linear-algebra

[copper pelican]

(For matrices that have column sums of s)

[dapper crypt]

Yea

[copper pelican]

Can you use the fact that the transpose of a matrix has the same eigenvalues as the original?

[dapper crypt]

I'm pretty sure this is induction

[little crater]

it didn't show up in our text yet but maybe?

[copper pelican]

Maybe it’s just induction indeed as Max notes

If the row sums are the same then I do think there’s a simple vector for the nxn case

[dapper crypt]

Like 99% sure... I can show u what I mean if ur still not sure later

[copper pelican]

I.e. a vector such that Av = sv

[dapper crypt]

Nvm I don't think it's induction as the induction hypothesis is really tricky

I know how to do it

@little crater hint: you need to use the fact the det(A-sI)=0 if s is an eigenvalue

Write out a general matrix of this form (using $x_{ij}$ notation) and remember the different operations you can do that don't change the determinant

[stoic python]

Max..

[little crater]

yeah I think I am going to do that

[dapper crypt]

I can give a further hint if u get stuck

[hexed urchin]

thanks!

[little crater]

nyann gave me a hint about taking the performing a row operation such that you you add all the rows above the last row to the last row

since we had (A-sI) and we knew that the columns summed to s

this results in the row entries in the bottom after performing the row operation

of s-s

=0

per entry

and since the row is now all zero we can get the conclusion the matrix isn't full rank .... non pivot columns ... (A-sI)x= 0 has non trival solutions... s is an eigen value for A

[dapper crypt]

Yea I mean I added all the rows to the top row ( doesn't change the determinant ) and realised it makes the top row all 0s. So expansion via the top row gives you 0×all the minor determinants so = 0

Talking bout rank is also a good way nice job 👍

[dapper crypt]

Anyone got some really useful or interesting facts about Eigenvalues, could be anything like Cayley Hamilton type stuff... or a lemme / q that was interesting

[wintry steppe]

minimal polynomial stuff is alright

[dapper crypt]

ty had a little read just now seems interesting, any other stuff is greatly appreciated

[little crater]

is there such that as a transpose matrix (for nxn matrices) such that BA=A^T and BA^T=A . I am going out on a limb and saying no but I am not sure. Maybe special cases but idk about in general. If we called A^T = C. We then have BA=C and BC=A. If we assme A!=C then I can't imagine that such a B would exist as these would evoke row operations and we would need both a row and column operations I would assume?

and I guess what I would be after a permutation matrix but I don'

t think that can exist in such a way

This would require both a row and column operation

so no I guess

[wintry steppe]

"is there such that as a transpose...."

is there a what?

what is your question?

[little crater]

does there exist a matrix that can always transpose an arbitrary nxn matrix or for that matter an mxn matrix?
Basically BA^T=A and BA=A^T

Anyways I don't believe there is such a matrix

Basically is it possible in general to construct a matrix that will transpose a given matrix.

[trail spear]

I am trying to learn linear algebra for machine learning but all the books that are saying linear algebra for beginners are not for beginners at all, any book recommendation?

[tranquil steeple]

no, there is no such matrix

[wintry steppe]

it would help if you wrote down what books you didn't like so people don't recommend them to you

[trail spear]

There was mathematics for machine learning from marc peter, linear algebra and optimization for machine learning. The other ones I am not sure I dont have them on my phone anymore.

Oh and Introduction to linear algebra 5th edition by Gilbert Strang

[tranquil steeple]

Maybe take it as a course or you lack the prerequisites?

[winter geode]

I wanna show that two homogenous systems of linear equations with two unknowns that have the same solutions are EQUIVALENT, but I’m dealing w a gap in logic

can I use the fact that they have the same solutions as basis for selecting scalars so that I can express any eq in one system as the linear comb. Of the eqs in the other system ?

That would complete the proof but idk if the fact that they “ have the same solutions “ works, and I’m not sure what does if so

[little crater]

for this question

do they just mean this

I know there are really only 3 distinct eigenvalues but the polynomial must have some sort of (x-1)^2 situation

or maybe it is this

okay sorry looking at the book it seems that both are okay

[polar marsh]

why is it that the minimal polynomial of a symmetric matrix is the following product: (x-l^1)* (x-l_2) ...*(x-l_s) with l_1, l_2, ... l_s the different eigenvalues of the matrix?

[wintry steppe]

(real) symmetric matrices are diagonalizable, and the minimal polynomial of a diagonalizable matrix is the product of (t - c) as c ranges over the distinct eigenvalues

[polar marsh]

okay but why is the minimal polynomial of a diagonalizable matrix this product?

[wintry steppe]

the shortest explanation is that the multiplicity of an eigenvalue as a root of the minimal polynomial is the size of the largest jordan block present in the matrix's jordan form, and in this case they're all just 1 x 1 blocks

[polar marsh]

Is this really the shortest explanation?

[wintry steppe]

it's the shortest i could come up with

without just saying "'minimal polynomial splits into ... iff diagonalizable' is a standard result in most books"

[polar marsh]

okay well thx for your help

[wintry steppe]

it's the shortest, but maybe not the most elementary

i'll type the easy proof in a moment

the polynomial $$p(t) = (t - \lambda_1) \cdots (t - \lambda_k)$$ ($\lambda_1,\dots,\lambda_k$ the distinct eigenvalues) divides the minimal polynomial $m(t)$. by picking a basis of eigenvectors, you can check that $p(A) = 0$, where $A$ is your matrix. by the uniqueness of the minimal polynomial, $p(t) = m(t)$.

[stoic python]

TTerra

[wintry steppe]

@polar marsh you might prefer this

needed a moment to recall the easy proof

[dreamy charm]

In Axler's Linear Algebra Done Right, the proof that dim U is <= dim V where U is a subspace of V is "Suppose V is finite-dimensional and U is a subspace of V. Think of a basis of U as a linearly independent list in V. Now use the fact that the length of a linearly independent list is <= the length of a spanning list to conclude that dim U <= dim V. But wouldn't it be reasonable to interchange the roles, i.e. think that U is a spanning list and V is a linearly independent list? And wouldn't that suggest the opposite?

[wintry steppe]

the converse is also true (if the minimal polynomial takes this form, then the matrix is diagonalizable), but the proof is a little trickier

[wintry steppe]

what do you mean by "U is a spanning list and V is a linearly independent list"? these are subspaces, not finite lists of vectors

be a little more precise

[polar marsh]

thx

[dreamy charm]

right but I'm just going by his logic

he defined a subspace U to be a finite list of vectors in the proof

[wintry steppe]

no

he took a basis of U

the proof goes as follows: if $\beta$ is a basis of $U$, then it is a linearly independent set in $V$, so its size $|\beta| = \dim U$ is less than or equal to the length of any spanning set in $V$, hence $\dim U \leq \dim V$.

[stoic python]

TTerra

[dreamy charm]

oh o kthanks

[robust owl]

So, for this: "A vector space A is infinite dimensional iff A has some infinite LI subset". Is the forward direction possible without choice or some equivalent? I see how to do it by well ordering theorem I think since I can pick x_{k+1} to be the least elt of A not in span{x_1,...,x_k} and build up my LI set that way.

[fringe fjord]

Which definition of "infinite-dimesional" do you have to start from?

[robust owl]

This is from Friedberg, Insel, Spence. They define a vector space V as finite dimensional if it has a basis consisting of a finite number of vectors. A vector space V is then infinite dimensional if it is not finite dimensional.

[fringe fjord]

Hmm, yeah, then I think you do need some form of choice.

[robust owl]

Part of my curiosity is because the next section after the section containing this exercise covers Hausdorff Maximal Principle, so my worry here is that either the exercise can't be done without some equivalent to choice, or there's any easy way I'm overlooking.

[fringe fjord]

https://math.stackexchange.com/a/1587031 gives a very rough outline of a forcing model of ZF with a vector space that has neither a basis (so in particular it's not finite-dimensional) nor any infinite linearly independent set.

[robust owl]

That's really weird and kinda cool lol

[little crater]

yo chat chat

so ummm

explain why det(A) is the product of the n eigenvalues of A

[wintry steppe]

but you already did

[little crater]

that is what I thought but my prof wanted me to mention I guess something about the fundamental theorem of algebra I believe.

[wintry steppe]

it's not like it asked you to prove that you can write det(A - \lambda I) like that

but if you want to prove you can, then the fundamental theorem of algebra is what you need

[little crater]

okay yeah I guess so

[tall void]

do you need lebesuge integrals to properly define a hilbert space?

the definition of a hilbert space is "a complete inner product space" that's it right?

[keen sierra]

you don't need integrals at all to define a hilbert space

just the basic theory of vector spaces and metric spaces

however, for the specific hilbert space L^2 you need lebesgue integration

[tall void]

ah okay

like

the hilbert space of square lebesgue integrable functions or something?

@waxen minnow

[waxen minnow]

thats what L^2 is yeah lol

ic

[zinc timber]

yeah

[errant mist]

What does M_n mean in the picture?

[subtle walrus]

square n times n matrices over the field F i guess

[errant mist]

yeah, but just wondering if the M stands for matrix in general. Then do I read it as some matrix algebra over the field in this case n x n matrices?

[subtle walrus]

that would be my guess

there are a bunch of different notations for this kind of thing

[errant mist]

ahh confusing)

We will then have in the corollary when choosing n x n matrices A that if A is diagonalizable then this reduces down to a special case of Jordan normal form where we end up with just the eigenvalues on the diagonal. Also C would not be unique here as there are many choices.

[dim epoch]

In this context just looking at it as a standard nxn matrix would make sense

[dim epoch]

jordan bases aren't unique so that's fine

in the process of computing it you choose certain vectors out of certain kernels/subspaces

but for the algorithm to give you your jordan form it doesn't matter which ones you choose so you can end up with different change of basis matrices

[errant mist]

understood, thanks

[dim epoch]

[modern stone]

Anyone here ? I have a question

[dim epoch]

[errant mist]

@dim epoch also wondering about what the notation [T]_epsilon means for the corresponding theorem?

[spice field]

I guess it means the matrix representation of T with respect to \epsilon

[errant mist]

yeah, I was thinking about that I was just afraid I was missing something with the brackets notation. I have never seen it used

[dim epoch]

yup should be just eps* T *eps^-1

[spice field]

so the condition means nothing but the characterisc polynomial splits on the field F?

[dim epoch]

i mean the theorem seems weird to me

since the jordan form would just be a diagonal matrix

unless im getting something really wrong here

[spice field]

in cases where F is the complex number field the theorem coincides with the usual Jordan normal form theorem

[dim epoch]

F doesn't have to be the complex numbers for that does it?

if you have a splitting polynomial over R that works as well

just that over C polynomials generally split

[zinc timber]

you need the char poly to split to have an upper triangular representarion, aka schur decomposition

[spice field]

no, but the condition that P splits on F is more general

as C is algebraiclly closed, every polynomial over C splits

[errant mist]

more polynomials split over C like x^2+1 = 0

[dim epoch]

yes but the polynomial splitting is already given as an assumption

[spice field]

yeah

[dim epoch]

so you don't need C for it to split

[spice field]

yeah, but only a subfield of C which is suffienctly large so that the given polynomail splits over it

[dim epoch]

hmm

im just confused on why the splitting polynomial is given as an assumption when going towards jordan forms

since the jordan form is usually used to get a form that's as pretty as possible if the polynomial does not split

hence we don't get a diagonal form

[spice field]

because if the polynomial doesn't splits, you cannot choose n many eigenvalues as they're the number of the roots of the char polynomial. Without that many eigenvalues, the proof cannot go on

[errant mist]

Dont we need it in order for the basis to exist? Thats why we assume the polynomial exist?

sorry meant splits

[spice field]

to make it digonal, one need 'Jordan blocks of size 1', or say, roots of char polynomial with no multiplicity.

[modern stone]

Can someone Help explain how to solve this problem

[dim epoch]

hm i guess we went about jordan forms differently

[spice field]

But if there's multiplicity, in general you get no diagonal ones but only those with 1 over them

[spice field]

precisely speaking it's because in the proof , first we fix an eigenvalue, then we choose a vector to build the basis. but if the eigenvalue we get is less then n, then we cannot get a basis but only 'independent set of k vectors' where k is less than n.

[dim epoch]

actually nvm im just dumb

ignore what i said

on the jordan form part

[errant mist]

yeah, if u get diagonal entries this reduces down to the trivial case aka the square matrix is
.

diagonalizable

@spice field thanks

[spice field]

another viewpoint to check if this is correct: it's well-known that every complex matrix has a Jordan form. So if Jordan form is diagonal, then every complex matrix can be diagonalized. That's obviously, incorrect.

[errant mist]

@spice field I'm also wondering about positive symmetric matrices which I know contain real eigenvalues. There is a proof where they show if lambda is the greatest eigenvalue then lamba > 0.

[modern stone]

Anyone can give me some tips how to solve this ?

[errant mist]

how can they just take the trace of A in this case?

[dim epoch]

yeah i got confused with some terminology and prerequisites

[errant mist]

Shouldn't it be the trace of D in A = X D X^-1 since its diagonalizable?

[dim epoch]

this isn't linear algebra, try one of the help channels

[modern stone]

Thanks

[spice field]

it's a good exercise to give a non-diagonalizable matrix. Try it?

For example, over C, size 2x2.

[zinc timber]

cool thing is trace is invariant under similarity transform so it does not matter

[errant mist]

so, you are assuming multiplying with an invertible vector?

[dim epoch]

your matrix is diagonalizable

so you have a matrix B such that B* A * B^-1 is diagonal

with the eigenvalues being on the diagonal

[errant mist]

ye that I know

[dim epoch]

then you can take the trace of that diagonal form

since traces are invariant under changes of bases

[errant mist]

but the matrices B, B^1 implied or they dont effect the trace?

[dim epoch]

yes they don't

[errant mist]

ahh, that clears things up

[dim epoch]

B* A * B^-1 is a change of bases

are you familiar with that concept

[errant mist]

yep

[dim epoch]

nice

then that concludes it

[spice field]

traces are nothing but the sum of all eigenvalues.

[dim epoch]

like ryu said, traces are invariant under changes of bases

[errant mist]

yes, I understand now. thanks again.

[spice field]

so if the greatest one is even negative, then the trace can only be negative. Which is contrast to the condition.

[dim epoch]

was this an exercise for me or fredrik

[spice field]

for ones who are interested...

hint: permutation.

[dim epoch]

what example did you have in mind with that hint

I would've taken the standard example ((0,1),(0,0))

[spice field]

for a n-dimensional space you take a basis consisting of n vectors. Then consider any non-identity permutation of the n-vectors as a set. Try to represent this permutation as a matrix with respect to the basis. Show that this matrix is not diagonalizable.

[dim epoch]

oh that's interesting

[errant mist]

I will think about that

[dim epoch]

my example is basically that one for n =2

[spice field]

Because matrices that can be diagonalized, are intuitively 'stretching' the plane (if n=2) in two directions. Any permutation of the two directions can never be a 'strecthing', because it disordered the space orientation. think about this.

[zinc timber]

@spice field [0,1 ; 1,0] is actually diagonalizable so I don't understamd your example

[dim epoch]

that doesn't represent a permutation tho

[zinc timber]

why not?

[spice field]

I mean, not all of such permutations are 'non-diagonalizable'. maybe I shall try to make some editting. I mean some of them are. In perticular, those changed the orientation are.

[zinc timber]

can you elaborate

[spice field]

For example, in plane. There are two axies, x and y. Consider a permutation that reverse x but keeps y. What's the matrix of it?

[zinc timber]

? [-1, 0; 0, 1]?

also reverses x is not a permutation term

I hope you know the definition of permutation matrix

[spice field]

And how it reserved the orientation? (in fact this is a notion from differential geometry)

[zinc timber]

you mean when sign is -1?

[spice field]

yeah I know that. So basically I was thinking not 2x2 matrix but 3x3

[zinc timber]

also works as CE

[spice field]

in 3x3 case, you have some ways to change the orientation by only the permutation.

[zinc timber]

I don't see why a permutation matrix can't be diagonalized. After re labeling you can always find boock of a n-cycle (123..n) with mini poly xⁿ-1=0 which is diagonalizable in ℂ

[spice field]

As in 2x2 case, any permutation is trivial, because there are only 2 of them.

[dim epoch]

was wrong, mb

[zinc timber]

orientation reversing 3-perms are as follows (12),(13),(23) all of which are very well diagonalizable

[pearl elm]

Need help with this one

[zinc timber]

what's troubling you?

[cyan zenith]

If A,B,C were real numbers, how would you go on with solving this equation? Then think how to apply that to matrices

[spice field]

maybe i'm thinking something stupid...

but I was thinking the permutation is related to some property. Maybe not diagonalization though

[zinc timber]

same thoughts. I do want to claim that perm matrices are diagonalizable in ℂ

[pearl elm]

the algebra going on here because of left and right distribution laws I would say is tripping me up.

[zinc timber]

you don't need to distribute

remember A'A=I

[pearl elm]

Probably cuz I got rid of the initial Identity term

[zinc timber]

A' is the inverse of A

how about you do CC'(A+X)B'=CI=C so (A+X)B'=C?

that's the first step

[pearl elm]

can you fix your notation a little so im less confused

thanks

I being I_n

right?

[zinc timber]

$(A+X)B^{-1}=CI_n =C$

[pearl elm]

i feel stupid lmao

[zinc timber]

Here a movie of how it works

[spice field]

an easier but not-so-interseting example is rotation matrices, or in general, anti-symmetric matrices.

I was assuming they are related to permutation (in fact not) so I was doing an incorrect generalization.

(over R)

[zinc timber]

0 is anti symmetric tho

[spice field]

I mean, non-trivial

[zinc timber]

they are diagonalizable over ℂ btw

[spice field]

yes

they don't have real eigenvalues but have imaginary ones

[spice field]

Corollary: a trivial element satisfies all properties.

an obeservation: say if A is diagonalizable. If x is such that Ax is non-zero, then A^2 x is again non-zero.

So if some M is such that M^n=0 for some n, but M is not 0, then M cannot be diagonalized.

and one type of such M can be given by upper triangular matrices with diagonal entriss all zero.

for such M of size nxn, we have M^n =0. But M is not necessarily 0.

[dusky epoch]

matrices of the kind you just described have a special name

theyre called nilpotent

[spice field]

yes, I know that

thanks for that but I was just trying to explain why

[spice field]

this is because after being diagonalized Ax is basically x with its coordinates scaled

[errant mist]

Does there exist a symmetric version of a stochastic matrix?

[mighty cargo]

hint: consider 1/n for an n \times n stochastic matrix

[errant mist]

yeah if all entries are 1/n then its symmetric

[mighty cargo]

also trivially ||identity||

[errant mist]

what if both columns and rows sum to one. I believe this is a matrix i.e both left and right stochastic matrix

[mighty cargo]

yup, doubly stochastic

[errant mist]

oh, ok thx

[peak lodge]

is it possible to figure out A if you only know

$A\left(\begin{matrix}1\-1\3\end{matrix}\right)=\left(\begin{matrix}1\0\0\end{matrix}\right)$

[stoic python]

Vince Tafea

[peak lodge]

where A is a 3x3 matrix

[wintry steppe]

there might not be a unique A, but you can definitely find one.

[little crater]

You could construct an A of many :v

[wintry steppe]

one example is $$A = \begin{pmatrix} 1 & 0 & 0 \ 0 & 0 & 0 \ 0 & 0 & 0 \end{pmatrix}.$$

[stoic python]

TTerra

[little crater]

TTerra I know about the standard matrix representation but can you use any arbitrary basis and figure out the corresponding matrix A for a given transformation T or must we use the e1,e2,...en vectors?

for some transformation T

[wintry steppe]

each basis will give you a matrix representation

the matrix representations corresponding to two different bases are related via conjugation by the change-of-basis matrix

"e_1, ..., e_n" and "standard matrix representation" are not well-defined things in general vector spaces

if you're in R^n, it's fine

[little crater]

Oh I guess standard matrix representation is implicit that the vectors input follow the standard basis of columns of the identity matrix?

[wintry steppe]

yes

"standard" meaning "with respect to the standard basis"

[little crater]

I see

[little crater]

when I have a repeated eigen value in my matrix (say it is triangular) does that guarantee that it must be a root with a power of ^n for however many times it shows up?

like say I have the eigen value 5 repeated twice

does that mean my characteristic polynomial has (5-λ)^2 in it

[old minnow]

ye

[copper pelican]

Yeah pretty sure that’s the definition for a repeated eigenvalue

Note that this definiton applies to any nxn matrix, not just triangular ones

[deep flicker]

why is it that for rotating in a 2D plane, we use

M = { cosθ -sinθ   . {x
      sinθ cosθ }     y}

If I want to rotate vector v0 15 degrees, and given that x0 = cosθ, y0 = sinθ, is it not the case that vector v1 the translated vector has coordinates of x1 = cosθ + 15, y1 cosθ + 15 ?

[wintry steppe]

"translation" is not linear

by "translate 15 degrees" do you mean rotate 15 degrees?

[deep flicker]

sorry I meant rotate

ohh the matrix allows for translation too ?

[wintry steppe]

no

[placid zephyr]

in landaus second volume, he defines axail and polar vectors, but with his definition I do not understand how an axial vectors components dont change sign under reflections, as the third cartesian basis vector can be obtained by crossing the two polar basis vectors e_1 e_2 and its coords def change under reflection

[wintry steppe]

you cannot represent translation by matrix multiplication, it is not a linear map

rotating $(\cos\theta,\sin\theta)$ by some $\theta_0$ (in radians, because we don't use degrees in math) amounts to multiplying by that matrix with $\theta_0$ in it: $$\begin{pmatrix}\cos\theta_0&-\sin\theta_0\\sin\theta_0&\cos\theta_0\end{pmatrix}\begin{pmatrix}\cos\theta\\sin\theta\end{pmatrix} = \cdots = \begin{pmatrix} \cos(\theta+\theta_0) \ \sin(\theta+\theta_0)\end{pmatrix}.$$

[stoic python]

TTerra

[wintry steppe]

the ... are for you to fill out

it's just matrix multiplication and trig identities

[deep flicker]

I can do the multiplication!

Let me find the identities now

from multiplying it, it would be
x = xcosθ - ysinθ
y = xsinθ + ycosθ

[wintry steppe]

theta_0, but yeah

now plug in x = cos θ and y = sin θ and use the "angle addition formula" or something

[deep flicker]

I'm not sure what you mean

I'm trying to simplify it down to

hmm I see formulas for Cos(A+B) = cosAcosB - sinAsinB

and similar for sin(A+B)

[wintry steppe]

$(x, y) = (\cos \theta, \sin \theta)$ was your assumption. you have
\begin{align*}
\begin{pmatrix}\cos\theta_0&-\sin\theta_0\\sin\theta_0&\cos\theta_0\end{pmatrix}\begin{pmatrix}x \ y\end{pmatrix} &= \begin{pmatrix} x \cos\theta_0 - y\sin\theta_0 \ x \sin\theta_0 + y \cos\theta_0 \end{pmatrix} \&= \begin{pmatrix} \cos\theta\cos\theta_0 - \sin\theta\sin\theta_0 \ \cos\theta\sin\theta_0 + \sin\theta\sin\theta_0 \end{pmatrix}
\end{align*}

[stoic python]

TTerra

[wintry steppe]

you assumed that (x, y) = (cos θ, sin θ) here

[deep flicker]

just because I thought it does when using a magnitude of 1

[wintry steppe]

you could of course write (x, y) = (r cos θ, r sin θ), factor out the r, and do the same computation i just wrote

[wintry steppe]

i'm just clarifying what i meant here since you were confused

[deep flicker]

how are you making this simplification

[wintry steppe]

?

read the first line

(x, y) = (cos θ, sin θ)

[deep flicker]

ohhhhh

clever

I actually did input that in the calculator

[wintry steppe]

you should do this without a calculator

[deep flicker]

I was just using some real values

a 15 degree rotation for example

[little crater]

is the square root of a matrix a thing?

Like we had a question to come up with a method for it but it was specifically for the case of a diagonalizable matrix and having non-negative eigenvalues

but I just mean in general

[deep flicker]

@wintry steppe and so given that this

equals (cos60, sin60), for theta = 60

is it not always the case?

[wintry steppe]

if theta is 60 and theta_0 is 15 then no

also, please work with radians

nobody uses degrees

[deep flicker]

yes I'm sorry I got no clue about radians really

but the point I'm making is

why do we represent the matrix as such

when we can do x = cos(A+B)

y = sin ( A + B)

[wintry steppe]

because we like to write linear transformations as matrices

this is one of the main objectives of linear algebra

[little crater]

a lot cleaner to have a form T(X) that takes in a vector and outputs some vector

which i guess will encode the same stuff but in a nice package

[wintry steppe]

it's definitely a thing. you may not always be able to find one, but figuring out when you can (plus what properties a root has, and so on) is one part of linear algebra

usually you study square roots of operators when looking at operators on inner product spaces

[little crater]

So is the approach I took different from the standard notion of square root on a matrix or is it more of a specific case?

[wintry steppe]

wtf does "standard notation of square root on a matrix" mean

[little crater]

okay

[wintry steppe]

a square root of a matrix A is a matrix B with B^2 = A

[little crater]

okay

well on one question I applied it like the following (assume the matrix was diagonalizable)

M^k=PD^kP^-1

so M^(1/2) = PD^(1/2)P^(-1)

and defined D^(1/2) as taking the square roots on the main diagonal entries

but idk what the approach would be in general for something that can't be diagonalized

[wintry steppe]

you should be more careful than just plugging in k = 1/2

[little crater]

we assumed eigenvaleus were non-negative

[wintry steppe]

"M^k = PD^kP^{-1}" is something that holds for positive integer k

please be more careful than just plugging in something that's not an integer

anyways

[little crater]

I know

[wintry steppe]

making sure.

[little crater]

My point is we were told to come up with a method (question on assigment)

well "conjecture"

[wintry steppe]

what's the specific question?

(screenshot please)

[little crater]

(c)

I was able to verify (for B^2 = A)it worked but was more so curious is if there is a general way of computing roots of a matrix as this seems to be a specific case

[wintry steppe]

i don't know if there's a nice general method

square roots of operators are usually looked at when you study inner product spaces and their operators (e.g., a positive operator has a square root), but i don't think you get much more than just a coordinate-free version of what you've done here

[restive crane]

if i am doing a proof and I am using the fact that because two subspaces are the same, then their dimension is as well, do I need a sub-proof (or just include a proof) of the dimensions being equal because their subspaces are as well

[little crater]

Yeah we have not touched inner product stuff yet (next week) but we won't eve be doing inner product spaces

[wintry steppe]

that's so trivial you should just say it and leave it

[restive crane]

okay good lol

thanks!

and also the euclidean subspace R5 has dimension 5 right

[wintry steppe]

yes, R^5 has dimension 5

(1, 0, 0, 0, 0), (0, 1, 0, 0, 0), (0, 0, 1, 0, 0), (0, 0, 0, 1, 0), (0, 0, 0, 0, 1) is a basis

[little crater]

tbf this was a 6 week course and we lost 2 days + this is a first course in linear algebra (Still seems what we covered vs how much the book has is a big gap)

[wintry steppe]

i should remark that a matrix need not have a square root, and if it does, it doesn't have to be unique

you have to impose some conditions to be able to find a root, let alone a unique one

[deep flicker]

@wintry steppe Last thing I promise. Just in summary is the reason we wouldn't just do x1 = cos(A+B), y1 = sin(A+B) because we basically want things in matrix form to comply with the rules of linear algebra and all the reasons surrounding why that's useful for linear algebra and that wouldn't be understood without actually looking deeper into linear algebra or is there a simple(r) explanation?

[wintry steppe]

"comply with the rules of linear algebra" makes linear algebra sound like a ruthless dictator lol

but pretty much

you could just write the result like that, but it's more illuminating if you write the operator (of angle addition) itself as a matrix and study that

for then you can apply all the usual linear algebra stuff to study it

[deep flicker]

Thank you for your help, I'll keep working on my understanding of this stuff

[old minnow]

Our dear and beloved supreme leader, Linear Algebra

[wintry steppe]

So given a Matrix A, I get how to find the Column space, you just row reduce and the pivot spaces, are the column space, how do I find the null space?

it says its the solutions to Ax=0 so does that mean I put it into another matrix, basically adding the zero vector to the end of A?

[little crater]

The column space is the span of your pivot columns in your matrix A. You determine your pivot columns in a matrix by row reducing and finding the columns that have have pivots (ie pivot columns).

The null space is is solving Ax=0. We can get a basis of the null space by row reducing our matrix a and finding the non-pivot columns. These non-pivot columns will be the free variables. We then can interrupt this matrix as a linear system and write out the non pivot column variables in terms of the free variables. Once you do that you have parametrized your vectors in terms of your free variables. You then just take the vector themselves and place them in your set and call it a basis of the null space of A. it is possible there are no free variables and hence your null space is zero dimensional with no basis vectors. @wintry steppe

[wintry steppe]

right! thank you.

[little crater]

It happens that since we are looking for Ax=0 then row reducing doesn't have any downsides with regards to solving this as the row operations don't change the null space (basically if we have matrix U that is row equivalent to A such that it is the row reduced version Ux=0 and Ax=0 have the same solution set)

[wintry steppe]

would you mind helping me with 23? I am having trouble with the answer they provide

[little crater]

This isn't true for Ax=b as row reducing will impact that arbitrary b . So Ax=b and Ux=b will have different solution sets even if they have the same dimensionality

[wintry steppe]

oh! so I dont use the RREF matrix

[little crater]

well you do

[wintry steppe]

but the original entries?

[little crater]

it is just you use your orginal vectors

[wintry steppe]

okay

Im just confused how its [4, -5. 1, 0 ] and [-7,6,0,1]

[little crater]

you are looking at the Nul A

you do see the size of your matrix right?

3x4

[wintry steppe]

yes its 3x4

[little crater]

that means any x

has to be 4x1

that column vector x

in Ax

[wintry steppe]

right

[little crater]

are you confused on the size or the actual basis?

[wintry steppe]

the actual basis

[little crater]

okay

well we know that row reduction doesn't change the null

[wintry steppe]

right

[little crater]

it is also important to by able to recall what this matrix is saying

so

looking on the right

[wintry steppe]

4x1+5x2+9x3-2x4=0?

[little crater]

yeah that type of hting

thing*

but also realzing by looking at the pivot columns and non pivot columns

it tells you the free variables

[wintry steppe]

right

[little crater]

Looking at the row equivalent matrix we can see we have 2 pivot columns and 2 nonpivot columns

[wintry steppe]

okay

[little crater]

those non pivot columns tell us that x_3 and x_4 are free variables

[wintry steppe]

and x_1 and x_2 are basic variables?

[little crater]

yes

we now want to parameterize those vectors in terms of x_3 and x_4

[wintry steppe]

ohhh okay.

I see that x_1 has x_2 in it.

so I just put what x_2 is equal to into that space?

[little crater]

rewrite that matrix in terms of a system of equations

it would also be better if you completely row reduced the matrix to rref as it is easier

or else you need to do back subbing

[wintry steppe]

x_1+2x_2+6x_3-5x_4=0
x_2+5x_3-6x_4=0

[little crater]

well actually you should probably finish row reducing before this step

[wintry steppe]

oh I thought it was finished

my bad

x_1-4x_3+7x_4=0
x_2+5x_3-6x_4=0

[little crater]

okay now move your free variables over

so that you have

x_1 =....

x_2=...

[wintry steppe]

x_1=4x_3-7x_4
x_2=-5x_3+6x_4

[little crater]

and now recall that x = [x1 x2 x3 x4]

(column vector)

[wintry steppe]

yeah I remember now thanks a lot!

its actually really simple once you get it

Completely row reduce, and write into a system of equations. then youll get x number of vectors for how many basic variables you have...and like you said parameterize it 😄

[little crater]

here is one example I had that shows more so what exactly happens or maybe truly what you are doing

mainly the last bit

all you did was just rewrite your basic variables in terms of the free variables and then you split up your solution into those indivual vectors for however many free variables you have

the #freevariables also tells you how many basis vectors you have

and you can get the basis just by taking those two parameterize vector

[wintry steppe]

yes thank you!

[wintry steppe]

Learning it on your own is pretty tough. I took a class and that was hard on its own

I do have to ask what is the significance of the col space and null space?

[steel parcel]

Anyone ?

[wintry steppe]

"this is not an equation"

[little crater]

Col space tells you the span of you matrix column vectors which gives you know of if every Ax=b has a solution or not

And for it to have an inverse (assuming it is square) you need to know the col space spans that entire vector space

And there is a whole relationship between the column space and null space

At least regarding their dimensions and the number of columns a matrix has

Speaking about spaces

[next slate]

Linear Algebra Done Wrong is great

[little crater]

Is it bad my class never went over the row space or left null space

We only did null space and column space

And while you just apply transpose and carry out similar computation it just seems strange

[restive crane]

So, when you have a linear transformation and you want to determine it's corresponding matrix, you just apply the standard basis vectors to the transformation and the output is basically the columns of the matrix right?

So if you have another basis and you want to determine a matrix for that basis for a linear transformation, can you similarly just apply the basis vectors to the transformation to get the matrix?

[little crater]

I think I asked a similar question and I believe the answer was yes. But then that matrix would only make sense with vectors who use the relative coordinates of those basis vectors as well

The normal approach is standard matrix representation with e_i vector which are as you said the standard basis vectors

[restive crane]

Okay because the reason I ask is this question

I was not too sure how to do this but is the approach mentioned above okay?

like applying the basis vectors given to a projection onto the y axis to build a matrix

[barren crystal]

ye it seems good so far and is it a good start to read just by itself or best used to supplement another course?

[little crater]

Oh I'm not familiar with the projection stuff so I don't think I can help much

Haven't gotten to that

[restive crane]

It's alright, I've asked all my friends and they all have given a different answer to this question lol

[little crater]

What does on the y axis mean (0,y,0)?

[next slate]

I think you can read it by itself

[little crater]

So the 2nd entry in this case

[next slate]

Of course you can always use other books and keep building up

[restive crane]

Like the y-axis in R3 so yes you are correct in (0, y, 0)

however y can be anything (not only one)

at least that is my interpretation

[restive crane]

If anyone can help with this question, that'd be great!

[wintry steppe]

projection onto the y-axis is given by sending (x, y, z) to (0, y, 0). figure out what this does to the basis vectors, express the results as linear combinations of the elements of your basis, and throw the coefficients into a matrix accordingly

[restive crane]

By basis vectors you mean standard basis right? And then express the result of the projection by elements of the basis given to use in the matrix

[wintry steppe]

no, you're asked in the question to work with a specific basis

i mean that basis. the basis β.

β is not the standard basis

[restive crane]

okay sorry its late here, but just to confirm, use the given basis vectors and apply the projection to them. Then the result of the projection needs to be expressed by the basis vectors whose coeffcients need to go in a matrix

[wintry steppe]

something along those lines, yes

if you don't know how exactly, here's how the matrix representation of a linear operator is defined

if $T\colon V \to V$ is a linear operator on a vector space $V$ with an ordered basis $\beta = (v_1, \dots, v_n)$ (i use $v$ instead of $e$ so as not to make you think of the standard basis), then $$T(v_j) = \sum_{i=1}^n a_{ij}v_i$$ for some scalars $a_{ij}$. the matrix $(a_{ij}){1 \leq i, j \leq n} =: [T]\beta$ is the matrix representation of $T$ with respect to $\beta$.

[stoic python]

TTerra

[wintry steppe]

(the standard basis is only a thing in R^n. linear operators make sense in more general contexts.)

use this definition to complete the problem

(a more general definition would involve a linear map between two vector spaces and bases of each, but i didn't want to write this much)

[barren crystal]

alright thx

[golden mulch]

can someone explain the Gauss-Siedel method tome

to me

[wary girder]

Im learning vector space now, and I understand that we can apply different concepts of linear algebra to other vector space. What I dont understand is that why we define inner product with a weighting function, but there's no such analog in dot product

it's a bit hard to rationalize it, so im looking for an intuitive way of understanding it

[fringe fjord]

Can you explain "define inner product with a weighting function" more? The usual definition of what it means to be an inner product has nothing in it I would call a "weighting function".

[wary girder]

this is one of the definition i foound

a lot of other sources also include the weighting function

some dont

[zinc timber]

they are useful when you study Hilbert spaces and complete basis. Legendre polys, Chebyshev polynomials etc

projection, normality, gauusian quadrature, approximations use these immer products extensively

[wary girder]

i see isee

but is there an origin as to why we include the weighting function in these scenarios but not the others

[quartz compass]

yeah pops out of sturm-liouville theory

[wary girder]

icic, thanks a lot

[formal sparrow]

Hi. I was reading about HHL algo in quantum comp, the one used for solving system of linear equations. I think I understand it all but for this statement in the green circle. Could someone explain it to me?

Thank you

[slender patio]

Don't know what's HHL but it seems that they just wrote the vector b in that spectral basis

Since it is a basis, there exist coordinates b_j such that b = (that sum)

(the motivation of choosing that basis being the last equation)

[formal sparrow]

Thank you very much!

[somber loom]

can someone explain how a transformation can have several right inverses, but can only have a single left inverse, or even no left inverse at all?

[wintry steppe]

-a map can have multiple right inverses (existence is equivalent to surjectivity, uses choice if you care about foundational issues), and there's a unique one when the map is injective. an example of a map with multiple right inverses is R^2 -> R, (x, y) -> x. two right inverses are given by x -> (x, 0) and x -> (x, 2x).
-similarly, a map can have multiple left inverses (existence is equivalent to injectivity), and there's a unique one when the map is surjective.
-to find a map with no left inverse, simply find one that isn't injective

to answer your questions in order @somber loom

[little crater]

Say you have an Ax=b and you can't solve for that b with that given system, is there such thing as minimizing the system so that it is close to each value in that b vector as possible?

basically if Ax=b doesn't exist, how do I find the next best solution Ax=b' where b' tries to stay as close to the original b as possible

although at the moment I can't thing of a clear definition for "as close to original b as possible" but maybe you understand what I mean

[wintry steppe]

so you'd want to find the vector in the subspace im(A) which minimizes the distance to b

hmm

[little crater]

something like that I assume I didn't have clear definition of this b' but that seems logical

[wintry steppe]

the orthogonal projection of b onto im(A) will be the closest vector to b in im(A)

if b' denotes that orthogonal projection, then Ax = b' is certainly solvable

but after solving this, you might ask what you could infer about the original system

[little crater]

is this at all related to least squares by chance? (don't know anything about it other than my lin alg class is going to dicuss it sometime next week)

[wintry steppe]

i don't know, but i was actually about to bring it up

since i know that has to do with systems and minimizing something

[wintry steppe]

I didn't think of it, i was gonna say make the calcul but its much more elegant that way X)

dually you might ask what the minimal solutions x to a consistent system Ax = b are

i don't remember exactly how to compute it, but i know they exist and use inner product shenanigans. it should be in friedberg's book as an appendix somewhere

[dim epoch]

oh i remember covering that in my numerics class

okay i dug it out and we only did that for normed vector spaces

[little crater]

Well thanks, I'll have a look around these topics. I feel like the approach will depend on how I define "closeness"

[dim epoch]

hmm i feel like you might need some more structure to rigorously speak of distance

but maybe i'm just not aware of some other way of doing it

[little crater]

maybe a %error on each b entry is to be minimized when that cumulative %error is the least?

although % is weird depending on the direction taken (i guess you say you are up or down % from the original b entry value.)

[wintry steppe]

i'm assuming brandon's working with real-entried matrices

so there's already inner products to work with

but in generality, you need to be in a normed space or an inner product space to rigorously speak of distances and angles and what not

("normed space" is distances, "inner product space" is distances plus angles)

[little crater]

Well tbh I am not working with anything but sure we can say real-entried matrices. The thought just occurred to me when watching a video and I was just curious if there was a name or topic on that issue.

[dim epoch]

if it's about R^n i can give you a method

uuh let me try to find the english word for "Lineares Ausgleichsproblem"

let me make sure that i'm not spouting bullshit first, is something along the lines of this what you're looking for? @little crater

[little crater]

ngl I am not sure what this means.

[wintry steppe]

that looks like what i was thinking of

"x* minimizes |Ax - b|"

[little crater]

Oh

[wintry steppe]

if Ax = b is a consistent system, then it's the (a?) smallest solution to Ax = b ignore this, i misinterpreted argmin.

but if it is not a consistent system, then it merely minimizes |Ax - b|

[dim epoch]

yeah, this concerns itself with systems that aren't solvable exactly

[little crater]

and minimize in this context refers to what exactly? The size / length of that vector x*?

such that it would be a close to zero in length as possible

[dim epoch]

it minimizes the distance

[little crater]

okay

[dim epoch]

you get a vector that sends your approximate solution to the one you want and takes the norm of that

hmm my class assumed that A is of full rank for the solution to that problem tho

just saw that

actually nvm

this is the equation that solves it

[little crater]

is it possible that you end up with multiple solutions?

[dim epoch]

if A is of full rank your solution is unique

if it isn't it might not be unique

didn't think i'd ever use something from my numerics class

neat

[wintry steppe]

"didn't think i'd ever use that" is something i've said to myself a lot on this server

[dim epoch]

lmao

sadly i don't think i can provide you with a proof of that that isn't german

[wintry steppe]

simply assume we speak german

[dim epoch]

[quartz compass]

I think it follows by just taking the derivative and setting it to 0

[dim epoch]

let me look it up in the numerics bible

okay so basically it's about wanting Ax-b to be orthogonal on Im(A)

then it follows fairly easily

Bild(A) is Im(A)

[quartz compass]

yeah what I said can't work I'm assuming positive (semi?) definite lol so would only give a special case

[dim epoch]

btw one thing to note is that this picture assumes using the 2-norm

since it gives a nice geometrical and significant statistical meaning

but generally you can use other norms on R^n for that as well

[little crater]

Well I will put that on the list of "things to look up". Thanks for the insight

[dim epoch]

you're welcome

[robust owl]

So, I need to show for T a linear transform from V to V, W a subspace of V and T_W the restriction of T to W, that W being T-invariant implies T_W is linear. But I don't understand why we need the T-invariance assumption? Given any x,y in W and c a scalar we know cx+y is in W also by closure. Since, T is linear we already know T(cx+y)=cT(x)+T(y), where T(x)=T_W(x),T(y)=T_W(y) and T(cx+y)=T_W(cx+y) by definition of T_W. Am I missing something?

It seems like this ought to guarantee us that T_W is linear regardless of whether W is T-invariant.

[wintry steppe]

you need to be able to restrict T to a map from W to W

you can obviously just restrict it to W and get a linear map W -> V. but you care about linear operators, so you want to know whether or not this map has its image contained in W.

once you know that, you can also restrict the codomain and get a true map from W to W

[robust owl]

Sorry if this is silly but why do we need T to be restricted to W?

Actually wait

Okay, so in my set theory book the restriction of f to a subset A of the domain of f didn't actually have any restrictions on the codomain of f (besides being a subset of the codomain of f obv). But in FIS the restriction of T to W is a map from W to W.

Is this just a convention difference? Lol

[wintry steppe]

you're right, restricting the domain of functions usually doesn't affect the codomain

FIS is abusing the terminology a little, and calling the map W -> W obtained by restricting both the domain and codomain of T the "restriction"

when it is actually the restriction of T to W, with its codomain restricted afterwards

[robust owl]

I see

I think I was also reading it kinda too sloppily. FIS only even defines the restriction of T to W when W is a T-invariant subspace of V.

[lusty falcon]

how is the operation of this

[robust owl]

Are you asking how they got the bottom line? It's from doing vector addition and scalar multiplication.

(1-t)(1)+t(2)=1+t and so on for the other two components.

[lusty falcon]

thanks :333333

[dreamy charm]

how would you verify that the example of a linear map from R3 to R2 is linear?

How would you show additivity and homogeneity?

[copper hinge]

Show T(av + bw) = aT(v) + bT(w) for scalars a and b and v,w in R^3.

For sanity check let v = (v_1, v_2, v_3), same with w.

Also it’s R^3 to R^2, not what you said.

[lusty falcon]

name of book

[dreamy charm]

Thank you

[dreamy charm]

Axler’s linear algebra done right

[wintry steppe]

hey! anyone got a second about subspace/basis problem?

I got the right answer, but the book gives it as a 2D vector, and I have 3D?

And I just don't understand why?

no one can tell you why if you don't tell us what you actually did

oh my bad

on #5 I put into a matrix and RREF it

I got a row of all 0s on the bottom

[dusky epoch]

what's the goal here

[wintry steppe]

To find the B coordinate vector of x

[dusky epoch]

uh huh

and your work?

[wintry steppe]

sorry for sideways

[dusky epoch]

,rccw

[wintry steppe]

is <0.25, -1.25,0> the same as <0.25,-1.25>?

[dusky epoch]

no, it is not.

[wintry steppe]

I didnt think so...

So is it because the bottom row is all 0s that it becomes 2D/

[dusky epoch]

let's doublecheck your arithmetic just to be sure

,w rref {{1,-3,4},{5,-7,10},{-3,5,-7}}

[wintry steppe]

I used a rref calculator tbh lol

[stoic python]

Results provided by WolframAlpha

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

Requested by Ann#0413

[dusky epoch]

okay

[wintry steppe]

thats pretty cool you can do that though

[dusky epoch]

you really shouldn't be using decimals unless explicitly instructed to, btw.

anyway

[wintry steppe]

oh okay

[dusky epoch]

you are solving a linear system of three equations in two unknowns.

since there are two unknowns, the solution will be a vector of size two, not three.

[wintry steppe]

oh okay thanks

there are only 2 unknowns because there are only two b vectors?

[dusky epoch]

the two unknowns are the coefficients on b_1 and b_2, yes.

[wintry steppe]

Thanks! Im still learning! going over the stuff a second time really helps

[winter geode]

how is this row reduced

isnt it by definition that any row with one non zero term has all other terms as 0

[wintry steppe]

can you write down the definition of "row reduced matrix"?

[winter geode]

an mxn matrix is row reduced if: 1) the first non zero entry in each non zero row of the matrix is 1. 2) each column of the matrix which contains the leading non zero entry of some row has all its other entries as 0.

oh

[wintry steppe]

by that definition this looks pretty row reduced to me

i think you got 2 confused

[winter geode]

yep

sorry fr clogging the chat with what seems 2 be a misunderstanding 😛

thx tterra

[wintry steppe]

don't worry

we all mess up definitions sometimes

[prisma socket]

I am not sure how to calculate it for matrix n>=3

it looks pretty obvious that the answer is somewhat a^n+b but i am not sure

[copper pelican]

Induction should work

[dusky epoch]

$D = aI - B$ where $B$ is the matrix with $1$'s on the subdiagonal and $-b$ in the top right corner.

[stoic python]

Ann

[dusky epoch]

$B$ is the companion matrix for the polynomial $x^n + b$.

[stoic python]

Ann

[hard drum]

Here as well you can expand along the top row, since you just get two triangular matrices whose determinants are easy to compute

[dusky epoch]

or that

[prisma socket]

lol I just realized that it is a triangular matrix and for some reason I thought its hard to calculate cuz its not a diagonal matrix

but its the same determinant as a diagonal matrix

thanks

[dusky epoch]

it isn't strictly a triangular matrix

[fringe fjord]

But all the minors corresponding to nonzero elements of the first row are triangular.

[prisma socket]

yes, I was talking about the minors

[hard drum]

Though I suppose the minimal polynomial method is pretty nice too!

Well it's easy enough to see (i guess) that the minimal polynomial of B with 1 on the subdiagonal and -b in top right hand corner is x^n + b by considering the successive images of e1 under x -> Dx and then by Cayley Hamilton and degree reasons that must coincide w the char poly up to signs I guess?

Nice

[restive crane]

For this question, to get the dimension of a subspace, I need to find a basis for the subspace and then see how many vectors are in the basis. To obtain a basis, would I just row reduce this matrix?

[fringe fjord]

That's one way.

You can also, less systematically, consider whether there are any non-trivial choices of a,b,c,d that lead to a zero vector.

Due to the last row we must have d = 0.

Then due to the second row we must have a = 0 too.

Then there's just -3b+6c = 0 and b-2c = 0 left. But both of these are clearly satisfied as long as b=2c. So we can exclude one of the "b" and "c" vectors from the basis.

[restive crane]

okay, but if we just row-reduced, the dimension would've just been the number of free variables right?

[fringe fjord]

Yes, that should give the same result.

[edgy shale]

Can someone please translate this to english?

When its aids “set of continuous real-valued functions” does it mean that it literally a collection of continuous functions that map values from 0 -> 1 to 0->1?

Also I never seen the notation R^[0,1] only R^n, so I’m not entirely sure what the hell that means either

[hard drum]

Given sets A and B, by A^B we mean the set of functions B-> A. R^[0,1] is a special case of this notation (and indeed so is R^n if we view n-tuples as maps from an n element set into R, but I digress...)
I'm not sure how to "translate it to English" besides expanding it - we're talking about the set of (all) continuous functions [0,1] -> R

Not just "a" collection and what you suggest is the set of continuous maps [0,1]->[0,1], whereas here they mean continuous maps [0,1] -> R

[edgy shale]

A^B basically means if a function f(x) maps a element in set A to set B it is consider to be in A^B?

[hard drum]

No, it's the set of functions B -> A, not the other way round.

[edgy shale]

so for example $R^{[0, 1]}$ can be thought of a function $f$ that takes elements in $[0, 1]$ to some real number $R$?

[stoic python]

ModernNoob

[edgy shale]

pls dont give up on me

[fringe fjord]

No, R^[0,1] is the set of all such functions.

It is a vector space if we take the "obvious" ways to add functions (f+g is the function that maps x to f(x)+g(x)) and multiply functions by constans (c·f is the function that maps x to f(x)·c).

The sentence you originally quoted says that the subset consisting of those function that are also continuous, is a subspace.

[edgy shale]

so $R^{[0, 1]}$ is a set of function $f$ that maps $[0, 1]$ to $R$? and so that set of functions can be considered a subspace of $R$?

[stoic python]

ModernNoob

[fringe fjord]

Yes to the first part. No to the second.

The vector space here is $\bR^{[0,1]}$ itself. \ The subspace is ${ f \in \bR^{[0,1]} \mid f\text{ is continuous} }$.

[stoic python]

Troposphere

[edgy shale]

oh so $R^{[0, 1]}$ is a sets of function $f$ that maps $[0, 1]$ to $R$ and the set of continuous function $f$ includes only those function in $R^{[0, 1]}$ that are continuous

[stoic python]

ModernNoob

[fringe fjord]

Right.

Minor correction: R^[0,1] is the set of functions [0,1] -> R.

[edgy shale]

also, is it possible to graph a vector of $R^{[0, 1]}$ like how we can put a vector of $R^2$ in a coordinate plane or is it just an abstract object (not sure how to put it)?

[stoic python]

ModernNoob

[uneven raptor]

graph a function from [0,1] to R

that's a vector in this vector space

[edgy shale]

so literally just f(x) = x, {0 < x < 1}

[uneven raptor]

haha no

(it's undefined at the endpoints but you get the idea)

[edgy shale]

oh yea lmao, ty

so i learn that vectors arent just arrows today

[edgy shale]

I want to go back to what potato said in the parenthesis. So $R^n$ also can be a considered the set of functions $f$ that maps $[a_0, ..., a_n] \rightarrow R$?

[stoic python]

ModernNoob

[fringe fjord]

More precisely $\bR^n$ can be considered to be the set of all functions ${1,2,\ldots,n} \to \bR$. Note that the domains of these functions consist only of the integers from $1$ through $n$.

[stoic python]

Troposphere

[wintry steppe]

first I had no idea \mathbb is interchangeable with \mbb, now I see \b works as well. wtf?

[tribal willow]

most likely it's macros

[wintry steppe]

$\bR^n$

[stoic python]

Renegade

[tribal willow]

i stand corrected i suppose

[wintry steppe]

ahahha

wait, you're the guy I see in #real-complex-analysis right

[tribal willow]

im there sometimes

[limber sierra]

it is

it's part of the default texit preamble

it worked for @wintry steppe because presumably their preamble is set to the default

,preamble

[stoic python]

Your current preamble. Use ,texconfig to see the other LaTeX config options!

No custom user preamble set, using the default preamble.

% Required to support mathematical unicode
\usepackage[warnunknown, fasterrors, mathletters]{ucs}
\usepackage[utf8x]{inputenc}

% Always typeset math in display style
\everymath{\displaystyle}

% Use a larger font size
\usepackage[fontsize=14pt]{scrextend}

% Standard mathematical typesetting packages
\usepackage{amsfonts, amsthm, amsmath, amssymb}
\usepackage{mathtools}  % Extension to amsmath

% Symbol and utility packages
\usepackage{cancel, textcomp}
\usepackage[mathscr]{euscript}
\usepackage[nointegrals]{wasysym}

% Extras
\usepackage{physics}  % Lots of useful shortcuts and macros
\usepackage{tikz-cd}  % For drawing commutative diagrams easily
\usepackage{color}  % Add some colour to life
\usepackage{microtype}  % Minature font tweaks

% Common shortcuts
\def\mbb#1{\mathbb{#1}}
\def\mfk#1{\mathfrak{#1}}

\def\bN{\mbb{N}}
\def\bC{\mbb{C}}
\def\bR{\mbb{R}}
\def\bQ{\mbb{Q}}
\def\bZ{\mbb{Z}}

% Sometimes helpful macros
\newcommand{\func}[3]{#1\colon#2\to#3}

Page 1/2 · Jun 20, 2022, 3:09 AM

[wintry steppe]

oh, perhaps. forgot I customized my preamble haha

,preamble

[stoic python]

Your current preamble. Use ,texconfig to see the other LaTeX config options!

No custom user preamble set, using the default preamble.

% Required to support mathematical unicode
\usepackage[warnunknown, fasterrors, mathletters]{ucs}
\usepackage[utf8x]{inputenc}

% Always typeset math in display style
\everymath{\displaystyle}

% Use a larger font size
\usepackage[fontsize=14pt]{scrextend}

% Standard mathematical typesetting packages
\usepackage{amsfonts, amsthm, amsmath, amssymb}
\usepackage{mathtools}  % Extension to amsmath

% Symbol and utility packages
\usepackage{cancel, textcomp}
\usepackage[mathscr]{euscript}
\usepackage[nointegrals]{wasysym}

% Extras
\usepackage{physics}  % Lots of useful shortcuts and macros
\usepackage{tikz-cd}  % For drawing commutative diagrams easily
\usepackage{color}  % Add some colour to life
\usepackage{microtype}  % Minature font tweaks

% Common shortcuts
\def\mbb#1{\mathbb{#1}}
\def\mfk#1{\mathfrak{#1}}

\def\bN{\mbb{N}}
\def\bC{\mbb{C}}
\def\bR{\mbb{R}}
\def\bQ{\mbb{Q}}
\def\bZ{\mbb{Z}}

% Sometimes helpful macros
\newcommand{\func}[3]{#1\colon#2\to#3}

Page 1/2 · Jun 20, 2022, 3:10 AM

[wintry steppe]

oh yeah, it's the common shortcuts

[azure ether]

I want to get better at Linear Algebra, I have an elementary linear algebra course I took back in 2019 but to be honest I don't remember most of it ( for example, I forgot what a basis vector is). Anyone have any good recommendation for online sources/courses or a textbook (I remember the one I used back then was not that great).
Thanks in Advance.
Note: I am interested in Deep Learning and Computer Vision which is where I will apply all my Linear Algebra knowledge in!

[keen sierra]

you can always check out linear algebra done wrong, which has the virtue of being free and pretty good: https://www.math.brown.edu/streil/papers/LADW/LADW.html

among conventionally published books, the one by friedberg, insel and spence is probably the best introduction for most purposes

[copper pelican]

There’s a list of common linalg recommendations (and a brief review) in #book-recommendations too

[errant mist]

If one wants to take the largest coordinate of a vector such as an eigenvector does one typically use infinity norm for this?

[wintry steppe]

What do you mean by largest coordinate?

[errant mist]

element in the vector which is the maximum

[wintry steppe]

Well I guess infinity norm work then

[errant mist]

thanks

[tranquil steeple]

and if it is an eigenvector, then the actual maximum value depends on your scaling of the vector.

[errant mist]

yes, in this case its a symmetric matrix then there must exist an orthonormal basis.

[fringe fjord]

There's the caveat the the infinity-nor might also be giving you the magnitude of the smallest coordinate.

[errant mist]

@fringe fjord could you elaborate?

[fringe fjord]

The infinity-norm of (0.6, -0.8) is 0.8 even though the largest coordinate is 0.6.

(This may or may not be a problem for you in the context you're going to use it in).

[errant mist]

oh, right thats true. good point. Actually, that clear things up quite a bit as one may take the infinity norm then take the absolute value of that value to construct a largest element in a given vector.

[wintry steppe]

If −7<x≤6
determine a and b for the inequality :
a<3x+8≤b

[wintry steppe]

Can anyone help me out with this?

[quiet salmon]

I don't think this is the appropriate channel for your question, try #prealg-and-algebra

[wintry steppe]

Okay mate

[quiet salmon]

guys I'm having a hard time verifying that $U = {(x_{1}, x_{2}, x_{3}, x_{4}) \in \bR^{4}: x_{3} = 5x_{4} + b}$ is a subspace of $\bR^{4}$ if and only if $b = 0$. how does this work?

[stoic python]

texaspb

[quiet salmon]

$b \in \bR$

[stoic python]

texaspb

[fringe fjord]

Is it the "if" or "only if" direction that's giving you trouble?

[quiet salmon]

tbh both because I don't know how to start it 😢

[dusky epoch]

do you know in general how to verify that something is a subspace?

[fringe fjord]

Okay, then start with the "if" direction.

[quiet salmon]

well I have to see if 0 \in U, and if it's closed under addition and scalar multiplication right?

[fringe fjord]

Yes.

[dusky epoch]

yes

[quiet salmon]

ok but how do I get a confirmation of that when he gave me this rule

x3 = 5x4 + b

is what I don't get

[dusky epoch]

can you write out what the zero vector is?

[quiet salmon]

yeah it's like (0, 0, 0, 0)

[dusky epoch]

is it ""like"" (0,0,0,0) or is it exactly that?

[quiet salmon]

ooh

in this case it would be

(0, 0, 5*0 + b, 0) right?

[dusky epoch]

no, that is not the zero vector.

my last question may be considered somewhat of a trick question by some; all i wanted was for you to be more precise in your wording.

the zero vector is indeed (0,0,0,0). and what happens when you plug that vector into the equation x_3 = 5x_4 + b?

[quiet salmon]

b = 0

hmm

that makes sense

[dusky epoch]

note however that this is going in the opposite direction to what tropo suggested

namely we proved that if U is to be a subspace of R^4, then b has to be 0

[quiet salmon]

ok that's the if part

[dusky epoch]

no, it's the only if part.

[quiet salmon]

wait really

[dusky epoch]

it may be somewhat confusing.

that's why personally i prefer to speak of directions in an iff as => and <=

and we proved the => direction

[quiet salmon]

now I should suppose that b = 0

ok that's perfect, thanks for the help

I think it's gonna be tricky to check if it's a subspace

supposing b = 0

[velvet moss]

how so

[dusky epoch]

it should not be tricky

[velvet moss]

it’s pretty clear

[dusky epoch]

just walk thru all three parts of the definition

[quiet salmon]

okie

[wheat sorrel]

Linear algebra is DUMB

[dusky epoch]

[neon laurel]

I've found AB =BA=8I

Idk what to do further

Pls help me

[fringe fjord]

This means that B^-1 = (1/8)A.

[neon laurel]

How?

B^-1 = A/8 right?

[fringe fjord]

No.

After edit: yes.

[neon laurel]

Right nw?

What I've to after that

I've to solve for variables

[fringe fjord]

Notice that B is the coefficient matrix in the system you're being asked to solve so the equation is BX=[4,9,1]^T.

[neon laurel]

BX = C

Right?

X = B^-1 C

[fringe fjord]

Sure, if you call [4,9,1]^T C.

[neon laurel]

??

[neon laurel]

Right?

Or A/8 × C

Am I Right?

[fringe fjord]

Yeah.

[neon laurel]

Thank you so much dude ;)

[thorn siren]

am i allowed to ask a linear algerbra question here or should i only use a helper room?

[wintry steppe]

you are allowed to ask here

[thorn siren]

A trader has two types of coffee, one coffee costs SEK 13.50 / kg and the other coffee costs SEK 10.50 / kg. Of these two coffees, he wants to make a coffee blend that weighs 50 kg and can be sold for SEK 11.30 / kg. How much coffee should he take from each variety to meet this?

[wintry steppe]

what are your thoughts

what have you tried to begin with?

easy

he might just be asking for us to do it for him

[tranquil steeple]

100% Lindvalls

[wintry steppe]

why not ping them and ask

if it's their homework, they should do it themselves

[fringe fjord]

(Just because the ingredients cost 11.30 kr per kg of finished product doesn't necessarily mean it can be sold for that price.)

[eager burrow]

ah it's the exercises from das kapital

[velvet moss]

this is what I’ve tried so far

I know I need to show c1=c2=c3=c4=0

or that the matrices are linearly independent

but I have no idea how to do that because they are matrices

[fringe fjord]

For this particular purpose, a matrix can be thought of as a 4-element vector written funnily.

[velvet moss]

that works?!?

how does that make any sense

[fringe fjord]

Because you're here viewing the spaces of matrices as a vector space, meaning that you remember how to add matrices and how to multiply a matrix by a scalar, but you're not caring about multiplication of two matrices.

[velvet moss]

oh ok

thank you

[green trench]

If you have a set of vectors (v1,v2,v3…vn) if you prove a part of it like (v1,v2,v3) are linearly dependent would that mean the entire set is linearly dependent?

Cause if the first three have non trivial solutions then the rest could all have solutions = to 0

And since the first 3 are non trivial it wouldn’t matter for the rest

?

[dusky epoch]

misusing the word "solutions" here

but yes, any superset of a linearly dependent set is also linearly dependent

[copper pelican]

I think you can prove it directly without needing induction

Think about how the transpose operations behaves with respect to sums and products of matrices

[dusky epoch]

alternatively, you can show that Ax = 0 for all x ∈ R^n

[versed parrot]

what is the determinant of a block matrix if all the component matrices have determinant 0?

[wet stratus]

Doesn't tell us anything. Let (e1, e4, e3, e2) be the 4x4 matrix with the standard unit vectors as columns. Then each 2x2 block matrix has determinant 0 but the matrix is invertible.

On the other hand for example for the zero matrix every determinant is 0

[dusky epoch]

@versed parrot are you dealing with an arbitrary block matrix or maybe a block triangular one or something

[wintry steppe]

Hi! are there any online courses on linear algebra?

[nimble kiln]

what level? Introductory?

[vague crane]

mit ocw has one iirc, there should also be a lecture series to go with strang's 'Linear Algebra and its Applications', and there's textebooks everywhere

[wintry steppe]

Yup

[wintry steppe]

I don't think I can use a textbook, I will try the mit ocw

[nimble kiln]

I used Khan academy, myself

[vague crane]

textbooks are the best way to learn em

just putting that out there

[wintry steppe]

I tried that too! They have 114 videos it was a bit intimidating

[wintry steppe]

I agree

[nimble kiln]

are you learning for a particular purpose or wanting to learn thoroughly?

[wintry steppe]

Not thoroughly

Just want to learn them at HS level

[nimble kiln]

https://www.khanacademy.org/math/linear-algebra

there's a lot less material here, but it gives you the basics

[wintry steppe]

Oh

[nimble kiln]

3Blue1Brown also has a great series on linear algebra, but its more of an overview than something to learn from

[wintry steppe]

I think this will be sufficient

I think Khan academy's course is more aimed towards HS students

[nimble kiln]

I'm starting on a couple books currently (one after the other, not both consecutively) as part of a reading list a youtuber I know of put out. First one is Linear Algebra Done Right, which has corresponding videos

although for a first dive into LA, its probably a bit much tbh

[wintry steppe]

Yeah

@nimble kiln Just to be clear this linear algebra course is similar to an introductory LA course in undergrad right?

[nimble kiln]

the Khan academy one?

its probably a bit stripped back, but I'm not entirely sure

[wintry steppe]

Oh

[nimble kiln]

its enough that you can move on, unless you need some quite abstract stuff

[wintry steppe]

Thank you! I appreciate your help

[nimble kiln]

if you do ever want to go deeper, LADR

[vague crane]

LADW is better

[nimble kiln]

LADW is a thing

[wintry steppe]

What is that?

[vague crane]

ignore that

if you ever wanna go deeper

[wintry steppe]

What is LADR

[vague crane]

check this post out

#book-recommendations message

[velvet moss]

I’m honestly struggling on how to even begin this one

so I can let T be a linear operator on V

then the chain of transformations are as follows

W -> U^-1 -> V -> T -> V -> U -> W

[dull spoke]

Prove bijection

[copper pelican]

Also note that we’re looking at a function that sends an operator to an operator

The objects are the operators themselves

[velvet moss]

I believe I got it now thanks for the help guys

[wintry steppe]

as long as you're okay getting the strangest possible treatment of anything even vaguely related to determinants

[dusky epoch]

determinants are EEEEEEEVVVIIIIILLLLLLL tho arent they

[vague crane]

https://www.axler.net/DwD.html

[torn stag]

no determinants are not evil

unless you think change of variables for integrals is evil

and exterior algebra

[vague crane]

tell that to axler ¯_(ツ)_/¯

[torn stag]

yes I think it's good to introduce det as a multilinear map

it makes it clear where the formula comes from

axler doesnt seem to do that

[velvet moss]

I’m struggling to understand how a dual basis is defined

can someone enlighten me?

[torn stag]

@velvet moss $\omega_i(e_i) = 1$, $\omega_i(e_j) = 0$ for $j \neq i$.

[stoic python]

IlIIllIIIlllIIIIllll

[torn stag]

If you apply $\omega_i$ to a vector $v$, it picks out the component of $v$ associated with $e_i$.

[stoic python]

IlIIllIIIlllIIIIllll

[velvet moss]

ok I think I get it

[quartz compass]

I think of the dual basis as the simplest choice of basis you could pick

"take your basis vectors, put them in a matrix, ok now what's the inverse matrix to that look like..."

[velvet moss]

it’s similar to a transformation that brings each Ta_i = b_i

right?

[velvet moss]

I see

[torn stag]

The dual basis gives you a way to write down components of vectors

I guess it is the default basis used for the dual space

[green trench]

if a triple scalar product is not zero for three R^3 vecors does that mean they are linearly independent?

[quartz compass]

yup

it's the same thing as taking the determinant

[hexed urchin]

I'm stuck with the basis for the range on this one.

I decomposed the right matrix as follows:

I then thought that this would be a basis for the range which is the same span as the standard bases for R^2 so I just said that the standard bases for R^2 would suffice for the range of T. Does that make sense or should I have looked at it differently?

[hexed urchin]

I know this was already answered, but the triple scalar product gives a volume, so if you imagine there being 0 for the triple scalar product it kind of begs the question, "how can the vectors be situated such that the volume can be zero"

[green trench]

on the same line through the origin?

[hexed urchin]

That's possible. You may also get 2 vectors that span out a surface, but if even 1 of your vectors is solely lying on the surface, you could end up with no volume regardless

like a sheet of paper. Some surface area, but still no volume

[green trench]

i see

[hexed urchin]

yes - the third adds no new information to the span so you might as well represent it as a vector in R^2

[green trench]

and that would create a free variable

RIGHT?!

[hexed urchin]

yeah because in that case you would be able to set up a 3x3 matrix with each vector being a respective column and some combination of two vectors would be able to make the third zero

[glacial terrace]

The norm of a matrix A is the maximum eigenvalue of A?

||A|| = max(v_i) if v_i are the eigenvalues of A?

||A|

[wintry steppe]

$$\begin{pmatrix} 0 & 1 \ 0 & 0 \end{pmatrix}$$ has $0$ as its only eigenvalue, but its operator norm is $1$. you need to assume something like "$A$ is symmetric"

[stoic python]

TTerra

[wintry steppe]

for if A is symmetric, you may assume it is diagonal after picking an orthonormal basis of eigenvectors

in which case the computation is straightforward, and you have what you want

@glacial terrace

[glacial terrace]

Interesting, thank you!!

[little crater]

Okay I am going to be honest I don't have any clue how to approach this other than it being a theorem in my book :/

the lecture was all over the place so it wasn't well explained

[quartz compass]

I'd start here $(Ux)\cdot (Uy) =x^TU^TUy = \cdots$

[stoic python]

Merosity

[little crater]

Okay

Is that because

v dot u would be v^T *u

like that is how the dot products work with regarding a single column vector