correct

for example the first column should be (0, 1, 0, 0, ..., 0, 0), since f acts on the first element of the basis by sending it to the second

[fair wren]

what i don't get it

[wintry steppe]

then you should review how matrix representations are defined

[fair wren]

hm?

[wintry steppe]

what about what i wrote was unclear?

[fair wren]

i mean how did u even calculate the first column?

[wintry steppe]

i calculated it using the definition of matrix representations

i don't understand what this picture means without further context

[wintry steppe]

i still don't understand it, sorry

[fair wren]

yeah i see what u mean

so M(f) = (0 1 0 .. 0) / (-1 0 ... 0) / .. / (0 .. 0 1) / (0 ... -1 0)

right?

[wintry steppe]

are these the columns? because if so, that's correct

[fair wren]

something like this

right?

same structure for other n-tuples

[wintry steppe]

i don't understand the picture, but what you wrote just before it looked right

[wintry steppe]

i just want to make sure this is listing the columns (in which case it is correct)

[fair wren]

lemme draw it

[wintry steppe]

you don't need to draw it, i understand what you are writing

[fair wren]

no?

[wintry steppe]

that is not correct

[fair wren]

why

[wintry steppe]

because it's not

[fair wren]

ye

i see

sec

[wintry steppe]

that would be the matrix with respect to the basis (f(e_1), e_1, ..., f(e_n), e_n)

[fair wren]

@wintry steppe

[wintry steppe]

that is correct

[fair wren]

finally

[fair wren]

https://cdn.discordapp.com/attachments/978467303299616819/978853878286155786/btr.PNG

slightly better?

[wintry steppe]

sure

[fair wren]

for which values a,b,c is A diagonalizable

so det(A - yI_3)
= (1 - y)² (c - y)
= 0 iff y = 1 or y = c

how would i extract the eigenspaces for these eigenvalues

for example y =1

(A-I_3)X = 0

X = (x,y,z) in R^3

i get

(ay +z) / (bz) / ((c-1)z) = 0

this gives a solution (1, 0 , 0) = x_1

is this right @wintry steppe ?

[wintry steppe]

that is indeed an eigenvector

[fair wren]

for y= c it gets complicated

[wintry steppe]

that it does

[pallid rampart]

So a matrix is diagonalizable iff it has a full basis of eigenvectors. As you already noticed the eigenvalues are 1 and c. Since we know eigenvectors of different eigenvalues are linearly independent so it makes sense to treat the case of c=1 and c≠1 differently. First if c=1 then to be diagonalizable you'd need 3 independent eigenvectors, meaning the null space of A-I must have 3 basis vectors (and you continue from here). For the case of c≠1, you're already guaranteed an eigenvector with eigenvalue c, so you need 2 eigenvectors with eigenvalue 1, meaning the null space of A-I must have 2 basis vectors and you continue from here

[serene solstice]

How 2 do 2b?

[wintry steppe]

maybe use part a

[serene solstice]

Tried to. It didn't work.

[slender patio]

Try to compute $A(x)^2$ with the formula given in a)

[stoic python]

Twentycents

[serene solstice]

Okay. Did exactly that, and still scrambling at higher powers of $A$.

[stoic python]

Jonathan Phan

[vestal magnet]

Would appreciate any help with this 🙂

[molten pilot]

sorry but I have no idea what that means

maybe if you showcase what you've tried and explain precisely what you don't get people will be more willing to help

[wet stratus]

well for starters, how did you define the sqrt operator on matrices. can you just check that P satisfies that condition?

[vestal magnet]

We did that using eigenvalues

like sqrt(eigenvalue)

[wet stratus]

well sqrt(eigenvalue) would be a number

I assume here you still have a matrix

can you please work with us and give us the complete definitions of what you are working with

[vestal magnet]

They just gave us an example of how to find S if we know S^2 = T
sorry

and for that we used eigenvalues

[wet stratus]

and e_i are eigenvectors of T which span V ?

[vestal magnet]

e_i are the vectors for the some B orthonormal base, and the eigenvalues are lambda_i

was trying to work with inner product but maybe I did something wrong cuz didn't get the desired result

[wet stratus]

wait your definition of non-negative included P^*=P right?

so $T^T = (UP)^UP = P^ U^ U P = P^* P = P^2$

[stoic python]

Denascite

[wet stratus]

if your course had defined the sqrt to just be a matrix which squares to what we want we would be done

now if $Pe_i = \sqrt{\lambda} e_i$, then clearly $T^*T e_i = P^2 e_i = \lambda e_i$ so if $\sqrt{\lambda}$ is an eigenvalue of $P$, then $\lambda$ is an eigenvalue of $T^*T$. Sadly that doesn't yet show the other direction

[stoic python]

Denascite

[wet stratus]

we would be done if we could for example show that P has n eigenvalues

[vestal magnet]

yeah

[wet stratus]

or that there are eigenvectors of P which span V

maybe you have shown something like that in the past in your course?

using your definition of non-negative?

[vestal magnet]

our full definition was:

1. T=T* and T eigenvalues are real non-negative or
2. There's S which T = S*S or
3. T=T* and for each v in V, <Tv,v> >= 0

[vestal magnet]

can't think of a case we used in the course anything from those

sorry

[wet stratus]

you have never had some sort of spectral theorem?

your course is weird

[wintry steppe]

May I ask a question

[dusky epoch]

[wintry steppe]

Is my understanding of this text from the book correct?

[dusky epoch]

yes

[wintry steppe]

Thank you ^^

[torpid moat]

How does one show the following:
Given $T$ is a bounded linear operator, if $\innerproduct{\psi}{T\psi}=1$ for all $\norm{\psi}=1$ then $T=I$.

[stoic python]

thestonethatrolled

[steep sandal]

Please help in solving question 6

[zinc timber]

is f linear map?

[steep sandal]

Don't know

[zinc timber]

I'm assuming it's not then

[steep sandal]

Ok

[zinc timber]

for simplicity, why don't you look in lR instead of lR^3

can you think of a function f such that |f(x)| = |x|

there are few obvious choices

[steep sandal]

Can u explain??

[zinc timber]

explain what?

I'm asking you to think of some function of R to R such that |f(x)| = |x|

which is not linear

[steep sandal]

Ohhhkkk😅😅

Didn't read that

[zinc timber]

(say f is linear then f(x) = cx, so |f(x)| = |x| = |cx| means |c| = 1 or c=+-1)

this is not the example I am interested in

[steep sandal]

A wilo be orthogonal

Will*

Got it bro

[zinc timber]

[steep sandal]

@zinc timber if u have doubt in it I can send u my solution

Do anyone know how to check if two matrices are similar
Plz don't say just the definition ik that but it will be very lengthy

Then

[dusky epoch]

compute their JNFs and see if they're the same

[steep sandal]

Bruh

I think we have something to do with eigen values

Got it on Google @dusky epoch

[steep sandal]

Nvm got it

Is there anyone who is free tomorrow or today's night

I really need your help

Bruh I m just a beginner can u explain in simple language

[wet stratus]

hmm. where is my mistake then. consider f(x) on R such that f(x)=x if 2n+1 <= |x| <= 2n+2 and f(x) = -x if 2n < |x| < 2n+1 each for some n(so essentially we are either flipping or not flipping each interval (n, n+1) depending on which endpoint is even). f is not linear but preserves the absolute value

[steep sandal]

Yes

Yes

Yes

Yes else we have to define that

Ohh Nice

[zinc timber]

f(x) = |x|

it's not linear yet |f(x)| = |x|

[winter harbor]

Oh shit, yeah

[zinc timber]

same for R3

[winter harbor]

We have to impose restrictions in order for a norm preserving map in general to be affine

Damn

https://en.m.wikipedia.org/wiki/Mazur–Ulam_theorem

Ok, what I said before was pretty dumb

But works for inner product preserving maps tho.

[zinc timber]

ye so affine map with the condition f(0) = 0

[steep sandal]

I assumed f=Ax to be true and got A to be orthogonal

[zinc timber]

everything is included except f linear

[steep sandal]

But don't know how to show first part

[steep sandal]

Anyone??

[steep sandal]

.

[zinc timber]

^^

[steep sandal]

🤣

[torpid moat]

👀

[steep sandal]

Where are you guys from??

[zinc timber]

Here's a sexy proof $\ip{x, \phi x} = 1 = \ip{x, x} \implies \ip{x, x-\phi(x)} = 0$ for all $|x|=1$

@torpid moat

[zinc timber]

try to show <x, x- \phi(x)> = 0 for all x

[wet stratus]

well that part is easy. but how do you show from that that this implies x=phi x

[zinc timber]

left as an exercise

|| <x, y> = 0 for all x in V then y=0 ||

[wet stratus]

btw is there a list of commands we can use in this server? like the \ip ?

well but y depends on x

[zinc timber]

^ || just take x=y, then <y, y> = |y|^2=0 so y=0 ||

[zinc timber]

true

[wet stratus]

and it's only true for all x with norm 1

[zinc timber]

not a problem since you can always do x/|x|

gimme a minute

$\ip{Tx-x, Tx-x} = \ip{Tx, Tx} - 1 = \ip{Tx, Tx} - \ip{Tx, x} = \ip{Tx, Tx-x}$

[tribal willow]

oh wow

ive always been using \braket LOL

[zinc timber]

ip is a custom command

[tribal willow]

oh

[zinc timber]

you can use $\innerproduct{a}{b}$

[tribal willow]

hmm

i see

[zinc timber]

@torpid moat is our space real or complex?

it it's complex then <x, Tx-x> = 0 means Tx-x == 0 or Tx=x

don't think it's true for reals, T(x, y) = (x-y, x+y) then T(x, y) *(x, y) = x^2+y^2 = 1

[cunning violet]

dunno what kind of linear algebra calculation that is

[wet stratus]

that's not how you subtract two matrices

[cunning violet]

oh i think i got it lol , its matrix a - matrix b

yeee 😄

15min thinking got me nowhere, once i post it i see the mistake lol

[steep sandal]

Can anyone help me in proving this theorem

[zinc timber]

start with {} and keep adding vectors to it

until you hit a point where you can't include any more linearly indep vectors to it

[steep sandal]

Can u send a picture??

[zinc timber]

no

[fringe fjord]

You'll need to prove that the list stays linearly independent all the way and that you cannot continue adding vectors forever.

For the latter part, perhaps you already have a proof that a set of n+1 vectors in R^n is automatically linearly dependent.

[quaint mango]

I need some help with finding a solution to the matrix equation AX = B - 2X. I know how to do normal ones like AX = B, but when they throw more than 2X in I kinda get confused

[fringe fjord]

First add 2X to both sides.

Then on the left you'll have AX+2X which is the same as AX+2IX which is the same as (A+2I)X.

[wintry steppe]

Deadlines in 2 hours, just gimme the answers.. I'm exhausted from working whole day

[fringe fjord]

And now the equation has the form you know, just with a slightly different matrix.

Deadlines in 2 hours, just gimme the answers..
We're not going to help you cheat.

[quaint mango]

I'm gonna process that one a bit. I understand kinda what you mean but I'm gonna write it down on paper quickly

[fringe fjord]

(You didn't see my sign error!)

[wintry steppe]

Broski then explain it idc tbh 🤕

[zinc timber]

7

[wintry steppe]

🤣

57 mins actually 🥺

Fuck

[quaint mango]

just quickly so I understand. AX + 2X = B - 2X

[fringe fjord]

No.

AX = B - 2X.
Add 2X to each side and you get
AX + 2X = B

[quaint mango]

and that can be written to (A+2I)X = B

[fringe fjord]

Yes.

[quaint mango]

thanks for your help. Might be back in a few minutes haha, but gonna give it a try now

[quaint mango]

am I supposed to first calculate the matrix (A + 2I) and call it matrix d, am I supposed to get the inverse of this then and move it so X = (D^-1)B?

[fringe fjord]

Yeah, that is one way that will work.

[quasi delta]

I got a question. I was looking over Determinants, and was wondering what they are. They give area of a parallelepiped figure, show if a matrix is invertible, and is the product of the eigenvalues, but i don't understand the idea of a determinant anyone got some input?

[quaint mango]

so first calculate 2 * identity matrice, then just add matrix A I guess?

[torpid moat]

how did you go from here to <x,x-Tx>?

[torpid moat]

yeah, it's complex

[zinc timber]

I didn't

just calculate <x, Tx-x>

[fringe fjord]

The determinant function has a lot of nice properties that can look quite different, and it's the combination of having all of them at once that makes it particularly useful. Until you've gotten a fair number of them internalized, it might be better just to think of it as "that one weird function that somehow does all the amazing things" than to try to stick to a single more concrete intuition for what it does.

[quasi delta]

well, i got through my linear algebra class, and now im trying to remember and understand that black box

[fringe fjord]

Then my best advice is to think of the determinant as the thing that has all those properties.

[tribal willow]

whats the significance of parentheses around i in the superscript

[wintry steppe]

it means you should treat it as an index instead of as an actual matrix power

(this isn't standard notation for "thing in brackets in superscript" it's just what it means here)

[tribal willow]

ty

[wintry steppe]

is it true that $\langle x+y, x+y\rangle = \langle x, x\rangle + \langle x, y\rangle + \langle y, x\rangle + \langle y, y\rangle$? why?

[stoic python]

mate

[proud island]

Hey guys I need help in vectors

here is my question

A small plane, A, flies at an azimuth of 283 at 200km/h: Another, B, flies at an azimuth of 071 at 140km/h. Determine the direction and magnitude of the vector speed of A relative to B.

[winter harbor]

It is, it follows from the bilinearity of <•,•>.

[wintry steppe]

do you assume x and y are real numbers?

[winter harbor]

Uhhh, <•,•> is a bilinear form on a vector space V, right? So I am implicitly assuming x and y have to be vectors in V.

[wintry steppe]

okay thank you. i believed only left-linearity holds.

[ocean galleon]

With A = LU factorization, what is the relationship between L and U? The lower triangular and upper triangular

Like is one of them the inverse of the other, transposed?

[little crater]

can someone explain to me how what I did even shows that T is a linear transformation? I mean I found the mapping

shouldn't I need to check T(0) also?

I feel like I would have to use the definition of a linear transformation but I know this problem doesn't require that

also I notice the they just list entries so would it be wrong to say they are vectors?

like to say T(x) = T(x1, x2)

that wouldn't be correct, would it?

i guess my question is are n tuples and vectors interchangable?

I feel like

(x1, x2) != [x1 x2] but idk

[wintry steppe]

any function R^n -> R^m given by matrix multiplication is linear

there's no point checking that T(0) = 0, that follows from the definition of "linear transformation"

[little crater]

sure but we don't know it is one

[wintry steppe]

yeah the point is to prove that it's one

i know

[little crater]

so I would assume I should check T(0)?

[wintry steppe]

no

"T(0) = 0" is not part of the (standard) definition of a linear transformation

you do not need to check it

[little crater]

I suppose but what would I say if they had a constant in the T(x1,x2) = (..., ..., 1, ...)

[wintry steppe]

then it wouldn't be linear

[little crater]

yeah

[wintry steppe]

because all linear maps satisfy T(0) = 0 (as a corollary of the definition of "linear map") and that map does not

[little crater]

idk this question just seemed very "algorithmic" like I understand that the matrix itself would be linear transformation but whatever. But regarding the notation are the T(x1,x2) is that okay to just say T(x)?

like how they are just written as tuples

[wintry steppe]

you can pretend tuples and column vectors are the same if it bothers you

many authors do

[little crater]

yeah i just didn't know because my other question seemed to wish between it

[wintry steppe]

it's just easier to write tuples in-line rather than make the formatting ugly and write big ass column vectors each time

[little crater]

okay nvm

I checked back in the book

seem that is exactly what it means but just didn't know if such rules apply in general regarding tuples/points and vectors

[fair wren]

how do i determine which columns -vectors- is linearly dependent?

i've done the reduce row method

on the rank of A

is this already reduced or do i need to go further?

i found that rank(A) = 2 already but idk which column i had in A -before being reduced- was linearly dependent.

[hollow finch]

the pivot columns give you the ones that are linearly independent. so since your rref matrix has pivots in the first and second column, then the first and second column of your original matrix forms a basis for the column space.
nonpivot columns are linearly dependent with the pivot columns. the third column of your rref matrix tells you that col3=-2col1-2col2 so its linearly dependent

[fair wren]

can you tell me what are the pivot/nonpivot columns in myexample?

[hollow finch]

like i said, columns 1 and 2 are pivot columns

a pivot column is a column with a pivot in the rref form. and if it doesnt have a pivot when in rref then it's not a pivot column

id probably google "pivot column" if you're not sure what it is

[fair wren]

@hollow finch since this doesn't contain a leading 1 then this the nonpivot column thus its the linearly dependent

right?

[fair wren]

So only the first two vectors -columns- are linearly independent

thus those two form a basis for the column space of the matrix