linear-algebra

[teal grotto]

hmm. will have to read about hahn-banach. havent heard of it yet

i guess if V = F[x], then it kind of works, since polynomials are defined in a nice way

but yea in the general case, i can't really think of anything

[weak needle]

Without axiom of choice, there are infinite dim vector spaces whose dual is zero

[teal grotto]

do you have an example in mind?

[weak needle]

What do you mean?

[teal grotto]

like an example of an infinite dim vector space whose dual is zero (without aoc)

[weak needle]

There was a pretty explicit construction, but I can’t recall it off the top of my head, sorry

[teal grotto]

no worries

[raw pilot]

posting here, hope that's okay:

A 3x3 matrix w/ Nul(A) = Span([x,x,x],[x,x,x])
Assume a vector v = [x,x,x] is a solution to Ax=b
Ax = [x,x,x]

Find 3 vectors different from v which are also solutions of this equation

I'm not sure how to tackle this. I understand that we know what 1 b is, and that the nul (A) lets the Ax=0. how do i piece this together to find the other solutions? are the other 3 vectors A?

[half ice]

Span([x,x,x], [x,x,x]) is an odd thing to write. Did this paste correctly?

[raw pilot]

i replaced the values with x

[hard drum]

for concreteness let's let u,w be linearly independent vectors in Nul(A) so Au = 0 = Aw and Nul(A) = Span(u,w)

consider taking linear combinations of the vectors you've already got (i.e. u,w,v)

(or for example, do you know the proof that if a matrix A has Nul(A) = {0}, then the solution to Ax = b is unique? Why does that fail here?)

[raw pilot]

ill give it another go with these tips, thank you!

[hard drum]

np

[zealous junco]

can someone explain why X is in the range of V means that X = VB, note here the b_i are unit vectors so they can't just be any matrix B

Also for context $X = \sum_{k=1}^r c_k a_k b_k^\star$ where $a_k$ is also vector of unit norm

[stoic python]

Anticipation

[zealous junco]

c_k is a scalar

and the $\mathcal{T}(u)$ there refers to the toeliptz matrix of vector u

[stoic python]

Anticipation

[zealous junco]

nvm i got it, i think this b_i is not a unit vector

[zealous junco]

Ok, another question from above: Let W a LxL be positive semidef and B as NxL. Let's say B is full column rank. how do you show there's always an E, positive semidef in NxN so that W = B* E B

[short magnet]

Let V be a finite subspace of R[X] the space of polynomials with real coefficients. Show that V contains basis wherein all polynomials are of distinct degrees.

Any hints for tackling this question?

[zealous junco]

i think one way is to argue like this. First say V has dimension n, then list out a finite basis of V. Each vector must be a finite degree polynomial, so there must be a finite maximum degree over all basis vectors, say N. For each vector e.g. x^2+3x+2, form it as a row [0.....0 1 3 2] in a matrix in R^{nxN}, then row reduce that matrix and since the pivots are at distinct column, you got your basis

lol probably theres an easier way, something like a contradiction, maybe..

im not sure about exact definitions so correct me if im wrong, but i think u can say V is isomorphic to a n-dimensional subspace of R^N, or something like that. In that case, you would be done already

[short magnet]

Hmm, perhaps that might be the case. I'll try that approach, hope it works ^^. Thanks!

Would induction on the dimension of V be somehow possible?

for dim V = 1 it's obviously true

suppose it's true for dimV = n and let's prove it for n+1

and then I'll try to write V as a direct sum

perhaps by taking a polynomial P with maximum degree from V

then let S supplement Vect(P) in V

and apply the hypothesis to S with dimension n

so we get a basis of S with distinct degrees

and add P to it?

why would that basis not contain a polynomial of the same degree as P?

[zealous junco]

sorry, i don't quite understand what you mean in the argument. I think induction on dimension of maximal degree seems awkward here

[short magnet]

no

induction on the dimension of V

[zealous junco]

ahh

[short magnet]

and then in order to apply the hypothesis we take the polynomial of maximum or minimum degree out

in an attempt to get a space of dimension n

[zealous junco]

ok i think i see your point. I'll think about how it can work

[short magnet]

okay, thanks

: )

[zealous junco]

Ok, I think i got it. So you have an additional vector right

[short magnet]

yes

I want to show the proposition is true for dimV = n+1

so I want to write down V as V = S + Vect(P)

(direct sum)

[zealous junco]

Ok let V be spanned by v1,…,vn ,a

[short magnet]

yes

[zealous junco]

Apply the hypothesis on v1,…,vn, then order the degree of v1,…,vn as deg v1<…<deg vn

Then for a, suppose it has degree same with one of the v1,…,vn

[short magnet]

ah

I got it

[zealous junco]

Then just remove that degree from a by subtracting vi from it

[short magnet]

yep

[zealous junco]

Yea

[short magnet]

or

[zealous junco]

There’s at most n steps basically

[short magnet]

if they are unitary

then deg (a-vk) <= deg a - 1

so a - vk is in the span of v1,...,vn

thus a is also. Which is absurd!

so it follows that the degree of a is distinct from the others!!

thanks : )

[zealous junco]

Yea that works too

[short magnet]

there is also a second part to this question, which is the existence of a basis with polynomials of same degree : )))))

anyways, thank u ^^

actually, it follows easily from the first one

just add the polynomial with the maximum degree to all the others

shrugs

[acoustic holly]

sup !

Suppose i have an n x n matrix P such that P commutes with all other n x n matrices

i.e for all A, AP=PA

i have to prove that P is a scalar matrix

i managed to prove that P is necessarily diagonal

but im having trouble showing that all the diagonal entries of P are the same scalar

here btw our matrices are defined on a random field K

ah nvm i got it

multiplying from the left by a diagonal matrix will multiply the i'th row by the i'th diagonal entry

multiplying from the right by a diagonal matrix will multiply the i'th column by the i'th diagonal entry

since AP=PA then all the diagonal entries are necessarily equal

and boom i have my result

[hard drum]

One way to do this is to use induction and consider the matrix with 1 1 on the top row and 0 elsewhere as that shows that the top two entries on the diagonal are the same

[forest quiver]

Yo do you have a quick second ?

Oh wait you are not that person nevermind

sorry

I will just post this here

This is something I wrote for my paper on linear models in Demographics

I mean this is something you see in Linear algebra

but is it a linear model?

There is an exponent...

Is it fair to include this idea in a paper about LINEAR models ?

[last holly]

still linear

i think there was a thread about this earlier

you can convince yourself with induction (a bit overkill), if a transformation is linear, the same transformation applied twice is linear (base case)

as it continues to be closed under addition, multiplication, and has a 0 vector

[forest quiver]

Yeah that's what I am doing

I am plotting some points from this rn

and they are all in a straight line

But they aren't the same distance apart

that's interesting

[nocturne jewel]

Ax=b is a linear system
AAx=Ab is also a linear system, just whose co-efficient matrix is A^2 and constant vector is Ab
repeat for all of N

you can have exponents and still be linear

[forest quiver]

But I was thinking

After applying the same linear transformation many times, I got a straight line with my output vectors

Like I know it makes sense its a linear transfromation

but is this always the case?

aren't there some transformations that rotate vectors 45 degrees clockwise for example?

Like then I won't get a straight line

I will get a circle depending on the vector I put into the transformation

am I making sense or is this just BS

[wintry steppe]

I believe you get a straight line because your columns add to 1

[forest quiver]

That actualyl makes sense

like intuitively

but I can't explain it

[wintry steppe]

Something to do with how population doesn’t vary, only proportions do

What are you plotting, anyway

[nocturne jewel]

Looks like you're doing Markov chains.. which yes, the vectors are probability vectors

[forest quiver]

Wait this is markov chains?

[last holly]

steady-state vector yea

[forest quiver]

Wow what have I stumbled into

[last holly]

it's common for them to be introduced early just as a taste

[forest quiver]

population change

but I keep the ratio between people who emigrate and people who stay the same in both of my locations

[agile sky]

you're not exactly dealing with markov chains as A does not satisfy the necessary conditions to be a transition probability matrix, unless that's just a mistake

[forest quiver]

Good cuz idk what that is

[agile sky]

It's a stochastic process that utilizes similar ideas to what you're working on, if you like this sort of stuff you should look into. I don't think it would be necessary to have a lot of background in probability theory to understand it and apply it in situations like these

[forest quiver]

Yeah I am sure I will look into it

Thanks

[rustic panther]

Anyone knows how to do this?

[dusky epoch]

https://www.wikiwand.com/en/Smith_normal_form

this help?

@rustic panther ive managed to figure it out for this matrix, in case the wikipedia article didnt help

[rustic panther]

got stuck I'm not even sure if I'm still on the right path @dusky epoch

[dusky epoch]

you should not be getting any fractions in here so unfortunately no youre not on the right path

there's a crucial difference between normal row-reduction and reduction to smith normal form:

in the latter, you're allowed to do column operations

[rustic panther]

Yeah I realised I can't just blatantly go R_1 = 1/4 R_1

this isn't RREF

Jordan form changes if I take fractions like that I think

[rustic panther]

@dusky epoch how do I proceed with ECO here

[dusky epoch]

subtract multiples of cols 1 and 2 from each other until you get a 2 somewhere, then use that 2 to zero out the middle row

then switch some rows around

[zinc copper]

Top 10 anime math problems

Oh darn you removed the pic

[rustic panther]

Haha my lens was blurry

[worn hawk]

If I have det(A) and det(B) , is there a way to compute det(A+B)?

[marble lance]

No

[wintry steppe]

https://math.stackexchange.com/questions/673934/expressing-the-determinant-of-a-sum-of-two-matrices#1937052

[marble lance]

Note thay the methods there are for special cases or where you also know what A and B are

[whole lark]

In probabilities : having « i » events mutually independent isn’t the same as saying that each two are independent ?

[wintry steppe]

So $\bigcap_i E_i = \emptyset$

[stoic python]

The Library of Babel

[wintry steppe]

Is this what you mean by i events mutually independent

[fickle citrus]

That's kinda...not ideal notation

[wintry steppe]

What is usually used

[fickle citrus]

You need $i\neq j$ typically

[stoic python]

ShatteredSunlight

[fickle citrus]

Anyway all those implicitly assume countability I think

[wintry steppe]

Yeah this is true for finite sets of events

[fickle citrus]

A bit clunky but essentially you should have some basic $A\neq B$ for sets $A$ and $B$

[stoic python]

ShatteredSunlight

[wintry steppe]

But I don't think you do

I think my notation is clear enough, although maybe nonstandard

[fickle citrus]

ok sure, yeah I think your notation does a lot of assumptions

[wintry steppe]

Fair

[fickle citrus]

Like, are collections of 3 allowed, yes no?

[wintry steppe]

Oh god I give up

[fickle citrus]

It's simpler to introduce more notation

So just $A$, $B$, $A\neq B$ then intersect is empty

[stoic python]

ShatteredSunlight

[wintry steppe]

I don't know if we're on the same page

According to Wikipedia the generalized intersection can be written like I did

[fickle citrus]

That's not my issue

[fickle citrus]

But anyway, back to this, it is not the same

[fickle citrus]

My issue with this is that it's vague, although I don't think it quite answers the question too

[wintry steppe]

I was just asking if that's what they meant by i events mutually independent

[wintry steppe]

Is your issue just with finiteness?

[fickle citrus]

That cannot be what they mean by mutually independent because mutual independence has a very usual definition that does not mean disjointness of events

[wintry steppe]

Oh yeah I'm just confused

That's mutually exclusive

[fickle citrus]

Yup

[wintry steppe]

Anyway the answer is here

https://en.m.wikipedia.org/wiki/Independence_(probability_theory)#Pairwise_and_mutual_independence

[fickle citrus]

If I were to just add, I'd either use the top (so there's an end), have it very clear what $i$ can be or literally use $\cap_{i\in S}$ for some $S$

[stoic python]

ShatteredSunlight

[fickle citrus]

Basically dangling $i$ are not good 😦

[stoic python]

ShatteredSunlight

[wintry steppe]

Yeah you're right that is clearer

[fickle citrus]

👍

[lusty sphinx]

Hello!

could you help me to solve this
determine whether the following lines are parallel or secant and check it graphically

help please, I need it today

[tawny harbor]

#❓how-to-get-help

[limber sierra]

this question is better suited to #prealg-and-algebra or #precalculus

or one of the 10 questions channels

that said, as a rule, find the slope; same slope = parallel

negative reciprocal slope = perpendicular, im not sure what secant means here

(probably a mistranslation?)

[lusty sphinx]

yes, probably

[blissful vault]

Does this check out, and what I could I do better writing wise?

I accidentally over indexed the eigenvalues by adding 5

[frosty vapor]

yeah the extra 5 yes

and each coefficient a_j is nonzero anyways because eigenvectors are nonzero

so the part where u let lambda not be in the set could be simplified/made more clear to just mentioning this as well

oh u said its not possibke ok

[blissful vault]

Ya I was trying to exclude the possibility of eigenvalues outside of 0,1,2,3,4

[frosty vapor]

thats good

[blissful vault]

I feel like this problem would be way more straight forward if I worked with matrices and got a characteristic polynomial.

[frosty vapor]

hmm

[blissful vault]

for some reason I was blanking out when working with the system of equations. I guess I was hung up on how each system of equations was equivalent to the next. Might be just my own problem lol. I feel like it is easier to see how things are equivalent when working with matrices.

[frosty vapor]

yeah it would be more direct yes

im like blanking a little bit but the last few parts seem unclear to me

[blissful vault]

Is it the part where I say "the final system of equations is equivalent to a_lambda being non-zero and aj = 0 for j not equal to lambda"?

[frosty vapor]

yeah that kinda sounds weird

cus like the eigenvectors are nonzero anyways

its not that finding zeroes means that the other values are Not zero

they were nonzero to begin with

[blissful vault]

Agreed. I'll think this over and rewrite that last part in the morning.

[frosty vapor]

yeah everything up to that looks fine

but the conclusion seems odd

but good

glad to read over this lol its been a while

i think also you might want to show how u get the eigenfunctions from the eigenvalues each(?)

[blissful vault]

Do you mean for each eigenvalue how I associated it with the corresponding set of eigenvectors?

[frosty vapor]

yea

[blissful vault]

I think I did hand wave that a little. Should I do a proof for each eigenvalue : 0,1,2,3,4?

[frosty vapor]

yeah probably it should be a quick computation u dont need to be too crazy about it

[blissful vault]

Thanks a lot for the help. Peace! I'm going to head to bed.

[frosty vapor]

np

[zealous junco]

let $W \in \mathbb{C}^{L\times L}$ be positive semidefinite and $B \in \mathbb{C}^{N \times L}$. When is it the case that for any $W$ there is an $E \in \mathbb{C}^{N\times N}$ positie semidefinite so that $W = B^\star E B$

[stoic python]

Anticipation

[zealous junco]

i assume B needs to be rank max(N,L) but would like explanation, thanks

(positive semidef here also means symmetric)

[zealous junco]

Solved

[short magnet]

What would be the best way to quickly calculate the dimension of the space of cubic splines (i.e. we fix a subdivision of [0,1] : 0=a_0 < ... < a_n = 1, and let f be of class C^2 such that it's restriction to any interval [a_k, a_k+1] is a polynomial of degree three. This f is said to be a cubic spline !). I have tried to construct an isomorphism (polynomials of degree three can be totally defined by 4 points from which they pass using Lagrange interpolation) but that extra smoothness condition, which requires continuity for derivatives, is really troublesome.

Hermite interpolation seems like it'd be a solution, but it's a bit more involved

I found a proof of sorts whilst googling but it uses terms I'm unfamiliar with like contraintes and stuff.. when I'm just looking for an elementary proof; i.e. I'm looking to construct some sort of isomorphism, but I can't think of one..

what would be the quickest way to calculate the dimension of that vector space?

[zealous junco]

I think its n+3 dof, or in other words dimensions

unless I'm misunderstanding, the first subinterval has 4 degree of freedom (dof) and the C^2 condition means the points of connection btwn intervals makes each next spline have only 1 dof. you can see this by the requirements that the function's left endpoint value and first, second derivatives must equate the ones for the previous function's right endpoint

consequently if you require C^3, the dof (dimension) will reduce to 4 as expected, because the only cubic that joins C^3 with a previous cubic is the function itself.

[short magnet]

that makes sense..

I guess I'll just justify it instead of proving it rigorously. That'd be enough :""3

thanks for your time and answer though, I appreciate it greatly!

[steel moon]

hey

[raw pilot]

so assuming we let the vectors in the nul span = u and w respectively

i took a linear combination of uwv and got

and i dont know where to go from here

[dusky epoch]

what's that matrix meant to be?

[raw pilot]

thats the reduced matrix from u,w,v

the v being given, and u,w the vectors in the span

[dusky epoch]

and... why would you need that?

[raw pilot]

had help earlier that suggested i should do that

[dusky epoch]

all you're looking for here is any vectors of the form $$\bd{v} + \alpha\bd{u} + \beta\bd{w},$$ where $\alpha$ and $\beta$ are (nonzero) constants of your choosing

[stoic python]

Ann

[dusky epoch]

that's... it, really.

no need for any matrix-bashing.

also, who and where suggested it?

[raw pilot]

so like this

[dusky epoch]

yeah sure

do that another two times with different alpha and beta, and you're done

[raw pilot]

i mightve misunderstood the suggestion, it was earlier in this channel

[dusky epoch]

care to link the message?

[raw pilot]

#linear-algebra message

[dusky epoch]

yeah that did not entail any matrixbashing

[raw pilot]

my bad haha my

[raw pilot]

and thank you so much for the help i got it right 🙂

[hard drum]

I assume T is a linear operator from V to W, where V, W are vector spaces ?

if so, yes, the kernel is defined as a subset of V

But it also happens to always be a subspace of V which you can prove (directly / by subspace test)

[spare crystal]

In the solution to this problem

why can't the minimal polynomial of M be x^2 + x + 1?

[vivid atlas]

Can somebody explain "No set of m-dimensional vectors can have
more than m mutually linearly independent columns, but a matrix with more than
m columns may have more than one such set."?

[glacial mango]

You can have a 2x4 matrix where the first two are an l.i. set and so are the last two (m=2).

[wintry steppe]

a real polynomial of odd degree has to have a real root, and the roots of the minimal and characteristic polynomials are the same

so the minimal polynomial of M has to have a real root

x^2 + x + 1 doesn't

something like that

gut tells me there's a simpler reason

[spare crystal]

oh i see that makes sense

yeah i guess the solution skims over it so maybe there is a simpler reason? but that seems simple enough

[wintry steppe]

yeah it's just the lack of explanation in the solution makes me think i could be overthinking it

[wintry steppe]

So like , if you just did T: R^4 -> R^5 would that mean you technically just made a vector pop into the 5th dimension mathematically. I'm having a very hard time visualizing this lmao

Like from my understanding T: R^2 -> R^3 would basically just make a line on a piece of paper pop into 3D suddenly.

[sonic osprey]

They're functions, not really pictures

Like, you can consider the function T: R^2 -> R^3 that sends everything to zero, in other words, T(v) = 0 for all v in R^2

(you can and should confirm that this is linear)

[wintry steppe]

Oh yeah true, well lets say its not the zero vector

[sonic osprey]

This isn't a line and is only a point in R^3

[wintry steppe]

or a point

[sonic osprey]

My point is that this is true for maps from R^4 to R^5 too

The image doesn't need to be 4-dimensional inside of R^5

It can be a point, a line, a 3D space or 4D

[wintry steppe]

like if it was [ 5, 10 ] - > [ x , y , z ] wouldnt it appear visually to suddenly pop into that dimension

[sonic osprey]

I don't think that really makes sense

It is true that a function from R^2 to R^3 takes something with two dimensions and outputs something with 3 dimensions but

I don't think I would call it popping into that dimension

[wintry steppe]

Right, but visually when you see a vector that isn't the zero vector in R2 it just looks like a line, then it would appear to pop out in a 3D direction after being transformed instantly. It wouldn't slowly rotate into 3D unless you had an animation

[sonic osprey]

I think its weird to think of it like that

It's just a function that takes in something 2D and spits out something 3D

[wintry steppe]

Ofc you would your a math person

jk lmao

Also yeah I see what you're saying that 5D could include lines and planes and all kinds of stuff. I was imagining like the same object just suddenly appearing in each dimension due to a LT

Like a 2D apple suddenly morphing into a 5D one

[sonic osprey]

I mean my earlier point is that a 2D space like R^2 doesn't necessarily get mapped to a plane in R^5

like the example earlier, linear transformations can squish things down so that 2D space becomes a point

[wintry steppe]

https://www.youtube.com/watch?v=0t4aKJuKP0Q

Kinda what I was trying to explain , basically these are just animated transformations right

[wintry steppe]

Ohh yeah thats true, but you're also thinking of it "squishing" which ... visually is an animation also I believe

[sonic osprey]

I mean, I don't know if I would call those transformations in the usual sense

All the things are a function of time so that for every time t, you get something in R^4

[wintry steppe]

Thats true. I think you're right that the way its generated isnt what I was talking about HMMM

okay so what about this

The cubes are still all there, but they are instantly transformed into 4D visually

even though they moved position

[sonic osprey]

These are just 3D slices of 4D space

It seems what they're doing it just looking at [w,x,y,z] and varying z to give you the different slices

[wintry steppe]

No, in the game if you throw something at that "empty" space it would detect collision, its still there in the 4th dimension

we just see slices in 3D

[sonic osprey]

Oh I was thinking of a different part of the video

where he moves the slider up and down with the spheres

[wintry steppe]

Oh no lmao

[sonic osprey]

But kinda the same thing is happening here

What they show is a single 3D slice of 4D

and when the collision happens, the cubes move out of that slice

[wintry steppe]

Yeah... I feel like this is also still a bad example cause of the way the games engine accounts for time, and the fact its a 3D environment trying to visualize a 4D object

lmao

[sonic osprey]

I'm not really sure what you're trying to give an example for

[wintry steppe]

I honestly dont even know , I Just learned about transformations and had a weird visual idea of it lmao

[sonic osprey]

https://www.youtube.com/watch?v=kYB8IZa5AuE

maybe something like this could help

but it only really talks about transformations from R^2 to R^2

[wintry steppe]

Yeah i was trying to find one where he goes across dimensions but I dont see one lmao

I think he mostly focuses on transforming within the same dimension which is understandable

[lyric dawn]

hello, i think this is true but i dont know how to prove it. let say i have a matrix L of size k x n. where k > n. lets say the rank of L is p. necessarly p < n.

i want to show that there exist a matrix L' of size p x n such that ||Lx||_2 = ||L'x||_2 for all x

[zealous junco]

u can try using QR decomposition, and R is upper triangular so the dependent rows of L becomes zero

And Q preserves length

[dawn mesa]

Do you have to visulize linear algebra to be able to understand Linear algebra?

[last holly]

Yes and no

I don't know anybody personally who can visualize 6 dimensions freely

But building intuition in 1, 2 and 3 helps reason about the rest

Use geogebra and other tools to help

Strang says he's really bad at visualizing

But I'd take that as professional modesty

(he may mean, there are people out there that can do it much better than he can)

[flint pivot]

personally I don't but I also know how to visualise it so I don't know how it would seem to someone who doesn't

[next dune]

is knowing linear algebra past SVD important for industrial work in data science

theres a numerical linalg class that im thinking about taking that goes over fundamental matrix types, rounding error, condition and stability but i dont know if it would be practical for me

[nocturne jewel]

I mean... certains things yes, however once you get into vector spaces it's near impossible to visualize imo

[dusky epoch]

yes

[wintry steppe]

guys does this means same thing? u and v are vectors

[ocean sequoia]

Yea

Both refer to the dot product

[wintry steppe]

ty

[glacial mango]

?

[jagged marsh]

can three 2 dimensional vectors (three vectors that are in the same plane) be linearly independent assuming non of them are colinear?

aka, can you make any two dimensional vector with a linear combination of two non-colinear vectors?

[glacial mango]

No. The maximum number of l.i. 2D-vectors you can have is 2 (pretty sure).

[glacial mango]

Yes, this should be easy to prove using the determinant.

[nocturne jewel]

slight nuance, however the bottom one is a specific example of the top one

[fickle citrus]

Won't be much different until the generalisations come in

[ocean sequoia]

^ figured where they are at wouldnt matter

also @jagged marsh you can prove that without using the determinant @glacial mango

row rank must be equal to column rank

[hard drum]

This is v fundamental - a key theorem used in even defining dimension in the first place is that linearly independent sets are no larger than spanning sets, and hence any set of 3 vectors in R^2 is linearly dependent as {(1,0),(0,1)} is spanning

[ocean sequoia]

I think you meant linearly dependent

normally wouldnt nitpick that but given thats a bit fundamental to what the question was is figured I would

This is v fundamental - a key theorem used in even defining dimension in the first place is that linearly independent sets are no larger than spanning sets, and hence any set of 3 vectors in R^2 is linearly dependent as {(1,0),(0,1)} is spanning

[hard drum]

Ye lol typo

Should've just left out the end of the sentence and then it works lol

[ocean sequoia]

yea i figured again not trying to be a pendant just v important here

think your explanation was very good

[hard drum]

ofc dw! Cheers

[ocean sequoia]

[hard drum]

[wintry steppe]

guys i'm starting university soon and i'm doing review from last year

https://cdn.discordapp.com/attachments/839266177251409960/879131164357570560/unknown.png

does anyone know how to solve these type of problems ? i forgot them 😭

im desperate lollll

[last holly]

Do you remember matrix multiplication?

[wintry steppe]

yea

[ocean sequoia]

do you remember how to solve system of equations?

[wintry steppe]

nope

[ocean sequoia]

what about guassian elimination

[wintry steppe]

a little bit

actually

i remember doing bad in that last year lmao

[ocean sequoia]

all good

for the first one we have (1,0) = (x,y)(1,2)

[wintry steppe]

yep

[ocean sequoia]

ok we need to solve for x and y

do you know how to do that

[wintry steppe]

i forgot 😃

[ocean sequoia]

well what would you guess be

[wintry steppe]

img uessing u convert the matrices into an equation and solve for it

like

[nocturne jewel]

The matrix representation of a transformation is to make the columns of the matrix the output vectors in the same order as the input vectors

[ocean sequoia]

wait is that what he is doing? am im helping him wrong here?

[nocturne jewel]

it wants the matrix representation then to determine if f is injective

[ocean sequoia]

oh ok

my fault

i thought it was just solving y = ax

[nocturne jewel]

no..

read the question prior to helping

[ocean sequoia]

yea I just skimmed it wow my fault

[wintry steppe]

wait im confused now

[ocean sequoia]

im so sorry

[nocturne jewel]

so for example, in 1 you have the basis {[1,0]^T, [0,1]^T}

we know f([1,0]^T) =[1,2]^T

so you will make the 1st column of the matrix the output co-ordinate vector

since [1,0]^T is the first basis vector

[wintry steppe]

wha

[nocturne jewel]

what?

"wha" means nothing in regards to getting help

[wintry steppe]

what does this mean

[nocturne jewel]

you know matrices have columns?

[wintry steppe]

yea

[nocturne jewel]

the 1st one, will be the output from the transformation of the 1st basis vector

[wintry steppe]

OH

[nocturne jewel]

(in the basis of the output space, but it's just using canonical bases rn)

[nocturne jewel]

likewise the 2nd column will be the output from the 2nd basis vector being transformed

etc

for the dimension of the input space

[ocean sequoia]

do you understand what you need to do Mahiii

[wintry steppe]

hold on

OH

YEAH

I DO

THANK YOuu

[nocturne jewel]

so for example, what's A for 1?

[ocean sequoia]

^

[ocean sequoia]

@wintry steppe

[wintry steppe]

for one would it just be like

(1 2
2 4)

?

as A?

@nocturne jewel

[nocturne jewel]

yes

$f(x)=\begin{bmatrix}1&2\2&4\end{bmatrix}x$

[stoic python]

Mosh

[wintry steppe]

what does it mean when a function is one on one?

[restive raft]

every output of the function is associated with a unique input

[ocean sequoia]

in other words if f(x) = f(y) then x = y

[nocturne jewel]

It means that the kernel of the transformation (the set of vectors which map to 0) is trivial, so f(x)=0 only if x=0

However here we know we can write $Ax=0$ so we want only the trivial solution to that, which is a condition of invertibility on A

[stoic python]

Mosh

[nocturne jewel]

so f is one to one (injective) iff A is invertible

[raw pilot]

we have Nul(A) = {[2, -2, 1, 0], [-1,1,0,1]}

im not sure where to go from there

[weak needle]

If the zero vector along with the two vectors you mentioned are all in the nullspace, can you think of good candidates for w,z,w’ and z’? (Hint:w’ may be obtained by translation via nullspace vectors (apologies for the previous wrong hint))

[raw pilot]

i got it, thank you!

[vivid field]

I’ve been doing some practice with rref form and I’ve been taught that I can only use addition when replacing rows, however I’ve seen many videos on YouTube where people are subtracting rows. Is this also allowed ? Apologies if this is a dumb question.

[last holly]

Subtraction is addition basically. It helps to think of it as addition though (addition of negative multiple)

It makes more sense when the row operations get represented by elementary matrices

Just more concise

It helps for proofs to break down things into additive and multiplicative inverses sometimes too

And computationally it's more concise too

[ocean sequoia]

have you seen that you can scale vectors still ala multiplication? if so its the same as multiplying by negative one

then adding

[rancid kindle]

Trying to prove W is a subspace of V

not sure what to do with the 2A_12 = 3A_21

usually we just have x_1 and x_2 in R

[half ice]

It's a rule that the matricies have to follow. To be a member of W you must have 2A{12} = 3A{21}

@rancid kindle

[rancid kindle]

where and how would I use that in my answer, is where i'm confused

[dusky epoch]

prove that if two matrices satisfy this rule then so does their sum

that gives you closure under addition

[half ice]

Might be easier to say that all matricies in W look like:
a 3t
2t b
For some real a,b,t

[flint pivot]

thats way easier than I thought wow

[dusky epoch]

yeah it is easy ive just seen a lot of ppl without much experience in basic proofs falter on these

[short magnet]

let E be a euclidian real vector space, $$ \alpha,\beta, \gamma \in \mathbb{R}, ; a \in E $$ Solve the equation : $$ \alpha \langle x, x \rangle + \beta \langle x, a \rangle + \gamma = 0 $$

[stoic python]

Der Gegenstand ist einfach.

[short magnet]

obviously, we could simplify it a bit by discussing if alpha = 0 or not and dividing by alpha

and by assuming a unitary

but I just don't know where to begin with this ..

[dusky epoch]

is E finite-dimensional

[short magnet]

yes

as it is euclidian

[dusky epoch]

do we have a basis for it on hand

[short magnet]

of course

with Gram Schmidt

[dusky epoch]

well then this reduces to $\alpha x_k^2 + \beta a_k x_k + \gamma = 0$ for $k = 1, \dots, n$

[stoic python]

Ann

[short magnet]

we even have an orthonormal basis

[dusky epoch]

yeah i'm talking about an orthonormal basis and the coordinates of x with respect to it

[short magnet]

I see

so now

we have an equation of the second degree

(a system of them)

that simplifies it greatly

thanks! : )

wait a second, this actually reduces to : $$ \alpha \sum_{k=1}^{n} (x_k^2)+ \beta \sum_{k=1}^{n}{a_k x_k} + \gamma = 0 $$

[stoic python]

Der Gegenstand ist einfach.

[short magnet]

which is completely different

a solution to that system (with gamma' = gamma/n) is a solution to your system of equations but the reciprocal is not necessarily true

[stoic python]

Der Gegenstand ist einfach.

Der Gegenstand ist einfach.

Der Gegenstand ist einfach.

[vivid field]

Could someone check to see if I did this correctly. Trying to determine values of h and k when the system has no solution, a unique solution and many solutions. Intuitively this is what I came up with.

[pearl turret]

Can someone explain what im supposed to do on proble, 19

[azure gate]

The system is consistent
So you have to find value(s) of h for which the system has a unique solution

[short magnet]

Okay, another question in euclidian spaces.. let $$ a_1,...a_n \in \mathbb{R} $$ and f such that $$ \forall x_1,...,x_n \in \mathbb{R} : s.t. : \sum_{k=1}^{n}{x_k^2} = 1 ; : f(x_1,...,x_n) = a_1 x_1 + \dots + a_n x_n $$

[stoic python]

Der Gegenstand ist einfach.

[short magnet]

find the maximum of f

the statement about the sum of x_k squared being one is the same as the norm(X) = 1

with X = (x1,...,xn)

we can even write f(X) = transpose(A) X

so f(X) = <A,X>

Cauchy-Schwartz tells us that f(X) <= norm(A) = sqrt( sum of a_k squared)

but that is just a bound that isn't attained (I think..)

how do I find the maximum of this function

[grim bridge]

cauchy schwartz always attains its bound

when the two vectors are colinear

[short magnet]

yes

so A/norm(A)

gives us the answer

sorry, was a bit stupid there :"")

[grim bridge]

there's an absolute value missing in your inequality though

[pearl turret]

The book says the correct answer is h ≠ 2

and i see how t oget that

but i dont get what it means

[azure gate]

For h=2 the system has no solution

So any other value makes it consistent

[pearl turret]

what does consistent mean maybe I'm stuck on that

atleast one solution

so basically in that one itrs

1x + hy = 4 and 3x + 6y = 8

?

there are values of x and y that make that true for the constants given

[azure gate]

By consistentency we basically mean that the system has at least one solution

[pearl turret]

Are these matrices just systems of equations from elementary algebra?

[azure gate]

Yes

[limber sierra]

(though they might be over another field)

[valid wigeon]

https://learnopengl.com/Getting-started/Transformations#:~:text=Because vectors are,and 3D space.

im a bit confused on this part

what does it mean by "position vectors"

are they different from normal vectors?

[hard drum]

often if a point is assigned a set of coordinates, say, (0,0,2),we might call the corresponding vector (treating R^3 in this example as a vector space) the position vector of that point

[nocturne jewel]

position vector just means "measured" from the 0 vector

[lyric dawn]

its perfect, thank you !

[dull dome]

@dusky epoch pinging

[dusky epoch]

yeah

okay so

a basis is basically... a more general version of a coordinate system

if you worked with R^3, your vectors are almost always expressed in the standard basis {i,j,k}

or {i,j} in R^2

a collection of vectors forms a basis of your space if every vector in the space can be expressed in terms of your collection, with no redundancies

i hope this makes some sense

[dull dome]

I get it.
But what I need to understand is why we go to do this?
I mean what is the motivation for defining basis, sub-space

[dusky epoch]

as i said

bases are generalized coordinate systems

idk how you'd like the concept of subspaces to be motivated however

[last holly]

Why is it useful to learn sets then?

It's useful to have structures to reason about collections of vectors.

Useful is an understatement here

[dull dome]

That kinda makes sense

[vivid field]

I started my linear algebra course 2 days ago and unfortunately it’s all online no lectures and just textbook use. I’ve noticed that it’s taking me a long time in the process of putting augmented matrix’s of a number of systems in to row form or row reduced form. It’s not the arithmetic but rather knowing how to just takes the most efficient steps to do so. Is there a specific order in doing this or am I just very slow at this.

[last holly]

consider this like learning a new language

i'm a native spanish speaker and can grasp a lot of what portuguese and italian speakers say, but if i wanted to try to be functional in them i'd need to really dig in to the frame of thinking of those languages

similarly, in the beginning it's enough to "look" at linear algebra from a distance and know what's going on vaguely. but to be conversant in it really is a whole set of different modes of thinking

in short, don't be discouraged. it's very normal. but also linear algebra is very useful so it's worth it

[vivid field]

Thanks so much for that explanation. I loved the analogy. I hope to develop that mode of thinking in my journey through this subject @last holly

[nocturne jewel]

what

??

1. dont randomly ping me/other users
2. I have no clue what you mean

can you post a picture of the notation?

[harsh halo]

Assume A is a 6 x 5 matrix and Ax = 0 has a unique solution. Show that the columns of A do not span R^5.

[nocturne jewel]

Ok...

[harsh halo]

This question statement is wrong I think because the columns of A are not in R^5

am I right?

[nocturne jewel]

the column vectors are in R^5

wait

yeah should be R^6

[harsh halo]

so this question is coming from my first year linear algebra class where we haven't even defined vector spaces or bases and only talked about matrices (and there correspondence with systems of equations and linear combinations), linear combinations, and just now span

[harsh halo]

so if this where R^6 instead, obviously we can't have 5 vectors spanning the whole space because it is of dimension 6

[nocturne jewel]

yes

[harsh halo]

but this is too much terminology for where my class is at right now

is there anything special about the setup for me to use

[nocturne jewel]

extremal properties of a basis says the cardinality of a set less than the dimension means it isnt a spanning set

[harsh halo]

yea but this is from a pset which assumes you don't know what a bases is (and dimension)

[nocturne jewel]

right ye

[harsh halo]

so I'm thinking there's some underlying structure to the setup that let's you do this more easily

"the setup" as in Ax = 0 having a unique solution where A is a 6 x 5 matrix

[nocturne jewel]

can you use/have you learned rank-nullity theorem?

[harsh halo]

no we haven't even covered the abstract definition of a vector space let alone linear maps

[nocturne jewel]

yeah rank-nullity just stems from learning the standard subspaces formed by a matrix

[wintry steppe]

Is it true that basis of kerT is a subset of basis of domain space ?

[ocean sequoia]

the kernel of a linear transform from V -> W is a subspace of V

[lavish jewel]

sounds about right

[torn hornet]

you can extend the basis of ker(T) to a basis of the whole space.

but the basis of ker(T) wont be a subset of any arbritary basis of the space

[ocean sequoia]

think about it this way T(v) and T(u) = 0 with u and v being vectors in the domain

you can use that to show T(u + v) = T(u) + T(v) = 0 + 0 = 0

and its also pretty obvious its closed under scalar multiplication

[wintry steppe]

Then the extended basis of kerT will it be a standard basis of that domain space

[torn hornet]

not necessarily

[wintry steppe]

Okay so there exist basis of domain space which is the superset of basis of kerT?.

[gray dust]

yes that's a rewording of

you can extend the basis of ker(T) to a basis of the whole space.

[wintry steppe]

Okay thanks !

[dark carbon]

how would i solve this

[lavish jewel]

i think you got the wrong channel

that doesn't look linear lol

[dark carbon]

idk wherre else to put it so

[lavish jewel]

so go to the algebra channel

[dark carbon]

im doing multivar calc

ended up witht hat nasty thing from lagrange multipliers

[lavish jewel]

then try the multivar calc channel

[versed forge]

guys how do u know if a matrix is commutable

[last holly]

try it

try multiplying a some matrices AB and then BA

see if they're the same

(unless you mean commutative under addition )

(should be simple enough to try with 2x2 matrices)

if you find one counter-example you're done

if you can't, see if you can generalize it to make sure it's solid

[versed forge]

ok

[raw linden]

Assume that we have m eigenvectors x1, ..., xm to A with associated eigenvalues λ1,…, λm. Assume further that the vector u is a linear combination of the eigenvectors, ie. that u = c1 x1 + ⋯ + cm xm where cj∈R. With the help of arithmetic rules for matrices, we can then calculate:

Au = ??

Help please I'm very tired of this problem

[dusky epoch]

do you know what an eigenvector is?

[raw linden]

Yeah

The main definition is Ax = lambda*x

[dusky epoch]

and you are told that we have these eigenvectors x_j, and their corresponding eigenvalues lambda_j, for j = 1 to m

would you not agree that $Ax_j = \lambda_j x_j$

[stoic python]

Ann

[raw linden]

I asked my teacher and she said if I solve this I will get an answer

[hard drum]

yes

use linearity

[raw linden]

shouldn't it be $Au = \lambda u$

[stoic python]

SuperXL07

[raw linden]

?

[dusky epoch]

no

u itself is not an eigenvector of A

and we have no lambda-without-an-index anyway

[raw linden]

first I answered that $Au = (lambda_1,..., lambda_m) (x_1,..., x_m)$

[stoic python]

SuperXL07
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[dusky epoch]

this is nonsense

[raw linden]

maybe I got it now let try then I will come back, one moment

[raw linden]

$Au = \lambda_1 x_1 +, ..., +lambda_m x_m$

[stoic python]

SuperXL07

[raw linden]

$Au = \lambda_1 x_1 +, ..., +\lambda_m x_m$

[stoic python]

SuperXL07

[raw linden]

is that correct?

[dusky epoch]

no

you are forgetting the coefficients c_j that you had started with

$u = \sum_{j=1}^m c_j x_j$

[stoic python]

Ann

[dusky epoch]

you will have $Au = \sum_{j=1}^m \lambda_jc_jx_j$

[stoic python]

Ann

[raw linden]

the answer have to match with the following conclusion:

This means that we can interpret multiplication with the matrix A as meaning that we scale the parts of u that are parallel to the respective eigenvector, and that the eigenvalues are our scaling factors.

are 100% sure that this is the right answer?

sorry to be annoying

[dusky epoch]

yes i am 100% sure

we go from c_j x_j to lambda_j c_j x_j

scaling doesn't kill the c_j that was already present

[raw linden]

I answered like that because in the book they wrote like following: Note that if x is an eigenvector of A corresponding to λ then so is kx, for each real number k ≠ 0. Self-vectors corresponding to a certain intrinsic value are thus not unique, but only the direction of the vector matters. Note also that we can interpret the matrix multiplication Ax as a scaling of the eigenvector x with the eigenvalue λ.

[dusky epoch]

okay look

you just completely ignored the coefficients that the x_j had before multiplication by A

you can't do that, alright???

[raw linden]

Yeah you have right

I'm very thankful for your generous help. Best regards

[boreal crane]

Thats a really cool exercice that was in my Exam

[wintry steppe]

is it possible to prove a matrix is unitary by calculating just the determinant? the determinant is 1 but i dont know if that's enough to justify it being unitary

[dusky epoch]

no it's not enough

you can have a matrix with determinant 1 that isn't unitary

$\bmqty{7 & 10 \ 5 & 7}$

[stoic python]

Ann

[wintry steppe]

what else do i have to do? is finding the determinant even necessary?

[dusky epoch]

it's not necessary, no

[wintry steppe]

ah, so it's just a second subquestion within the question i got

[dusky epoch]

to prove a matrix is unitary, you need to show that it satisfies the definition of a unitary matrix

[wintry steppe]

it doesnt really clarify it that well, thank you

[hard drum]

What definition of a unitary matrix are you using hazy?

[wintry steppe]

no idea

the question told me to prove it's unitary and then said "find the determinant"

so i thought i had to find some specific value on the det to prove it

turns out that's not the case at all

[hard drum]

It's pretty important to know the definition of smth if you're going to use it ig

[wintry steppe]

im willing to go with UU^H = I

[hard drum]

Ye finding the det doesn't help (altho it can prove a matrix isn't unitary)

Yeah exactly

And so the best way to check if it is unitary is just to calculate UU^H

[lavish jewel]

is the matrix square?

[hard drum]

That's what Ann meant by checking it satisfies the definition

[wintry steppe]

yes

[hard drum]

(Well hazy said it had determinant 1 so it must be square for that to make sense ig)

[lavish jewel]

just double-checking, cuz UU^H = I isn't really enough

[nocturne jewel]

Are unitary and isometry synonymous then?

cause that looks like the requirement for isometry

[somber bronze]

hello everyone! I was wondering if i could see the subspace of 2x2 matrices such as to have the second row equal to 0 (let's call it W) as the core of the following linear application:

[stoic python]

<eo

<eo

[somber bronze]

ops, remove that equal in the linear application, ker has c+d equal to 0

if anyone wants to help please mention me, thanks in advance!

[viral cargo]

@somber bronze Oh, so you're mapping 2x2 matrices to R?

(trying to understand your linear application)

It's just (a b | c d) -> c + d

[somber bronze]

i have to find the dimension of the subspace where 2x2 matrices have the second row equal to 0, i tried to find a solution using the linear application i wrote

[somber bronze]

yes i wrote by mistake that it is equal to zero but i changed my message later

[viral cargo]

Yes, your linear application's kernel is said subspace. But if you have a basis for it, you already have the dimension?

Also assuming you meant the sum of the second row is equal to 0, because the way I would read what you said is matrices where second row entries are 0

[somber bronze]

i meant the sum of the second row is equal to 0

I found the basis because from the nullity + rank theorem I know that the size of the starting space is equal to the size of the ker + the size of the image. Since W = 4 (2x2 matrix space) - 1 (dimension of R) = 3 I tried to see if a possible basis was the one I wrote

i just wanted to know if i answered correctly to the question of "find the dimension of W", and maybe if a faster way was possible

[viral cargo]

Well, this does verify the dimension of your basis

[somber bronze]

thanks for confirming!

[viral cargo]

Usually I would find a basis for the subspace then show it is a basis (span/linear independence)

then dim = card of basis by definition

[leaden tide]

I'm having trouble with this problem. We have $p$ a projector (= idempotent linear map) in a vector space $E$, and the two subspaces $F_1 = {v\in L(E),\exists u\in L(E), v = u \circ p }$ and $F_2 = {v\in L(E),\exists u\in L(E), v = u \circ (\text{id} - p)}$. The problems asks to show that these two subspaces are supplementary (= their sum is direct (intersection is {0}) + their sum is L(E) as a whole), but I can't show the first part, any pointers?

[stoic python]

Syst3ms

[rugged knot]

If a is invertible and c is nilpotent, is a-c always invertible?

[dusky epoch]

sure, you can write $a - c = a \cdot (I - a^{-1}c)$

[stoic python]

Ann

[dusky epoch]

a^-1 c is nilpotent with the same nilpotence index as c itself, thus I - a^-1 c is invertible

@rugged knot

[neat olive]

why wouldn't this be row-echelon form? every leading non zero number in a row below the first is to the right

the below is all true, no? so it has to be row-echelon, but webassign doesn't think so
-All nonzero rows are above any rows of all zeros
-Each leading entry of a row is in a column to the right of the leading entry of the row above it
-All entries in a column below a leading entry are zeros

[last holly]

That's weird

[half ice]

Row echelon isn't universally agreed, check your professor's notes. However yes, you agree with Google here.

[wintry steppe]

so why would D be the right answer here?

[robust pond]

were talking about in #chill fyi lol

i told ppl to come here but ya know

[fickle citrus]

I kinda don't agree with the wording, and I'd say solving a linear system is the same as finding parametric description. I however do agree with the 2nd part of D.

You can solve a system if it admits a solution, and you cannot solve a system if it does not admit any solution, and these occur
if and only if
you can parametrise a system if it admits a solution, and you cannot parametrise a system if it does not admit any solution

so I'm not sure what they want to refer to when they say 'true' or 'false'

[wintry steppe]

why can you not parameterise it if it does not admit a solution tho?

[fickle citrus]

The usual case for no solution is inconsistency, like x=0 x=1

so there is no real number that can satisfy both x=0, x=1

If there is no real number that can do it, there is no way to parametrise

[restive hearth]

What's the most efficient method to calculate the determinant of a matrix? I currently use Laplace expansion as a recursive function in my code but it runs at O(n!). I'm trying to implement row reduction but just looking around it seems to run at O(n^3), which is better than my previous implementation, but I think that better and more intuitive solutions do exist. There is also LU decomposition Anyone has any idea on what I can do?

[lavish jewel]

unless your matrix has further structure, stuff like LU, QR, or other decompositions are about as good as it gets

[restive hearth]

Ah I figured, I guess I could try making the workload lighter by hardcoding determinant calculations for smaller sized matrices

And maybe check for shortcuts like upper/lower triangular matrix diagonals

Thank you 🙂

[grand fog]

Hello, is the 3b1b and ocw enough to learn linera algebra?

[lavish jewel]

no, you'll need to read a book and/or attend a lecture at some point

that's only complementary content

they don't even cover any exercises beyond small, toy examples

they're more for people that have already acquainted themselves with the concepts beforehand

[grand fog]

what about for ML

[lavish jewel]

that depends on whether you wanna understand what you're doing or just use it

to only use it, you can read blogs and random junk

to understand it and make new stuff for new applications, you need to know what you're doing. then you'd need to study about linear algebra, statistics, and optimization

[grand fog]

Ah i see. Thank you

[grand fog]

What do you think about the exams and the problem sheet OCW offers?

[lavish jewel]

they can help practice

[short magnet]

if you're looking for a book then Linear Algebra Done Right by Sheldon Axler is good imo.. Although the exercises are mostly calculations and all, it's none the less well written and there are YouTube videos to complement your journey.

it also isn't'that long

I also think that OCW problem sheets aren't enough if you really want to get your feet deep into linear algebra

[oblique cargo]

any way to visualize higher dimension vector space or abstract vector space such as K[X] or M_n(K)

it might increase my problem solving speed at linear algebra exercices

[short magnet]

For M_n(K), I guess I just "visualize" the way it transforms basis vectors of K^n

it really depends on the context..

I don't really know how I'd visualize K[X].. I mean, I can't think of it as some sort of transportation or anything. You could certainly think of polynomials as sequences that are "almost nulle"

i.e. the sequence becomes null after reaching a certain n

OR. You could treat it as an euclidian domain

instead of considering the coefficients. You could think of it in an arithmetic way.. All polynomials are products of irreducible polynomials similarly to how all integers are written uniquely (with fixed order) as a products of prime numbers

this way of thinking extends well to rational fractions

I don't think that there is some sort of way to visualize them, or even a way of visualizing them that encapsulates their applicability.

it's perhaps better to continue treating them as abstract objects, and learn to treat them the right way at the right time. Which is something that can be done, I think, by working through many different problems ^^

[short magnet]

Michael Artin's Algebra has solid chapters on linear algebra. There is also Linear Algebra Done Right by Sheldon Axler. It's a way more approachable than Artin's... If you're good in french then : Xavier Gourdon's Les Maths en tête : Tome Algèbre is superbe!

Best thing is, you can couple it with his Tome d'analyse for some topology and analysis (functional and real analysis )

[deft apex]

If you are looking for a book wi5 a lot of problems I hear Anton is decent

[short magnet]

If you are okay with ones in french, just DM me and I'll hook you up with lots of problems to work through ^^

I think Axler is computation enough as it is

Yep

it goes into it

just in the end

basically, when the determinant is introduced early on, according to Axler, then we tend to use it to prove many statements and theorems

so Axler puts it aside

and uses different methods of proof

that don't use the determinant

and only introduces it in the last chapters

^^

I mean, if you're familiar with basic proofs (direct, induction, contraposition, contradiction ...) and have done some proofs

are familiar with some basic set theory and all that

I think you should be okay

let me take a quick look at the book to refresh my memory

I'll get back to you rn

Lots of undergraduate books initially review proofs, sets...etc

anyways, I think you should be alright

as long as you're armed with some solid background on set theory and proofs.

I mean, there isn't much to learn in logic and sets initially

it's better to just get a handle on the basics and refine your abilities in linear algebra, calculus...etc

otherwise you wouldn't have much to work with

much to prove..

what I'm saying is that you'll be doing more of them right now ^^

yeah

[grand fog]

Thx gusy

[torn stag]

The linear algebra book here is good: https://mtaylor.web.unc.edu/notes/linear-algebra-notes/

[lean acorn]

Hello I was wondering if anyone has any resources that talk about topics in linear algebra but through the lens of category theory? By topics in linear algebra I mean things like tensor product, bilinear/multilinear forms, direct sum, dual space, etc. For class I have to research how category theory can explain them or how what they fall under in category theory.

[half ice]

All entry books in category theory will contain these as examples

[fickle meteor]

I have a trouble for understanding this one. First, to clarify a typo, the $p_\alpha$ should really be $p$ at the end of the second line of the proof. And Second, my professor defines affine spaces to be cosets. But I don't know why all these affine spaces can share a same $p$ added to different subspaces? I know two cosets $u+P$ and $v+Q$ are equal iff $P=Q$ and $u-v\in P$, but how this is proven in this particular setting?

[stoic python]

Luna⭐

[pearl turret]

If I have this matrix in reduced echelon form, how can I tell if it is consistent and/or unique?

[half ice]

I'm finding Leinster is digestible @lean acorn

@pearl turret
Consider writing each line back into the equations they represent. The answer will hit you real fast if you do

[pearl turret]

not unique

but is that the only way?

or the most simple way

[half ice]

Well, now you should be able to do it by looking at the matrix. There's more variables than equations

[pearl turret]

oh

[half ice]

Unique solutions = every row has a pivot

[pearl turret]

i dont rly get what a pviot row is

or well we do pivot column

in our class

[half ice]

Sorry see the edit

Actually I'm wrong. Unique solutions is when you row reduce down to identity

[lean acorn]

Thank you! This is much easier to read than some of the other resources I found

[pearl turret]

Do you think you can help me with pivot positions?

Im asked to find the pivot positions on this

[half ice]

Gotta row reduce first

[pearl turret]

oh

well the above matrix is the row reduced one

but it says

[half ice]

Then a pivot is any column with only a 1

[pearl turret]

thats the given matrix at the top of the previous page

[pearl turret]

how do i repeat that to the

non reduced version

[half ice]

You don't haha

Gotta row reduce to see them

[pearl turret]

what is he asking for then

😮

[fickle meteor]

Ok i now get the answer. Because $p\in P_\alpha = p_\alpha + U_\alpha$, we have $p = p_\alpha + u$ for some $u\in U_\alpha$, and so $p - p_\alpha = u\in U_\alpha$ and so $p+U_\alpha = p_\alpha + U_\alpha$

[stoic python]

Luna⭐

[nocturne jewel]

The question is if that system has solutions or not

[willow crag]

Question, if eigen decompositions usually order the eigenvalues by size, must the the eigenvectors not be normalised for this order to matter to begin with?

[dusky epoch]

no

[lucid glacier]

You don't even need a notion of norm if doing regular decomposition

[willow crag]

Oh ok, but if utilised directly as a consideration of variance along the vectors (Like PCA/EOF), it's still the case?

I mean, if a vector isn't normalised I guess it has some kind of scaling that the eigenvalues do not account for, but perhaps that factor is equal among all the vectors

[tranquil steeple]

For many structured matrices the ordering of eigenvalues is a very hard problem (I am actually writing several papers on different approaches right now). We do not yet know how to order the eigenvalues to the true ordering, but we have a couple of approaches that seem to yield the same result. We have however not been able yet to scale it up to actually verify it. (this is not for general matrices, and standard solvers always sort the eigenvalues in non-decreasing order of the real part)

[tranquil steeple]

what do you mean? eigenvectors are not unique.

[willow crag]

Yea, but I thought that would then require the eigenvalues to be nonunique accordingly.
But from I can find, I think understand the vectors wrong, and I guess they should be seen as a direction/rotation rather as a part of the scaling

[tranquil steeple]

The eigenvalues are unique. Think of it like $Ax=\lambda x$ for a specifik eigenvalue $\lambda$, then you can take som scalar $a$ and have $aAx=A(ax)=a\lambda x=\lambda (ax)$ and then $y=ax$ is just some other scaling of $x$ and $Ay=\lambda y$

[stoic python]

Sven-Erik

[tranquil steeple]

You could keep $aA=B$ and $a\lambda=\gamma$ and have $Bx=\gamma x$ where $\gamma$ is an eigenvalue of this new matrix $B$

[stoic python]

Sven-Erik

[tranquil steeple]

And there are situations where people do not scale the eigenvectors with the 2-norm. I think for Markov procsses it is often used the 1-norm, and I work on an application where we need to "unnormalize" numerically computed eigenvectors with a very specific scaling

[willow crag]

Ah, ok, so it's all bound by the eigenproblem, like you described, right? So that's why the scaling doesn't really matter

I think maybe I overthought it a bit then

I didn't think that markov processes used eigendecompositions, but then again it's not something I've barely ever touched upon

[tranquil steeple]

for markov processes you usually are interested in the eigenvector associated with the smallest (or zero) eigenvalue

I do not know much about it either, but that is how I have understood it

(here is the only experience I have with it https://www.sciencedirect.com/science/article/pii/S0024379521002408)

Same thing for Fokker-Planck and Master equation discretizations

[willow crag]

Well you certaintly understand it better than I do, haha. On a sidenote, don't know if that's ScienceDirects way of doing it, but I like the way the equations are aligned with the paragraphs

But anyway, thanks for the help Sven-Erik. I'll get writing again

[tranquil steeple]

np 🙂 and there might be applications of like PCA and in ML etc where you need the eigenvectors to be normalized. I do not know.

[lavish jewel]

more simply, btw, the eigendecomposition of a matrix can be written as a sum of rank 1 matrices

and addition is commutative

so no, the order doesn't matter as long as you swap the columns and rows accordingly

[tranquil steeple]

Yes, the order can matter if you work with matrix sequences.

[lavish jewel]

sure, and also for actual computations of nitty gritty things

but just looking at a single matrix on a piece of paper...

[tranquil steeple]

Well, for example a circulant matrix, there is a proper true ordering of the eigenvalues (up to a circulant shift) and it is not in a monotone ordering

[lavish jewel]

wdym by proper ordering?

[tawny harbor]

Is there any case that we can get eigenvalue by norm of its eigenvector?

[tranquil steeple]

according to the spectral symbol

[lavish jewel]

you could still shuffle the eigenvectors and eigenvalues accordingly and get the same result

that's more convention that anything

[lavish jewel]

not directly, no. iteratively, yes. if you know the norm of the eigenvector and multiply it again by the matrix, the change in the norm will tell you the eigenvalue

[tranquil steeple]

Sure, but then it is according to the "rearranged symbol" (if we look at the circulant case) and then you lose a lot of information, which can be utlized in different algorithms.

[lavish jewel]

you didn't lose any info, it's just that all software is set up on the same convention

you make the problem more annoying, definitely, but not wrong

[tawny harbor]

What I did was power method, there I started with an arbitrary vector v_1, then basically v_n = A * normed(v_n-1)

[tranquil steeple]

As I said, "true" ordering, which is not the same as "correct" ordering 🙂

[tawny harbor]

And the norm of the convergent vector seems to be the eigenvalue

[lavish jewel]

what is normed(v_n-1) there?

like u make it unit norm?

[tawny harbor]

Yup

[lavish jewel]

then yeah

just be careful with the sign

[tawny harbor]

Can you reference me on that?

[tawny harbor]

I'm looking to read on why this occurs

[tawny harbor]

Sign of?

[lavish jewel]

of the eigenvalue

it could be complex, too, but if you take the norm, you'll only get the magnitude

[tawny harbor]

I see, thanks

[lavish jewel]

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

you can check the references there

the idea is that if a matrix has a dominant eigenvalue, it'll dominate the others after several multiplications

you can take a random vector and decompose it into a superposition of the eigenvectors of the matrix

[tawny harbor]

Yeah I've found it

Thanks

[lavish jewel]

when you multiply the vector by the matrix, the component parallel to the largest eigenvalue has its amplitude increased more than the others. by normalizing at each iter, you're making all the other components decay to zero, leaving only the largest one

it's super slow if the largest eigenvalues are very similar, and won't work if 2 or more of the largest vals are the same

[zealous junco]

it will still work tho, i think. just will converge into the subspace but not to ± some vector

[lavish jewel]

sure, that's what i meant

[zealous junco]

ah ok, i just wanted to check

[tawny harbor]

I'm not sure why people then use Rayleight quotient 🤔

[lavish jewel]

for what?

[tawny harbor]

http://mathreview.uwaterloo.ca/archive/voli/1/panju.pdf

For example here,

They use Rayleight to find corresponding eigenvalue

[lavish jewel]

precisely for the reason i stated earlier

this way gives you the correct sign/phase

[tawny harbor]

I see, cool

Thanks vm

[lavish jewel]

you can double-check yourself that this gives x^H M x / x^H x = lambda x^H x/ x^H x = lambda

whereas the iteration is rather | M x | / | x | = | lambda |

it's not needed tho

you could inspect a single component of the vector and just divide it by its previous value, due to how scalar multiplication works

but this way has an averaging effect

i'd guess it's more stable

[tawny harbor]

👍

[tranquil steeple]

You might want to check out this paper too: https://epubs.siam.org/doi/abs/10.1137/0720002?journalCode=sjnaam

Improving the Accuracy of Computed Eigenvalues and Eigenvectors

[lucid glacier]

I'm trying to make the concept of linear transformations on a normed space being 'close' to each other

I thought of defining a metric by using the norms on some basis (idek if it would be well defined tho)

Is there smth like this already that's well established

I thought about the operator norm but i'm not sure if it will give me.what I want

Also I mostly just care about automorphisms

[tranquil steeple]

maybe https://www.maths.manchester.ac.uk/~higham/narep/narep161.pdf

[astral hare]

I've come across a question that says give an example of T(transformation) such that T^4=-1, but what do they mean when they say a transformation is -1, Im guessing its a misprint and that they mean the negative identity, anyone know anything?

[lucid glacier]

some books/people denote the identity by 1

[lucid glacier]

the beginning seems to assume some matrix norm

[astral hare]

Yeah thats what I was thinking but they've been using I the whole time

[lucid glacier]

weird

[tranquil steeple]

I dont know anything about it, just heard about the types of problems in some talks

maybe it does not apply

[lucid glacier]

this isn't for an optimisation problem so probably not the direction I need

but thanks anyways

[tranquil steeple]

So, what are you trying to do? 🙂

[lucid glacier]

Eh well it's a bit roundabout but i'm trying to make precise the idea of shifting a convex/simplical cone in R^n a little such that it's totally irrational (it doesn't intersect any points of Z^n except 0), and show that if you change a cone in this way the sail (the boundary of the convex hull of the points of Z^n inside the cone) only changes far away from the origin.

[tranquil steeple]

Good luck 🙂 Never heard of sails

[lucid glacier]

thanks

it's a pretty niche subject

[west saddle]

So with doing gaussian

if I am following all of the rules correctly, no matter what path I choose, I should get a unique answer when it is in REFF?

I feel like I'm following the rules, but I can't seem to get the right RREF

[west saddle]

<@&286206848099549185>

I've done this problem a few different times, but i can't seem to get what i need to cancel out cancel out

[sonic osprey]

Just post the problem and what you've done

It is true that rref is unique though

[sick schooner]

Anyone know a great website for learning linear regression?

[stark acorn]

I dont get what I am doing wrong

https://www.symbolab.com/solver/system-of-equations-calculator/3x %2B 3y %2B 6z %3D 9%2C x %2B y %2B 2z %3D 3%2C 2 x %2B 5y %2B 10z %3D 15%2C −x %2B 2y %2B 4z %3D 6

this is soooo fustrating

i figured it out

stupied me

[wind shore]

is the b^Ty<0 term a dot product or matrix multiplication? also in general, a vector coming out from R^n, is it nx1 or 1xn?

[lavish jewel]

both

a row vector times a column vector is the same as a regular matrix multiplication

the resulting matrix is 1x1, so it's a scalar. you can also double check that if you write the sum, it's the same as a dot product

usually vectors are taken as columns unless stated otherwise

[wind shore]

gotya so I guess it's equivalent here in this pdf right? the term y^Tb is a scalar

[lavish jewel]

that just looks wrong

the A^T y they put there is undefined

and also the y^T b

they probably meant y in R^{m x 1}

[wind shore]

yeah seems that way

I guess that would only explains

@lavish jewel anyways appreciate the help

[desert rapids]

So here is a problem which I have solved, but I am still a bit confused with the line (in the problem statement), 'In particular, x,y,z,w are coplanar

Is it asking us to show that they have to be coplanar? This doesn't seem like a necessity

[hard drum]

It's asking you to prove the statement in the preceding sentence and is then remarking that this also means they are coplanar, i believe

[desert rapids]

Huh

So it is saying that they form a parallelogram iff they coplanar?

@hard drum

[hard drum]

no, but if they do form a parallelogram, they must be coplanar

[desert rapids]

Why is that?

@hard drum

I don't see that restriction in my calculations

[hard drum]

because a parallelogram is planar

[desert rapids]

My bad