linear-algebra

[stuck tendon]

Not sure what you mean ;o
u is in the kernel, since Pu = Iu - u(u^tu)=u-u=0, since u is a unit vector

[zinc timber]

null space of T- λ I

not null space of T

[wintry steppe]

That's what I mean lol

So how exactly do i show the nullity here?

It's a previous years exam

I have the answer, so I know it's 1

[zinc timber]

show null space is span{u}

[stuck tendon]

^, by showing that u is in ker(P), and knowing that ker(P) is a subspace, we have span{u} is contained in ker P. Now show the reverse containment

[wintry steppe]

It's the same as the solution lol

I'll keep thinking :)

Thank you for your help 🙏

Ahhhhhhhh

[zinc timber]

[wintry steppe]

This matrix is projecting to the Orthogonal space

[zinc timber]

yess

[wintry steppe]

So then kern p = Span u

[zinc timber]

yes

but try to argue that

[outer goblet]

If were trying to find a equilibrium vector E for a 2x2 matrix A for a markov chain, we can solve the equation EA=E

just set E=(a, 1-a)

but what would E be if A was a 3x3 matrix? or higher ?

[zinc timber]

eigen vector corresponding to ev 1

[outer goblet]

yeah but wouldnt solving it as en equation be easier?

[zinc timber]

you will get an equation

[outer goblet]

finding eigenvectors for matricies with so many decimals is pain when i dont have a calculator to solve for eigenvectors

[zinc timber]

don't have any other method

[outer goblet]

:(

[cursive ginkgo]

hi, i have A a n* n matrix so that : A^2(A − I𝑛)^2 = 0 (I𝑛 the n*n matrix with 1 on the diagonal)
knowing that (A − I𝑛)^2 ≠ 0 and A(A − I𝑛) ≠ 0.

I'm looking for the specter of A, I know that spec A ⊆ {0,1} but i don't know what to say else

[pliant kayak]

you mean the spectrum right?(It‘s just so that I can help you, not tryna correct)

I think you can follow, that A^2 has to be the 0-Matrix

But because A(a-In)!=0 is A!= 0

So I‘d say that A is nilpotent

Do you know what the spectrum is?

[wet stratus]

are you sure that A^2 has to be the 0 matrix?

unless I'm making a mistake we could for example have
A=
(1,1,0;
0,1,1;
0,0,1)
edit: I think I made a mistake here

or we can have
A =
(1,1,0;
0,1,0;
0,0,0)

so edit: ~~both spec A = {1} and ~~ spec A = {0,1} are possible

A={0} clearly not as then A is the 0 matrix

[pliant kayak]

I said that, because this (A^2(A − I𝑛)^2 = 0) has to hold right?

But still I'm not sure

[wet stratus]

yes but this doesn't mean that A^2 has to be 0

if 0 is not in spec A, then A is invertible and therefore from A^2(A-I_n)^2 we get (A-I_n)^2=0 which is a contradiction

so yeah spec A = {0,1} is the only possible option

[bleak elm]

What are some good resources for learn linear algebra ?

[dreamy iron]

#linear-algebra message

I am once again asking this playlist be pinned.

On a different note, Im looking like five practice problems proving such-n-such is a inner product, ranging in difficulty from easy to medium .

[wintry steppe]

proving something is an inner product
medium

[wintry steppe]

do you know of any more advanced linear algebra playlists? which cover topics like quotient spaces, dual spaces, tensor product, annihilators

[dreamy iron]

That is an excellent question

[zinc timber]

I can't understand the image

[ripe shale]

For A a square invertible matrix, we have Ax ≤ b, then does x ≤ A^(-1) b?

[dusky epoch]

what's your definition of ≤?

elementwise?

[ripe shale]

yes

[dusky epoch]

then no

take A = -I

-x ≤ b is not the same as x ≤ -b

[ripe shale]

So do you flip the sign?

[dusky epoch]

well in this particular case yes but in general Ax ≤ b does not imply any relation between x and A^-1 b

[ripe shale]

What if A is a positive matrix

[dusky epoch]

hm

not so sure

[wet stratus]

I know that we looked at M-matrices in numerical analysis for PDEs which have some properties which look similar to this. so I would assume that just being positive is not enough

some other key words were inverse monotone and inverse positive matrices

[woven zephyr]

well, in general the answer is going to be no

take any invertible A with positive entries and at least one negative entry in its inverse

then take x to be zero

[ripe shale]

My matrix has all positive entries

[woven zephyr]

right, your A has all positive entries

are you assuming the inverse also has all positive entries?

[ripe shale]

No my bad I didn't read the inverse part please continue

[woven zephyr]

anyway, from here it's simple, you can just pick b to be zero everywhere and one in a specific entry so that A^{-1}b has a negative entry, leading to a counterexample

[ripe shale]

What if b is positive as well

[woven zephyr]

then you can just do the same construction but add a very very small epsilon to each entry of b

and you still get a counterexample, although writing the proof is more of a pain

[ripe shale]

Hmm I see

Here's another question, is there any other positive orthogonal matrix besides the identity?

[woven zephyr]

yes

[ripe shale]

My thought is no because the determinant would be greater than 1

[ripe shale]

Counterexample away

Wait

Permutation Matrix

[woven zephyr]

oui

[ripe shale]

how about a positive orthogonal matrix with all entries non-zero

Now I'm pretty sure it's no

[woven zephyr]

no, and you just need to look at the definition for why

[ripe shale]

Oh yeah it's obvious

[stoic python]

meitar5674

[subtle gust]

Y'all

If a set of n vectors does not form a basis for R^n

Does that mean that those n vectors do not span R^n OR those n vectors are linearly dependent

[native rampart]

Both

[subtle gust]

I understand that a set of 3 vectors forms a basis for R^n if either of the conditions is true

[wet stratus]

3 or n?

[subtle gust]

N*

[native rampart]

They don't span R^n because the set is linearly dependent

[subtle gust]

Ah i see

There's just smthng that i'm failing to understand

Evertime it's explained to me i just don't get it

When we test a set of vectors for spanning

Why is it enough to show that the det of the matrix whose columns are those vectors is zero

To prove that this set does not span R^n

[native rampart]

Well the det=0 implies the matrix is not invertible

[subtle gust]

True

[native rampart]

Which is same as saying linear dependence of columns

[subtle gust]

Oh i see

What if we have 4 vectors tho

And we're checking spanning for R^3

The set could be lin dependent and spans R^3

[native rampart]

Well you remove the redundant vector first

[subtle gust]

We wouldn't know tho

Like if the det is zero

Then yeah sure the set is lin dep

But it may still be spanning R^3

So why do we conclude that it doesn't span R^3 if the det is 0?

[native rampart]

Well you don't do it like that

The test is always with n vectors in R^n

[subtle gust]

Not on my final exams tho 😭

Oh hol on 💀

I just realised

That a matrix with 4 vector columns

Would have no det

[native rampart]

Yes

It's not even square

[subtle gust]

What a dumbass i am

Sorry 💀

Yeah got it now

So if the det is zero (assuming the matrix is square)

The column vectors are lin dep

And there would be absolutely no way they span R^n

[wet stratus]

yes

[subtle gust]

Gr8

Tysmmmm @native rampart and @wet stratus

[wet stratus]

the problem with testing using the det is that it doesn't tell us how linearly dependent the columns are. they could still either span a space of dimension n-1 or they could span a space of dimension 1. either way the det is 0. also calculating the det for huge n takes ages

you could instead also do something like row reducing

[native rampart]

Actually

Efficient det algorithm is literally as fast as row reduction

[wet stratus]

yes sorry I worded it badly

row reducing also takes ages but it at least tells you more than just calculating the det

and can be done even ifthe matrix is not square

[vestal magnet]

Hey, all, how is everyone? I need to prove this:

$dimImT^* = dimImT$ as well as this $dimkerT^* = dimkerT + dimW − dimV$
Can't I just say that $dim(Im(T^)) = dim(Im(T)) dim(F) = dim(Im(T))$ and then use the fact that
dimImT + dimKerT = dimV?

[stoic python]

meitar5674

[native rampart]

Why do you think you can claim that @vestal magnet ?
T is a map from V to W while T* is a map from W* to V*

[delicate crow]

Not sure if this is the right place to ask im working on a question and i end up with this expression and I’m wondering if this simplifies into an expression or not

[dusky epoch]

isnt this (1 + z/n)^n

[delicate crow]

Oh right

Idk how I didn’t see that

Ty hahaha

[robust owl]

How do Insel/Spence, Hoffman/Kunze and Axler compare?

I'm taking a second course in LA next semester. So, I emailed my prof asking for a book to work through over the summer, he recommended Insel/Spence or Dummit/Foote (which seemed odd), but he also said he heard Axler was good.

I bought a copy of Insel/Spence and it looks way thicker and more detailed than Axler.

The prereq for this course is our first semester abstract algebra course, which basically covers all of the groups chapters of Fraleigh.

[wintry steppe]

they cover basically the same material with different points of view

axler restricts to real and complex fields, and avoids determinantal arguments

axler's view of stuff like the characteristic polynomial is pretty strange, since he avoids determinants

the treatment in FIS is much easier to use

[robust owl]

Are they comparable in terms of depth? Like, is one more elementary than the other or are they both roughly the same?

[wintry steppe]

pretty much the same

[robust owl]

Do you have a personal preference for any particular book at this level?

[wintry steppe]

FIS

i don't agree with axler's treatment of determinants and related stuff

it makes sense to try not to use them if you're going to do stuff like functional analysis (which axler does), but for introductory linear algebra it does not matter

[robust owl]

I see I see. I'll probably give FIS a shot. Thanks for the advice!

[past barn]

Would you guys recommend Serge Lang's Linear Algebra book for a first timer?

[zinc timber]

I wouldn't, at least

try #books #book-recommendations

[wintry steppe]

Gilbert strand got a good book

[outer goblet]

can someone explain to me what the anwser is for b

[inner wave]

Could someone explain the jump between the first system of equations to T(\alpha, \beta, 0, \gamma, \phi - \gamma)?

[outer goblet]

it says find A_0

what does [L_A]_B mean

is it just A multiplied by the matrix vector made of B basis

[inner wave]

and L_A is just the linear transformation of A?

[outer goblet]

yeah

[inner wave]

from R3 to R3?

[outer goblet]

yeah given by that matrix

[inner wave]

Never seen that notation before

Could you give a full translation of b

Maybe I could figure it out then

[inner wave]

but could someone still tell me this

[outer goblet]

the rest of the text says just to show that A_0 is invertible, unrelated to acutally finding it

Let L_A: R^3 -> R^3 be given by L_A(x) = Ax

find [L_A]_B

[inner wave]

Isn't that just a do-nothing transformation?

Since A and L_A are the same thing?

[outer goblet]

yes but _B is supposed ot mean A in beta coordiante system

[inner wave]

Oh this is change of basis stuff

Yeah not me man sorry

[outer goblet]

haha its ok

[inner wave]

Good luck

anyway you could help me with my problem?

[outer goblet]

i wish i could :(

[wintry steppe]

what is T

[inner wave]

Transformation

Check the picture, sorry for the lack of clarity. It's the transformation of the vector of size 5, right in the middle of screenshot

[wintry steppe]

i know it's a transformation, i want to know its definition

[inner wave]

Oh my bad one second

[wintry steppe]

when did u learn to use latex?

/where

[inner wave]

That's MathJax let me convert it

Old MathJax too, its from the website

This is scuffed ima screenshot

[inner wave]

Very top

Sorry I'm running on fumes today

[wintry steppe]

the line with T(\alpha, \beta, 0, \gamma, \delta - \gamma) is just them checking that the solution to the system of equations is indeed a solution

[inner wave]

But this question is about proving the surjectivity of the transformation.

[wintry steppe]

ok

that's what they're doing

[inner wave]

Yes

How did they determine that is a possible solution out of the set of infinite possible solutions

[wintry steppe]

it's impossible to know exactly how they landed on that solution

[inner wave]

Okay second question

[wintry steppe]

just looking at it, you already know what a must be, and c shows up a few times, so killing c by setting c = 0 simplifies things a bit

i guess that was their thought process

alternatively, you could write that system of equations in terms of matrices and do all the row reduction nonsense

[inner wave]

is the transformation surjective because the coefficient matrix of the system of equations above that line with the transformation of alpha, beta, etc. has an infinite amount of solutions?

[wintry steppe]

having just one solution is enough for surjectivity

but yes

[inner wave]

I think I am starting to understand

This just seems unintuitive

Thanks for your help

[outer goblet]

for markov chains

if a matrix is regular, it has an eigenvalue = 1

but can markov chain matrix have an eigenvalue of 1, and not be regular?

just checking if eigenvalue 1 is an eigenvalue of the matrix enough to determine if its regular

[slow wind]

Yes that is called the contrapositive

In logic whenever one proposition P implies another thing, Q, then the falsity of Q implies the falsity of P

Wait sorry I misread your question

If it's missing an eigenvalue of 1 then it isn't regular

but imagine a diagonal matrix with one 1, and the rest of the numbers on the diagonal 100

it wouldn't be regular, but it would have an eigenvalue of 1

[little crater]

"If A is an m × n matrix whose columns do not span R^m, then the equation A**x **= b is consistent for
some b in R^m"

I know this is true but I was wondering if it is good enough to say "let b = zero vector"

[fringe fjord]

It's a strange claim because the conclusion seems to be trivially true (as you notice) independently of the "columns do not span" assumption.

[little crater]

I am not sure how to start this problem

I believe it has some connection with

[little crater]

<@&286206848099549185>

[late heart]

A linear endomorphism with kernel 0 is an automorphism right?

i.e. f : V to V, V a vector space

I think it just follows by rank-nullity + dimension counting

just want to make sure I'm not being stupid

[robust pond]

does it work to just do like

Ax=3au_1 +2Au_2

[little crater]

i said let x = 3u_1+2u_2

[robust pond]

this gives x=3u1+2u2 as a solution

yea

[wintry steppe]

on finite dimensional spaces, where you can use rank nullity, yes. it's false in infinite dimensions (consider right shift on infinite sequences)

[robust pond]

so theres a solution

thats sufficient right

[wintry steppe]

i.e. (x_1, x_2, ...) \mapsto (0, x_1, x_2, \dots). this has kernel zero, but is not surjective

[late heart]

thanks!

[little crater]

Ax = A(3u_1+2u_2) = A(3u_1) + A(2u_2) = 3(Au_1)+2(Au_2) = 3y_1 + 2y_2 = b

[robust pond]

i dont get what youre showing here

[little crater]

me subbing in what let x equal

I said

[robust pond]

okay

so you found a solution but

[little crater]

let x = 3u_1+2u_2

[robust pond]

idk i guess im saying

it seems like youre done

[little crater]

yeah

I just didn't know if I did it right

[robust pond]

oh

[little crater]

I just applied

and seemed okay so I guess im done

[molten ivy]

How do I go about finding a unitary matrix P, such that PAP^-1 is diagonal for a given complex matrix A? Simply diagonalising the matrix doesn't yield a unitary P._.

[wintry steppe]

Anyone able to explain this last step?

I wanna replicate this, but I want to know how how this part under the square roots formulated

[limber sierra]

$(d_2 - d_3)^2 = d_2^2 + d_3^2 - 2d_2d_3 = d_2^2 + d_3^2 + (2d_2d_3 - 2d_2d_3) - 2d_2d_3$

[stoic python]

Namington

[limber sierra]

$=(d_2^2 + d_3^2 + 2d_2d_3) - 4d_2d_3 = (d_2 + d_3)^2 - 4d_2d_3$

[stoic python]

Namington

[limber sierra]

so $(d_2-d_3)^2 = (d_2+d_3)^2 - 4d_2d_3$

[stoic python]

Namington

[limber sierra]

take the square root of both sides.

as for why they did this: they wanted to express $d_2 - d_3$ in terms of only $d_2 + d_3$ and $d_2d_3$, since those were computed earlier in the question

[stoic python]

Namington

[limber sierra]

squaring something is a common way to make it positive, so to go from $d_2 - d_3$ to something involving $d_2 + d_3$, considering $(d_2 - d_3)^2$ is a natural first attempt

[stoic python]

Namington

[limber sierra]

and it turns out that, with some algebra, it works out.

[wintry steppe]

Tyty

[tranquil steeple]

Fun fact, that matrix A is diagonalized by DST-2 and eigenvalues are given by f(t)=4+2cos(t)+4*cos(2*t) tj=jpi/3 j=1,2,3

[wet stratus]

it's not always possible. A needs to be a normal matrix. otherwise you are out of luck

[molten ivy]

Well then it is one

The task is to find this unitary P so it has to exist

[stuck tendon]

find ONBs of each eigenspace and write the elements as columns

[subtle gust]

just a quick question y'all

any linear transformation is expressible as a matrix transformation times a vector right?

and the null space of that matrix transformation is the kernel of the transformation

and the column space of this matrix is the range

right?

[wet stratus]

are we assuming that the vector space is finite dimensional?

then ignoring that in general we only have isomorphic spaces etc, yes

[wintry steppe]

only once you've chosen bases of the domain and codomain. the matrix representation depends on the choices of bases, and under the induced isomorphism of your space with F^n, F^m, the kernel and range of your linear transformation get taken to the kernel and range of the matrix representation.

if you are in infinite dimensions, then there are no matrix representations

[sacred palm]

Hey anyone good with matrices here

just a curious question

how can one decide if the matrix is a point line, plane r3 or r4?

looking at the rows?

[quartz compass]

a matrix on its own is just a rectangle full of numbers, so it's not really any of those unless you have some corresponding way of turning it into one

[sacred palm]

When they ask which one does it span

Thats actually the question I think

Does the matrix span point line plane r3 or r4

Im trying to find a question but cannot find one

[quartz compass]

you can look at the span of the rows or columns, but I guess usually people look at the column vectors

probably better to post a screenshot of the actual question being asked

[sacred palm]

Okay i will post it here

?

Or in a ticket

And @ you?

[outer hazel]

2 -3 6 | 0
9 -5 4 | 0
0 0 0 | 0
does it mean it has infinitely many solutions. is it always the case for when all elements in a row are 0s?

[halcyon spindle]

Yes it has infinitely many solutions but it is not always the case when we have a zero row.
Consider an mxn matrix A where m > n. If the row reduced echelon matrix of A has n nonzero rows the m-n rows are zero. Therefore the homogenous system Ax = 0 will only have the trivial solution, so not infinity many solution. But if A is nxn matrix then yes a zero row corresponds to an infinitely many solutions for Ax = 0.

[normal loom]

are the eigenvectors the same
.5 1
1 2

[solar drum]

could someone remind me what this group represents? Here F is a field and p is some prime

[wintry steppe]

Everyone, thank you. Got a def b on my exam.

Tyty for your help

My friends are so sad rn, but I'm good!!!

Ty!!!

[fair wren]

fkin narc xd

[wintry steppe]

Z/pZ

integers modulo p

[solar drum]

thanks

[fringe fjord]

More precisely F_p is the unique field with p elements (which is indeed also the ring of integers modulo p). Note the the F is a fixed part of the notations, not the name of some different field that already exists.

[rain harness]

Hello. I think this is the right channel. I'm trying to a problem for a game called starbase which has in-game coding

I've been trying to solve a problem that masks you to imagine an aircraft flying. The aircraft (P) has the ability to pitch, yaw and roll about it's own axis. and measure it's current orientation. On the bottom of the aircraft is a laser that gives the exact distance of the laser to the ground representing the vector R. My question is asking whether it would be possible to find the vertical distance (D in the diagram) to the ground no matter the orientation to the plane assuming that Q is at (0,0,0) and is the point where the line / laser contact the ground. and P is at (x,y,z). It can also be assumed that the magnitude of R is known.
any assistance in solving the question or any pointers on how to solve and find a general solution it really be appreciated 😄

If just the pitch and roll parts of the question are considered it's relatively easy to solve using cosin unit vectors but the Yaw aspect is what i'm completely stuck on

[errant palm]

hi, i don't get why the basis for the row space and column space are what they are.

[quartz palm]

can someone help me get this system of equations into reduced row echelon form?
5x - 3y - 6z = -4
3x - y - 5z = 5
4x - 2y - z = -13

[halcyon spindle]

You have to first row reduced the coefficient matrix.
For A be the row reduced matrix of that system,

1. The first entry of the nonzero rows of A has to be 1.
2. For the column that corresponds to the first zero entry for each nonzero rows has every other entry for that column equal to 0.

Now for A to be in row reduced echelon form

1. Every zero rows is below all nonzero rows.
2. The column index that corresponds to first nonzero entry in each row has to be increasing from top to bottom.

While you do this you need to apply the same operation to the matrix (-4, 5, -13)^T.

Note if A is row reduced, you can easily get it to row-reduced echelon form by doing row exchanges.

Heres an example of a row reduced matrix.
$$\begin{bmatrix} 0 & 1 & 0 \ 1 & 0 & 0 \ 0 & 0 & 1 \end{bmatrix}$$.

[stoic python]

Plegasus

[halcyon spindle]

You can do a row with row 1 and 2 to get it to row reduced echelon matrix.

@quartz palm

[quartz palm]

ah thanks

i am having trouble getting to that format

can u walk me thru it

[halcyon spindle]

Ok let work through it together.

[quartz palm]

ok

[halcyon spindle]

Your system Ax = b can be matrix by augmented matrix
$$\begin{bmatrix} 5 & -3 & -6 & -4 \ 3 & -1 & -5 & 5 \ 4 & -2 & -1 & -13 \end{bmatrix}$$.

[stoic python]

Plegasus

[wintry steppe]

Whats recommended to take first, this class or calc3, both have calc 2 as a prereq?

[quartz palm]

either is fine

but i heard calc 3 is easier than linalg

[halcyon spindle]

Let row reduced it first, how should we start?

[quartz palm]

uhh multiply the first row by -1/3?

[halcyon spindle]

yep.

Then what?

[quartz palm]

multiply the 2nd row by 1/3

[errant palm]

😔

[halcyon spindle]

Yep, but that is going to take so long. Let try to get all the entries except for the 1 in the first column to 0.

So we went from $$\begin{bmatrix} 5 & -3 & -6 & -4 \ 3 & -1 & -5 & 5 \ 4 & -2 & -1 & -13 \end{bmatrix}$$ to $$\begin{bmatrix} 1 & -\frac{3}{5} & -\frac{6}{5} & -\frac{4}{5} \ 3 & -1 & -5 & 5 \ 4 & -2 & -1 & -13 \end{bmatrix}$$.

[stoic python]

Plegasus

[quartz palm]

hmm

[halcyon spindle]

Now you try to get all the entry in the first colum except for the 1 to 0 by adding to row 2 and 3 by multiplies of row 1.

With that done you can move on to the second row.

[quartz palm]

idk

[halcyon spindle]

Are you working this out on paper?

[quartz palm]

not rn

too tired

[halcyon spindle]

When you get the chance too, reread your notes or even ^ this and work it out.

[quartz palm]

ok

technically in our class we haven't covered rref but i just thought this was easier

[errant palm]

https://cdn.discordapp.com/attachments/540211747613704221/977002752179863582/unknown.png

hi, i don't get why the basis for the column spaces are what it says, can someone pls help?

[quartz palm]

what is theory 4.20

maybe knowing that can help

[tribal willow]

so basically if im understanding this right, T : V -> V and T(W) in W?

then where does the name “invariant” come from

can a subspace be "variant" in the sense that it varies?

[wintry steppe]

right, T(W) is a subset of W

that's the "i.e. if..." part

[tribal willow]

ye

i guess a better question is like

the etymology of "invariant"?

[wintry steppe]

W is "invariant" under T

well

i just restated the book

[tribal willow]

yea

but in the sense that "annihilate" means to send to 0, is there a similar reason for "invariant"?

[late tangle]

hello

[wintry steppe]

what about the usage of the word "invariant" here isn't clear?

[tribal willow]

i think im just trying to understand what it means for W to be "never changing"

[wintry steppe]

"T doesn't map W out of W" "T doesn't change W" "W is unchanged under T" "W is invariant under T"

[little crater]

Can someone comment on my solution for this question?

A. No, because if some vector y does not have solution for Ax = y, this means our column vectors of A don't span all of R^3. This also means we don't have a pivot in every row of our coefficent matrix (theorem 4). This then tells us that we must have some free variables as not all rows contain pivots. Now applying theorem 2, if we assume some z such that Ax= z is consistent, we know from theorem 2 that our z can only have infinite solutions because we have at least 1 free variable.

[tribal willow]

ohhh

ok that did it, thank you

[wintry steppe]

it's just a definition, you don't need to look too deep into it lol

[little crater]

(Theorems 2 and 4 I used)

[wintry steppe]

maybe another way to phrase it is "T fixes W" (note that this isn't "T fixes the elements of W"!!!!!!!)

[sweet sand]

oh, depending on how depth you are going to get multivariable calculus, if you split linear algebra in two parts you should take the second part with Multivariable calculus but definitively not util you know at least the first half.
but In my case I start with topology in Rn and then finish with differential in several variables leaving integral in several variables until Calc 4

[wintry steppe]

if your multivariable calculus class is basic (just stuff like partial derivatives and then a shit ton of integration), you don't need to take linear algebra before it. if it's more advanced (e.g. if the words "implicit function theorem" show up) then you absolutely should take LA first

[tribal willow]

my calc 3 course didn't cover stuff like implicit function theorem; is that something i would find in an analysis course?

[wintry steppe]

i dunno

maybe such a course would be called "multivariable analysis"

"analysis course" means a million different things

[tribal willow]

ah i just meant analysis course as an umbrella term for all courses with the word "analysis" in it

interesting though

[sweet sand]

well yeah you can find that in analysis but for example the book by Marsden mentions it and that book is not very analytical but rather practical(?).

[wintry steppe]

i die inside a little everytime someone uses the word "analysis" and isn't referring to a course that's not just the contents of rudin

[wintry steppe]

which book by marsden? i know the mechanics one

love that one

[sweet sand]

vectorial calculus by marsden & Tromba

[halcyon spindle]

Yes your solution is right.

[little crater]

Thanks for looking

First exam tomorrow and then we have a interview with our prof to discuss our solutions. cool thing is he seems very chill about how we get our solutions. He pretty much says it is open everything but whatever you put on your paper you better defend and be able to explain

Like these interviews are like 20% of our grade with the exam being 10%

Never had anything like that before but seems cool because it is sometimes hard to express what you want to say from a sheet of paper.

[halcyon spindle]

good luck!

[little crater]

We have not got into talking about linear independence and such but it would probably be easier to explain it like that.

[native rampart]

Wrong chat should be #calculus

[delicate crow]

oh mb i didnt realise

[late tangle]

is linear algebra hard? i'm going into it in 3 years and i wanna prepare

[native rampart]

If you just want a basic understanding of linear algebra,you can do it in like 2 months

[late tangle]

wow ty so much

i think i might take some online courses then

is y=mx+b linear algebra?

[native rampart]

That's a linear equation

[late tangle]

o mb

i know how to do best fit y=mx+b

how do you do

that thing

[wintry steppe]

affine

:^)

[native rampart]

This is just Lagrange interpolation

[late tangle]

just Lagrange interpolation 🙄

[uneven raptor]

is it? there are 4 points to fit but only 2 degrees of freedom

[keen sierra]

the word "best" isn't defined in the problem statement, but I assume a least squares fit would be intended

[zinc timber]

(T/F) A is diagonalizable then U is diagonalizable
, where U(B) := AB-BA

any ideas how I might approach?

[fringe fjord]

Without loss of generality, you can assume A to be diagonal.

But A cannot be a multiple of I, because then U(B) would be 0.

Pick the simplest A you can think of that satisfies these conditions, and calculate how U(B) depends on the entries of B.

Wait ... U rather than U(B) should be diagonalizable?

[zinc timber]

I have tried A=E¹¹ and E²²

that gives me digaonalizable operator

[fringe fjord]

Hmm, I mistakenly thought it was the concrete matrix U(B) that should be diagonal.

For U as an operator, the approach is definitely to work in a basis that diagonalizes A.

When you do the calculuations you will see that this basis also diagonalizes U!

[zinc timber]

that's my guess but this approach is kinda bruteforcy

[fringe fjord]

Perhaps, but it works and is pretty simple too.

[zinc timber]

it does but....

[fringe fjord]

Do you need an argument that works for infinite-dimensional spaces?

[zinc timber]

$A=c_i E_{ii}$ then $U(E_{jk}) = c_i E_{ii}E_{jk} - c_i E_{jk}E_{ii} = c_i \delta_{ij}E_{ik}-c_i \delta_{ki}E_{ji}$

[zinc timber]

sure

[fringe fjord]

wdym "sure"?

[zinc timber]

$=c_jE_{jk}-c_jE_{jk}$

[fringe fjord]

One of those c_i should be c_j

[zinc timber]

I would like an argument that works finite dim spaces

oop

[zinc timber]

0?

[fringe fjord]

So U(E)_{jk} = (ck-cj)E_{jk}.

(modulo possible sign errors)

[zinc timber]

what is the finite dimensional argument?

[fringe fjord]

That's what you have just done.

[zinc timber]

oh you were talking abt this one

$(c_j-c_k)E_{jk}$

~looks diagonal to me~

[zinc timber]

typo

[fringe fjord]

Eyup.

[zinc timber]

nah

only thing left is to show it's diagonal

oh

[fringe fjord]

But what your calculation shows is that the E_jk's are an eigenbasis wrt U.

[zinc timber]

ye basis os Ejk so it's doagonal

I thought it's a matrix itself

[fringe fjord]

U would be represented by an n²×n² matrix, if we vectorize its input and output.

[zinc timber]

ye

cool, tks

[wintry steppe]

is a line considered hyperplane?

intro to linear algebra

but i think a line can be a hyperplane if its in like 2d space

[wintry steppe]

Hey guys I got a question, i had a markov chain question. The teacher wanted me to show that the eigenvalue is 1 and what its eigenvector is. i iterated the steady state vector, and said that because its steady the max eigenvalue is 1 and that the steady state is the eigenvector.
The teachers solution shows him getting the actual eigenvalue of one and the other eigenvalues

and then ends the entire solution with the steady state vector
is the steady state vector a scaled eigenvector with eigenvalue of 1?

[jaunty thorn]

do you mean show that the eigenvalue of a stochastic matrix is 1?

[wintry steppe]

yes it was a markov chain question which was a stochastic matrix

[jaunty thorn]

just use the vector (1, 1, ..., 1), that multiplied by the stochastic matrix is (1, 1, ..., 1) so 1 is clearly an eigenvector

not a particularly satisfying proof but it works lol

because the rows of a stochastic matrix add up to 1

so you can figure it out just by inspection

[wintry steppe]

well he like went through the effort of getting the eigenvalues through the det

but that being said, i just figured okay, i got the steady state after 3 iterations of simple matix multplication, its got a steady state the max eigenvalue as one.

[jaunty thorn]

yeh thats a proper way of doing it

[wintry steppe]

is that steady state also the eigenvector?

because he scaled it in the solution from like the eigenvector from the polynomial he got

i made a silly mistake with svd 😦 i thought u held null of a since i knew it held row of a. sigh i feel dumb

def got a b

[jaunty thorn]

wait have you been given a specific stochastic matrix

or is he asking you to show that 1 is an eigenvalue for all stochastic matrices

[wintry steppe]

he gave me a specific one

[jaunty thorn]

ahh right if its small enough then yeah just compute it

[wintry steppe]

2x2 i just hate doing eigenvalues

and eigenvectors

[jaunty thorn]

lmao yeah relateable

good to be able to do though, the bastards come up absolutely everywhere

[wintry steppe]

yeah yeah ill use python

i spent like 2 hours trying to solve that tbh

was the last problem.

because it was the only one i couldnt like even know where to begin at

it was the first thing we went over in class, and i just sat there and plugged it out and said okay this is converging like this

and then i had an oh yeah the steady state shit

okay now im gonna spend time working on programming

thank you mabey

[jaunty thorn]

good luck and have fun

[rain monolith]

hi, im new here. I really need help with a question I have no idea how to approach, where can i post the question?

[tribal willow]

#❓how-to-get-help

[quartz palm]

i dont understand how they got from the last step to the current step

[wintry steppe]

helloo! can anyone explain to me aboute the absolute value and with functions so basicallly f(x)=|x| or f(x)= -|x| or f(x) = |-x|

oops wait wrong chat

[quartz palm]

wym explain

[wintry steppe]

so like graphing them

[quartz palm]

u just input points for them

so say for |-x|

for x = 1

it would be |-1|

[uneven raptor]

just divide the 2nd row by 0.8

[quartz palm]

no i mean like

how do u get 3.8 from R3 - 4R1

[wintry steppe]

thanks

[quartz palm]

np

[uneven raptor]

that row operation is applied to the matrix in the previous picture

which is not shown

[quartz palm]

ty sorry i just wrote the wrong number

[native rampart]

It's a cubic polynomial in the problem.
ok I guess it isn't exactly just Lagrange, because it can't have x^2 and x terms

[fair wren]

let E be a vector space with finite dimension, and f an endormorphism from E such that f² = -id

prove that that f has no real eigen values

[dusky epoch]

@fair wren do you still need help with this?

[fair wren]

yes im still trying to do something haha

i tried doing this

if i prove that f^n has no real eigen values

so is f right

so f has no real eigen values either

no?

[dusky epoch]

"if i prove that f has no real eigenvalues then f has no real eigenvalues either, right?" is this what you're asking?

[fair wren]

i edited

what do u think?

[dusky epoch]

you're overthinking it

[fair wren]

hm really

please tell me

[dusky epoch]

let's start from the basics

[fair wren]

❤️ sure ann-chan

[dusky epoch]

...we are not anywhere near close enough for you to -chan me.

[fair wren]

sowy

[dusky epoch]

anyway, can you tell me what the phrase "f has an eigenvalue λ" means?

[fair wren]

f has an eigen value if the matrix representing it M(f) = A, then there exists lambda such that det(A - \lambda * I_n) = 0

[dusky epoch]

jesus christ

no that's way too complicated

[fair wren]

oh damn

idk sorry

[dusky epoch]

besides, lambda cannot be put under an existential quantifier

[fair wren]

maybe

[fringe fjord]

It's true, but not in any sane development is it a definition.

[fair wren]

f(v) = lambda v?

[dusky epoch]

that's closer

[fair wren]

and lambda is an eigen value

[dusky epoch]

there's something you missed

[fair wren]

yes

v != 0

[dusky epoch]

there exists a nonzero vector v such that f(v) = lambda*v

[fair wren]

[dusky epoch]

now assume f has a real eigenvalue λ

whence derive the existence of nonzero v such that f(v) = λv

now think about what f^2(v) might be

[fair wren]

yeah and f²(v) = lambda ² v² = - v

[gray dust]

where did v^2 come from

its not even a thing that makes sense

[fair wren]

just squaring f(v)?

[fair wren]

its just a 1 x1 matrix squared

no?

[fringe fjord]

f²(v) here means f(f(v)) not f(v)·f(v))

[fair wren]

what what

[dusky epoch]

f^2 is the composition of f with itself

[gray dust]

powers of operators usually denote repeat composition

[fair wren]

im confused with this notation tbh

if its the second composition

then they should put ()

letting it like that is ambiguous

[dusky epoch]

no

[fair wren]

f^(n) means n-th composition of f

f^n is just f(x) * ... * f(x) n time ?

[dusky epoch]

this is an operator on a vector space

[fair wren]

its just confusing for me tbh

hm ok

[dusky epoch]

you can't multiply vectors by vectors

you can't

[fair wren]

there is no dot product then?

[dusky epoch]

dot product would return a number anyway

[fair wren]

hm ok

[fringe fjord]

And a function that takes a vector to a number cannot equal -id.

[dusky epoch]

so you would be unable to unambiguously interpret the "product" of three or more vectors

[fair wren]

hm

ok

[dusky epoch]

anyway, given that v ≠ 0 and f(v) = λv, what is f(f(v))?

[fair wren]

= lambda ² v

= -v

=> lambda² = -1

no real solution

contradiction then

[dusky epoch]

precisely

it is possible to generalize this a bunch

[fair wren]

so lambda ^ n = -1

hm?

[dusky epoch]

no, even more

[fair wren]

hm

tell me

[dusky epoch]

if an operator is annihilated by a polynomial then all of its eigenvalues are roots of the polynomial

i.e. if an operator A: V -> V satisfies p(A) = 0 where p is a polynomial then all eigenvalues of A are roots of p

[fair wren]

How do u get the polynomial p?

[dusky epoch]

what do you mean "get"

it's given

[fair wren]

hm?

[dusky epoch]

it's part of the premise

for you the polynomial was x^2 + 1

[fair wren]

how u got that hm?

[dusky epoch]

?

you were literally told that f^2 + id = 0

[fair wren]

This :d

[dusky epoch]

what is this if not the polynomial x^2 + 1 applied to f

[fair wren]

ah so u treat f as x and id as 1

hm

[dusky epoch]

have you seen the notation of a polynomial evaluated at an operator before

[fair wren]

no

[dusky epoch]

oh

[fair wren]

somehow u treated id as 1 idk why

[dusky epoch]

part of the definition of p(A)

zeroth power of any operator is the identity

[fair wren]

ah ok gotcha

[fair wren]

Im working on proving that the dimension of E is even

since i've proven that the eigen values for f are complex

so if the power of lambda was odd then for example y^3 = -1 there is a real solution y = -1

so the power of lambda should be even

thus dimE should also be even

[dusky epoch]

this is nonsense; the power of lambda is given to be 2

[fair wren]

wat

[dusky epoch]

your problem says f^2 = -id, no?

[fair wren]

yes

wym nonsense?

which part i said was wrong?

[dusky epoch]

what you're talking about seems to be entirely unrelated to the problem at hand

if the power of lambda was odd

who give a shit

the power of lambda is 2. it isn't odd.

[fair wren]

hm

so how do u further proceed from that to say the dimension is even

[dusky epoch]

from what

what you said?

[fair wren]

we can say that dimE = sum dimE_k , E_k are eigen spaces, since we have only 1 eigen value with multiplicity = 2, then dim E = 2?

[dusky epoch]

overthinking it.

[fair wren]

ikr

[dusky epoch]

also we don't necessarily have only one eigenvalue and its multiplicity is not necessarily 1

we wish to prove dim(E) is even

[fair wren]

its multiplicity is 2

[dusky epoch]

no its multiplicitly isn't necessarily 2 either

[fair wren]

Why

how do u know it isn't necessarily 2?

[dusky epoch]

i can give you an operator that satisfies f^2 + id = 0 with eigenvalues of multiplicity 69 if you want

but that's beside the point

[fair wren]

u was going to do smthg like lambda ^69 = -1?

[dusky epoch]

no

[fair wren]

and tell me it isn't necessarily a multiplicity of 2?

so whats ur point?

[dusky epoch]

my point is that your attempt is going completely in the wrong direction

just because f^2 + id = 0 does NOT, i repeat DOES NOT tell you anything about the multiplicity of either i or -i as eigenvalues of f

[fair wren]

okey so what is the direction i shall go thru?

[dusky epoch]

suppose towards a contradiction that dim(E) is odd.
then the charpoly of f will have odd degree (why?) and thus something can be said about charpoly(f) that'll contradict what's been proved previously.

[fair wren]

@dusky epoch

[fair wren]

check what i did up here maybe?

[dusky epoch]

no

you're confusing dim(E) with "power of λ"

[fair wren]

idk then

[dusky epoch]

do you want me to repeat myself or what

[subtle gust]

how would we prove EF and FE are similar

assuming E and F are both invertible

EF=InEFIn

tried every single combination lol

don't see how i can go from EF to FE

[dusky epoch]

$FE = F(EF)F^{-1}$

[stoic python]

Ann

[subtle gust]

🐸 🐸

how could i not notice

tysm

[subtle gust]

how would we prove that the adjoint of the inverse is the inverse of the adjoint

[dusky epoch]

show instead that adj(A)adj(A^-1) = I

[subtle gust]

how would we do that tho

[dusky epoch]

probably using the defn of adjugate

or using the fact that M adj(M) = det(M)I

[subtle gust]

adjugate?

are adjoint and adjugate used interchangeably?,,,,

anyways

[subtle gust]

i still can't see how we would go from this

[subtle gust]

to this

cuz i tried inverting both sides and i didn't reach anything useful

what i got was adj^-1(A)=(1/det(A))*A

[paper cedar]

What's the best way to input mathematical symbols and stuff into Discord?

[dusky epoch]

use TeXit to render latex tbh

[paper cedar]

Sorry, I am super new. Is there a channel I should go to to get setup with everything?

[zinc timber]

#latex-testing #latex-help #bots

[winter harbor]

If E is such a real vector space endowed with an endomorphism f : E -> E such that f^2 = - id, then you can define a complex vector space structure on E as:

(a+ib)*v := a*v + b*f(v)

Notice that the dimension of E as a complex vector space with the above scalar multiplication is precisely the original dimension of E as a real vector space. Now, use the fact that a complex vector space, when treated as a real vector space, always has even dimension.

[dusky epoch]

complexification moment

[winter harbor]

There's a more overkill solution

Using the fact that x^2 + 1 is irreducible over the reals.

The polynomial x^2+1 vanishes on f

thus dim(ker(f^2+id))= dim(E)

And the fact that x^2+1 is irreducible tells us that 2 = deg(x^(2)+1) divides the dimension of E.

So the dimension of E is even.

[dusky epoch]

[fair wren]

Cant agree more lol

I still don't understand neither Ann's answer or the one above

[dusky epoch]

well in fairness i didn't supply all the details but for whatever reason you didn't ask me to clarify anything

suppose towards a contradiction that dim(E) is odd.
then the charpoly of f will have odd degree (because its degree is equal to dim(E)) and thus something can be said about charpoly(f) that'll contradict what's been proved previously (all odd-degree polynomials have at least one root, thus f will have at least one real eigenvalue, but we already proved that it doesn't.)

[fair wren]

which theorem states that the dim(E) is the degree of charpoly of f

[fringe fjord]

One argument is to unfold the determinant of tI-f using Leibniz expansion. There will be one term that is a product of dim(E) factors of the form (t-<diagonal element>), and no other term has anything of the same or higher degree in t.

So not only does the characteristic polynomial have degree dim(E); it's leading coefficient is 1.

Or (-1)^{dim E} if your definition of the characteristic polynomial is the determinant of f-tI instead.

[little crater]

This look right?

The terminology might not be correct with the word "unique" but I meant as not being able to right it as a linear combination of the other

although I guess w or u could have flipped with my use of "unique"

[fringe fjord]

Yeah, the "v is unique" language makes no sense.

[little crater]

yeah kinda meant relative to the other vectors :v but it doesn't make sense with regards to w and u

"linearly independent" maybe?

although normally that is used I believe with a set of vectors and saying how none of them can be express with combinations of the others

[fringe fjord]

It would be more direct to say something like "the first three spans are all { (x,y,0) | x,y in R }, but span{u,w} doesn't contain anything with nonzero y-coordinate".

Speaking of linear (in)dependence is definitely better than "unique".

[little crater]

yeah that would make more sense

I guess I could also then show via substation with regards to the spans to show how they are not equal, let u = a w (some scalar a), then say span{u,w} != span{u,v,w} because span{u,w} = span{aw,w}= span{w} but I guess I would still need to show that span(w) was not also span(v).

Maybe start off by saying let v b linearly independent from w and u and that u = aw and then talk about the above stuff.

but seeing it wanted 1 example and not something in general I guess I can just use the actually vectors I provided as an example

[dusky epoch]

tI - f, not tI - E

E is the space

[fringe fjord]

whoops, fixed.

[high peak]

Solving Ax = b where A, b have complex elements would be the same as solving it normally (just w/ complex arithmetic) correct?

[halcyon spindle]

Yes.

[high peak]

thanks!

[winter harbor]

Complexification is a thing in linear algebra tho lol

[fair wren]

haha

[winter harbor]

Essentially, a finite dimensional complex vector space can be treated as the same thing as a real vector space endowed with an endomorphism J that squares to minus the identity.

This is a pretty simple idea that is used a lot in quite different contexts.

[wintry steppe]

👀

kahler moment

[winter harbor]

[wintry steppe]

[zinc timber]

[dusky epoch]

something something K[x]-modules

[fair wren]

what?

[dusky epoch]

which part of tropo's explanation did you not understand?

[pliant kayak]

Is someone in the mood to explain tensor product to me?

[winter harbor]

Yeah, I could do that.

But first

Could you explain in our own words what the tensor product of vector spaces is?

That should give me a rough idea of what is troubling you with the definition.

[pliant kayak]

It's in German, give me a second

(sending it so I can translate it)

It says that the tensorproduct of two Vectorspaces V1 and V2 over the field K is composed of a Vectorspaces V' and a bi linear function k:V1×V2 - > V' with the universal property that for every Vectorspaces W and every bi linear function phi: V1×V2->W exists an unique linear Funktion phi ':V' - >W, with phi' °k=phi
(I hope you can understand something)

I mean I get the definition, but somehow it confuses me anyway

[winter harbor]

Yeah so

The idea is precisely that the tensor product of two vector spaces turns bilinear maps into linear ones

In a canonical way

And that's precisely the reason why the tensor product exists.

This can be summarized as follows, for $V_{1}, V_{2}$ vector spaces over a field $\mathbb{K}$, the tensor product $V_{1} \otimes_{\mathbb{K}} V_{2}$ of $V_{1}$ and $V_{2}$ is the unique vector space (up to isomorphism) that satisfies for any vector space $W$
$$
\text{Bil}(V_{1} \times V_{2}, W) \cong \text{Hom}(V_{1} \otimes_{\mathbb{K}} V_{2}, W) \cong \text{Hom}(V_{1}, \text{Hom}(V_{2},W))
$$
The isomorphism:
$$
\text{Hom}(V_{1} \otimes_{\mathbb{K}} V_{2}, W) \cong \text{Hom}(V_{1}, \text{Hom}(V_{2},W))
$$
Is called the tensor-hom adjunction, and is in a sense what defines the tensor product (up to isomorphism).

[stoic python]

MISTERSYSTEM :urs:

[winter harbor]

Of course

This definition only tells us the universal property that the tensor product of two vector spaces satisfies.

But it doesn't tell us anything about how to construct one explicitly.

The next step would be to show that the tensor product of two vector spaces always exists by constructing it explicitly.

There are lots of ways to construct it

But you know

I think the best thing to have in mind would be that the tensor product satisfies this universal property.

Rather than knowing how to explicitly construct it.

[fringe fjord]

One way to look at the explicit construction it is that an element of the tensor product is a recipe for "what to do with a bilinear map from VxW once someone provides us with one*. At that time we can apply the map to various pairs of vectors, add the results, perhaps scale them. But we don't yet know which bilinear map we'll be given, so we don't know what its codomain is or what we can do with the results other than things that are possible in every vector space. And additionally we don't want to distinguish between recipes that are guaranteed to yield the same result for every bilinear map.

[pliant kayak]

Is this like the most important thing?

[fringe fjord]

It's literally the entire definition, so in that sense it is "most important". On the other hand, for the same reason that it eventually tells you literally everything there is to know about the tensor product, it doesn't necessarily provide much focus.

[fringe fjord]

In particular, don't become so focused on the universal property that you forget the map k and what it alone can tell you about the tensor product without the rest of the universal property:

• Each time you have a vector v in V1 and w in V2, you get an element k(v,w) of the tensor product.
• To an extremely coarse approximation the tensor product can be thought of as being made of these combined elements.
• The tensor product is a vector space, but its addition and multiplication operations need to work in funky ways in order for k to be bilinear.
• If c is a scalar, then c·k(v,w) must be both k(cv,w) and k(v,cw) -- and so these must equal each other -- but it cannot be k(cv,cw) because that is c²k(v,w), which is different unless c is 0 or 1.
• k(v,w) + k(v',w') cannot be the same as k(v+v', w+w'), because that would mean that k(v,w)+k(v,w) is k(2v,2w) = 4k(v,w) instead of 2k(v,w) like it ought to be. In fact k(v,w)+k(v',w') is not generally even in the image of the map k!
• However, bilinearity means that k(v,w)+k(v',w) = k(v+v',w).

[uncut yarrow]

how would you prove: $ \begin{bmatrix}
1 & 1 \
0 & 1 \

\end{bmatrix} $ is not diagonizable

[stoic python]

! matthewzz

[uncut yarrow]

is it because you have one eigenvector for the eigenvalue of 1 (which has multiplicity of 2)

[hard drum]

Yup, there is only one eigenvector (up to scaling)

[uncut yarrow]

also a scalar matrix is just

c * i (nxn identity matrix)

[zinc timber]

try to be more precise, you actually have infinitely many eigen vectors

you can geometric multiplicity is 1 where as AM is 2

[outer hazel]

I got x=0
y=0
And z=0
After using Gaussian elimination. Does this mean anything, or is that just the answer

[dusky epoch]

outside of context, this means nothing

[outer hazel]

Idk if im asking the right questions,
What does it mean when all 3 equations in a matrix equal to 0.

[dusky epoch]

not sure what kind of "meaning" you want to get from that

are you asking if systems of linear equations where all right-hand sides are 0 have any special properties?

[outer hazel]

Ah yes

[dusky epoch]

well, one particular property of any such system is that the solution set is a subspace of R^n (where n is the number of variables) and in particular such systems are always consistent because they always have the all-zeroes solution

there's even a name for these: homogeneous systems

[outer hazel]

I see
Thanks for the info

[dusky epoch]

this is #prealg-and-algebra not linear algebra

[proper cradle]

Suppose A is n*n symmetric matrix and positive definite then how i can write diagonal entries of A in terms of eigen values and eigenvectors based on jordan normal form?

[proper cradle]

help please

[hushed ocean]

Hi guys, does anybody know of a problem in computational linear algebra which frequently requires large chains of matrix multiplication?

Aside from computer graphics

[elder monolith]

can u me ?

[hallow arch]

What does the dash on top the A mean?

trying to figure out how to solve it

[fleet sun]

conjugate

[hallow arch]

thanks

[cinder meteor]

In the process of transferring all my linear algebra notes to obsidian for future reference, here's the graph of all the different concepts so far

this is about 5 hours of work, definitely should've done this throughout the year instead lol

[tribal willow]

here's my obsidian of lin alg notes

[zinc timber]

looks cool ngl

how does it work?

[tribal willow]

each node is an individual file, when you're writing you can link to other files

sorta like wikipedia

and then it draws a line between the two notes

[zinc timber]

cool

[wintry steppe]

grassman ring

based

[zinc timber]

hoffman kunge I suppose

[wintry steppe]

that's a nice balance of matrix computation stuff and theory stuff

[tribal willow]

yeah i like hoffman kunze approach

[zinc timber]

I like exercises

[tribal willow]

i hate exercises

[zinc timber]

https://tenor.com/view/treadmill-dog-puppy-bored-unenthusiastic-gif-16626885

[tribal willow]

is this calling me lazy

i wouldnt disagree im just making sure

[zinc timber]

nah

[cinder meteor]

We didn’t cover fields and rings in my class (it’s a high school linear algebra class) but I might try to figure that out on my own time

[tribal willow]

oh to have a high school linear algebra class

what a dream

[cinder meteor]

What are the nodes that start with ! for?

[tribal willow]

indices

[cinder meteor]

Ohhh gotcha

That might be a good idea for my notes

[tribal willow]

surprised you guys didnt learn fields tho

considering that like

[cinder meteor]

We did vector spaces

[tribal willow]

vector spaces are defined over fields

[cinder meteor]

Yeah I leaned vector spaces independently from fields completely

[tribal willow]

hmm

i guess it makes sense

don't really like

need to know what a field is to get what a vector space is

[cinder meteor]

It was defined to me by the axioms and once i heard abt the examples I had a pretty solid grasp of it

Absolutely blew my mind when I first learned that you could use polynomials and functions as vectors

[tribal willow]

tEcHnIcAlLy a field of scalars is a vector space axiom

[cinder meteor]

Yeah we got a pretty circular definition ig

Vector: an element of a vector space
Vector space: a set of vectors

[tribal willow]

a field is just a set with certain axioms

pretty sure the axioms of vector space and field are pretty similar

[cinder meteor]

I’ll check out fields rq

[keen sierra]

every field is a vector space over any of its subfields (including over itself), this fact is used heavily in field theory

[tribal willow]

oh damn

i didnt think about that

but it makes sense

[cinder meteor]

I’m assuming the circular plus and multiply signs are just generalized adding and multiplying for different fields?

[tribal willow]

i dont remember why i used oplus tbh

the only important thing is that there's operations addition and scalars

[cinder meteor]

I gotcha

[tribal willow]

it's bad notation on my part tbh

oplus typically represents direct sums in lin alg i think

[cinder meteor]

So does a field require you to be able to multiply two elements of a field? Because how do you multiply two vectors in R^2 for example?

[tribal willow]

dot product

or er

i guess

[cinder meteor]

But wouldn’t you want a multiplicative operation that returns another vector or does that not@matter?

[tribal willow]

you would need to define that multiplication

a multiplicative operation of vector * vector = vector would be cross product, i think

[cinder meteor]

I gotcha

[wintry steppe]

for R^2, you can multiply vectors like you do complex numbers

identifying $\bR^2 \ni (a, b) = a + bi \in \bC$, you get $$\begin{pmatrix}a\b\end{pmatrix}\begin{pmatrix}c\d\end{pmatrix} = (a+bi)(c+di) = ac-bd + (ad + bc)i = \begin{pmatrix}ac - bd \ ad + bc\end{pmatrix},$$ and that makes $\bR^2$ into a field

[stoic python]

TTerra

[wintry steppe]

(isomorphic to C)

[cinder meteor]

I gotcha

So for R^3 can you just foil some more with a stand in j?

[gray dust]

@cinder meteor plenty of things can be said for vector spaces without such an operation, but plenty can be also said for spaces with such an operation, see algebras

[wintry steppe]

the multiplication is not commutative, so you would not get a field

[tribal willow]

oh shit true

[wintry steppe]

you have a lie algebra, though

which is still nice

[tribal willow]

yes that is totally what i meant haha yes! i know what lie algebras are 100%

i am excited to learn lie algebra

it is somewhere down on my list of things i want to learn

[wintry steppe]

you should be

[tribal willow]

it's also supposedly really important for quantum mechanics?

[wintry steppe]

they're a lot of fun!

yes, i think so

[tribal willow]

dope

[wintry steppe]

i couldn't tell you exactly how though

[cinder meteor]

I’m just confused bc I thought vector spaces were a subset of all fields, and thus needed a multiplicative operation

[wintry steppe]

no no no no no no no no no no no no no

[gray dust]

compare the defs

[cinder meteor]

I just learned abt fields a half hour ago, gimme a break lmao

[gray dust]

a field is a vector space with itself as the base field

not all vector spaces are fields

[cinder meteor]

Ohh so the field isn’t a parent of the vector space

[gray dust]

in fact one must equip a vector space with vector multiplication to even ask if its a field

since a vector space doesnt have vector multiplication by default

[cinder meteor]

Ohh so the field isnt the elements of the vector space itself, but the scalars thst you do operations with vectors in the vector space?

Lmk if i have this wrong

[gray dust]

yes

eg R^n is a vector space over R

meaning the elements of R^n are called vectors

elements of R are called scalars

[cinder meteor]

So what field would the set of all second degree polynomials be? Would that also be R?

[gray dust]

in general a vector space requires two things, a set of vectors and a set of scalars (the latter called a field)

theres still poor terminology here

we say a vector space is OVER a field

also degree 2 polynoms dont form a vector space

but the polynoms of degree at most 2 do

and its a vector space over R

[keen sierra]

any field needs to be closed under multiplicative inverses, so any field containing polynomials also has to contain their reciprocals

the field of rational functions, for example

[gray dust]

i dont think thats what seen has in mind

we’re identifying vector space/field pairs

[keen sierra]

maybe he meant what vector space contains the 2nd order polynomials?

[gray dust]

yes, which is what my example does

[cinder meteor]

That’s what i meant, my bad

[gray dust]

but the polynoms of degree at most 2 do
to be concise the polynom coefficients are real

[cinder meteor]

So can you have two different vector spaces that share a set of vectors, but are over different fields?

(My bad for asking so many clarifying questions, fields are pretty new to me atm)

[keen sierra]

a vector space over some field is also a vector space over any subfield of that field

for example R^2 is a vector space over R, but also over Q

[gray dust]

its possible

another example is

C is a vector space over C

[cinder meteor]

That makes sense to me, since Q is a subfield of R

[gray dust]

C is also a vector space over R

[cinder meteor]

Because a number a in R can just be thought of as a+0i?

Thus making R a subfield of C

[gray dust]

the reason is it satisfies the vector space axioms, but thats not too illuminating

whats more illuminating is that the key behind checking the axioms is that R is a subfield of C

[cinder meteor]

I gotcha

[faint hearth]

How would I normalize a Vector based on a different vector?

Probably asked that wrong

[hallow vector]

any idea on how to start this one? I tried just putting the numbers in random order but couldn't find a combination

[dusky epoch]

4 6 10 13
6 4 13 10
10 13 4 6
13 10 6 4

[hallow vector]

what lol i've been staring at it for like an hour

how did you understand what kind of pattern the numbers needed so that the matrix is symmetric?

[dusky epoch]

i started with the first row and kind of filled it in row by row i guess

[hallow vector]

oh i see there's a diagonal pattern in the numbers now

also each mini 2x2 matrix in the 4x4 matrix is symmetric it looks like

[faint hearth]

Instead of normalizing based on 0, 0 I would want to normal based on another point, I. E. 50, 60

So if I normalized 400, 60 it would be 1, 0

[peak lodge]

say you have a matrix A, is there any tricks to find the determinant of A^3 ?

besides manually finding A^3 and then getting the determinant

[gray dust]

if u know detA then u can use det(AB)=detA*detB

so det(A^3)=(detA)^3

[peak lodge]

oh damn yeah i forgot about the determinant properties

thank you

[serene solstice]

How to prove that f is linear?

[subtle walrus]

using the definition of linear

[verbal wedge]

do I need to consider X^TX?

[dusky epoch]

do you know the definition of the frobenius norm?

[verbal wedge]

this one?

[verbal wedge]

I know of this one and the trace property of A^TA

[dusky epoch]

that first one will be enough

do you know why $\nrm{\bd{y}+\bd{z}}^2 = \nrm{\bd{y}}^2 + \nrm{\bd{z}}^2$ when $\bd{y}, \bd{z}$ are orthogonal vectors in $\bR^m$?

[stoic python]

Ann

[verbal wedge]

I'm not quite sure

I presume the proof is something to do with the inner product and how that's related to y^Tz

[dusky epoch]

...yes the proof does have to do with the inner product but also it's just pythagorean theorem generalized

[verbal wedge]

ah that makes more sense

thank you

[dusky epoch]

in fact $\nrm{\bd{X}}F^2 = \sum{i=1}^n \nrm{\bd{x}_i}^2$ as should be relatively clear

[stoic python]

Ann

[verbal wedge]

yeah, I got to that part, didnt realise the orthogonality implication though

I get it now tho thanks

[verbal wedge]

@dusky epoch if you don’t mind, I had another thought I didn’t quite get

[dusky epoch]

C has orthogonal columns doesnt mean C^T C = I, it just means C^T C is diagonal

[verbal wedge]

really? it says in an example that this is true

it's the next part to the previous question

I reduced it to equating the statement in the note I've written is true

[dusky epoch]

maybe they meant orthonormal instead

[verbal wedge]

yeah, it should be that shouldn't it

lets say it is orthonormal, does my reasoning still stand?

[dusky epoch]

it seems ok but im not 100% sure and dont have the energy to check it for sure

[verbal wedge]

no worries

[verbal wedge]

nvm I found the solution

[boreal wadi]

$x^Ty$ is a dot/inner product

[stoic python]

melling