linear-algebra

[heavy crown]

yes, and so is rank(A)

[dusky epoch]

yes

but you must have rank(A) ≤ n too

no matter what

a matrix cannot have higher rank than its number of rows or its number of columns.

[heavy crown]

Yes because rank(a)≤min(m,n)

Then

[dusky epoch]

that's it

[heavy crown]

It means m is smaller then n if rank(a)=m

Right?

[dusky epoch]

thats it

youre done

rank(A) = m

and rank(A) ≤ n

thats it

youre done, you got your m ≤ n, that's it

[heavy crown]

I understand

Thank you it makes sense to me now

[heavy crown]

If there is linear transformation T: V -> V I can include immediately that's it is isomorphism?

[lavish jewel]

a zero matrix is V -> V

[heavy crown]

oh right

And if they say T≠0

[lavish jewel]

still no

[heavy crown]

oof

[lavish jewel]

it would have to be invertible

[heavy crown]

okay

thank you

[lavish jewel]

hmmm invertible in a loose sense, to be fair

[dusky epoch]

why do you ask this

[heavy crown]

I had a question
Let T: V->V be linear transformation, dim(V)=2.
T≠0, T²=T, T≠id.

So I thought I could use this

[dusky epoch]

oh now it turns out dim(V) = 2

you cannot just leave out important details like this

what are you asked for? to determine whether or not T is invertible?

[heavy crown]

my bad sorry

They asked to prove T is not invertible

This is what I tried:

[dusky epoch]

well ok, i guess dim(V)=2 is not very important in this case

[heavy crown]

By contradiction assume T is invertible

[dusky epoch]

T invertible => T = id, contradiction.

is that your argument?

[heavy crown]

Let T(x)=y
Then T(T(x))=y (based on given)
Which also means T(y)=y
which is contradiction to T≠id

Yea

so that's good

The second part was where I struggled

Prove: there exists a basis {v1, v2} to V, so that T(v2)=0, T(v1)=v1

I can conclude from the dim that there exists such basis that consists of v1 v2 but I don't know how to continue

[lavish jewel]

idk if what you wrote above is enough

orthogonal projections do all of that

[heavy crown]

hmm

[lavish jewel]

which also goes hand in hand with hte second part

or maybe i'm wrong

[teal grotto]

@heavy crown, what are you trying to show again?

[heavy crown]

Prove: there exists a basis {v1, v2} to V, so that T(v2)=0, T(v1)=v1

[heavy crown]

And this is Given

[crude falcon]

given the vector space V generated by linear combinations of the elements {x,y,z}, consider C = {x,y,z} as the canonical basis of V

what does it means that C is a canonical basis opposed to a regular basis?

[wintry steppe]

doesn't mean anything

people like to call something canonical when it is for them the most natural thing to consider out of all possibilities

[teal grotto]

it should be the case that if $T^2=T$ then $T$ is invertible if and only if $T=\text{Id}_V$. one direction is clear. now suppose that $T$ is invertible. Then $T=T\circ T\circ T^{-1}=T\circ T^{-1}=\text{Id}_V$.

so since $T$ is non-zero and $T$ is not the identity, then $\dim\ker(T)=\dim\text{im}(T)=1$. If you need more clarification on this part, please ask. but you should be able to finish it from here.

[stoic python]

coycoy

[teal grotto]

when an object is called ‘cannonical’, it typically means that it it was the most natural choice that satisfied some property, or it is a natural object which satisfies some property, obtained without making choices. for example, there is a cannonical isomorphism from a vector space V into its double dual, meaning we did not have to choose a basis to obtain this isomorphism. conversely, there is not in general a cannonical isomorphism from V into its dual, because we need to choose a basis of V to obtain the isomorphism

[heavy crown]

thank you for detailed answer :)
You know maybe how to do the 2nd part of that question?

[teal grotto]

the second part is what i left for you to do. think about what it means for the kernel of T to have dimension 1. similarly for the image.

i can’t tell if that’s a good or bad reaction haha

[heavy crown]

hahaha let it be whatever you want it to be

[teal grotto]

i’m going to assume by the existence of the emojis T Y underneath the reaction that i have helped you figure out your basis

[heavy crown]

Your assumption is correct

[dire thunder]

a way to go is

T is not invertible means that null T is not just {0}

so there is nonzero vector v_1

also there is nonzero vector v_2 not in null T

v_1 and v_2 should be independent

hence T(v_1)=0, T(v_2) != 0

already a basis

you now just have to show that exist vector v_2 s.t T(v_2)=v_2

[teal grotto]

you have v_1 and v_2 swapped. and how are you concluding that T is not injective?

[dire thunder]

wym

it is given by assumption

T is not invertible and T is from V to V

[teal grotto]

not it’s not.

T is just not the identity and it’s not the zero transformation

[dire thunder]

i mean T: V->V is given and T not inverible already proved

[teal grotto]

ok. i see now. it seems kind of like a lot of contradictions needed to argue your points about v_1 and v_2 being in the kernel and not in the kernel respectively

[dire thunder]

no

if T:V->V is not invertible it cannot be neither surjective

nor injective

since for operators these conditions are equivalent to invertibility

[teal grotto]

true

solid argument then

[dire thunder]

well for part T(v_2)=v_2 we prolly can make an argument to eigenvalue but it is not guaranteed that T would have eigenvalue except than zero in this field

actually

[teal grotto]

yea i was trying to figure out an argument for that

[dire thunder]

i found on mse that eigenvalues of idempotent are either 0 and 1

[teal grotto]

it seems to make sense intuitively, just thinking about matrix multiplication where this works out

[stoic python]

Commander Vimes

Commander Vimes

[teal grotto]

nice

[dire thunder]

thus it is enough to take nonzero eigenvector not in null T

(if it exists and it should exist)

[teal grotto]

yea that was what i wanted username to conclude.

[dire thunder]

i mean prolly there is another way to prove this

prolly just by matrix multiplication

[teal grotto]

i mean, he would have to scale it by the inverse of the eval without the proof you have

i tried doing that earlier but it got kind of messy

i actually tried to find counter examples of this (matrices) when you have fields with different characteristics, but everything still seems to check out.

[heavy crown]

hey thank you for referencing and helping too. I understand this part

[heavy crown]

but this is quite problematic because they we don't know eigenvalues yet

[heavy crown]

so probably there is hm just the T(v_2)=v_2 is a problem

[dire thunder]

you may fuck with matrix multiplication ig

[heavy crown]

hmm

[teal grotto]

you could always just notice this fact without actually referring to or mentioning evals, like a lemma that you have.

[heavy crown]

yea youre right

Hey this should be a simple question but
Given: Let A be matrix of order nxn. A-I is invertible. A²=A
Prove: A=0

Here's what I tried:
A²-A = 0
|A(A-I)| = 0
|A| * |(A-I)| = 0

Then because A-I is invertible so |(A-I)|≠0. so |A| = 0
Is it enough to say A = 0?

(i don't think so)

[dusky epoch]

no it's not enough

you've overcomplicated it

you don't need determinants here at all

A(A-I) = 0

multiply both sides by (A-I)^-1

that's it

[heavy crown]

waw its that simple

yea idk I think too hard

thank you

[bitter cape]

Hi! I've got a question - I don't understand how to tackle this as it's in R4.

It goes "If P: R4 - > R4 the ortogonal projection in subspace H = Span{(1, 0, 1, 0)} of R4.

Present P(2, 0, 7, 9)" I'm not sure where to even start with this.

[teal grotto]

can you start by writing out the definition of P?

[bitter cape]

It's all that's in there. I suppose they put it there to represent a P of Polynomial

[lavish jewel]

P is just a name they gave to the transformation

start with the definition of an orthogonal projection

[teal grotto]

and <•,•> is the standard dot product

[stoic python]

coycoy

[teal grotto]

the notation P is literally just proj_H, meaning the orthogonal projection onto H

[lavish jewel]

that looks right

[bitter cape]

Problem is I'm not sure where to go from there

[teal grotto]

well it asks you to compute the projection of y=(2,0,7,9) onto H. so just plug that into proj_H and calculate

[bitter cape]

alright I'm gonna try here, thank you

[wintry steppe]

Suppose V and W are finite-dimensional and T in L(V,W). Prove that there exist bases of V and W such that with respect to these bases, all entries of M(T) are 0 except that the entries in row j, column j equal 1 for 1 ≤ j ≤ dim range T.

I can't solve this problem

How can I get started?

[lavish jewel]

what is M?

[wintry steppe]

The matrix of the linear map T

[teal grotto]

@wintry steppe think about $\ker(T)^{\perp}$ and $\ker(T)$ as sub spaces of V

[stoic python]

coycoy

[teal grotto]

ah wait. that’s close but we don’t always have inner products. unless your ground field is the real or complex numbers?

[wintry steppe]

I don't think so?

it's just some field

[teal grotto]

oof. if it’s an inner product space then that’s the right way to do it. back to the drawing board

[wintry steppe]

I'm not sure what you mean by inner product space

We've not covered inner products yet lol in the book

[teal grotto]

think of it as just a more general term for dot products

[wintry steppe]

No I know what it is

But I don't know how it affects the space etc

Have not learned that

[teal grotto]

oh. well it lets you say when two vectors are perpendicular/orthogonal and gives some notion of an angle

which would be very helpful for this problem

[lavish jewel]

you could use the proof of the svd for this

[wintry steppe]

Have not learned that either

[lavish jewel]

do you know that hermitian matrices are diagonalizable?

[teal grotto]

wow that’s a powerful theorem that looks really cool

[wintry steppe]

Don't even know what hermitian matrices are

[teal grotto]

me neither haha

[lavish jewel]

self adjoint?

wait, no

[teal grotto]

yes

[lavish jewel]

yes

yeah, self adjoint matrices

if you haven't see that either, i'm out of ideas. you could play change of basis shennanigans

[teal grotto]

u need an inner product space to even talk about those

we’re not given one sadly

all my homies hate change of basis

[wintry steppe]

Haven't learned about change of basis either

[lavish jewel]

oof

[teal grotto]

bruv

have you learned about eigen vectors and eigen values?

[wintry steppe]

No

I've learnt about maps, and matrix operations

I know lots of things about maps

[teal grotto]

okay good. the idea i’m thinking of will involve lots of maps if it works

[lavish jewel]

you can express a change of basis as a map without calling it change of basis 😛

[teal grotto]

we’re on the same page edd

[dusky epoch]

take a basis of ker(T), extend it to a basis of the domain, choose a suitable basis for the codomain based on that

[teal grotto]

bro why was i restricting myself to ker(T) perp omg

ann you’re a beast

[wintry steppe]

So u_1,...,u_k is a basis of ker(T)

Then u_1,...,u_k,v_1,...,v_l is a basis of V

Oh wait I think I see it now too

That is very clever

[stoic python]

coycoy

coycoy

[wintry steppe]

What about the remaining vectors? For example, say Tv1 is nonzero. So the first column should only have A1,1 in the top left, and other entries should be zero

But how do we make sure the other entries are zero?

[dusky epoch]

what remaining vectors

[wintry steppe]

The vectors we used to extend to a basis of V

v_1,...,v_l in my notation

[teal grotto]

the ones that make up the basis of the kernel of T?

[dusky epoch]

no, the other ones

[wintry steppe]

Tv_1 = A1,1 w_1 + ... + Am,1 w_m

[dusky epoch]

the first l vectors in your basis of W will be Tv_1, Tv_2, ..., Tv_l

that's how you ensure the 1s along the diagonal

[wintry steppe]

How do we make A2,1,...,Am,2 all zero

I know that ensures the 1's

But not the zeros elsewhere

[teal grotto]

hmm

@wintry steppe the standard basis vectors that died by getting mapped to the kernel of T will be the zero columns. just put those at the back of your ordered basis

[wintry steppe]

Yes obviously

But what about the zeros elsewhere?

[teal grotto]

those are the zeros elsewhere lol

[wintry steppe]

We want Tv_1 = A1,1 w_1, not Tv_1 = A1,1 w_1 + A2,1 w_2 + ...

Why is the rest of the column 0?

[teal grotto]

lets say i take in e_i and it gets mapped to some basis vector in the kernel of T, say v_i

then T(v_i) gets mapped to the zero vector in W

and that has to correspond to a zero column of the matrix of T

[wintry steppe]

Okay you don't get it

Let me try to explain it again

Assume that Tv_1 is NOT in the kernel

That means Tv_1 is nonzero

But Tv_1 = A1,1w_1 + A2,1 w_2 + ... + A_m,1 w_m by definition

However we want A2,1, A3,1,...,Am,1 to all be zero

We just want A1,1 to be nonzero

But we didn't guarantee that A2,1 A3,1 and so on are all zero

[teal grotto]

what do your w_i's represent again?

[wintry steppe]

w1,...,wm is a basis of W

[teal grotto]

but which part? one is from T and the other part is extended

[wintry steppe]

What?

Neither?

[teal grotto]

ah my mistake. thanks for ur patience i was having a bruh moment

[wintry steppe]

No problem

[stoic python]

SK

[wintry steppe]

Also, how can I determine the action of the operator on any given element?

[native rampart]

V is finite dim?

Feels like this would require choice in the infinite dimensional case

[fringe burrow]

Does dimension of a vector space depends on a field or not? For example if we look at V finite dimensional, is his dimension over C is equal to it's dimension over R?

[native rampart]

Yes

C over C is 1 dimensional

C over R is 2 dimensional

[fringe burrow]

nice thank you. C over R it's basis is {i,1} and C over C it's basis is {1}?

[native rampart]

Yes

[fringe burrow]

But is the reason for it is because we look at C over C as simply scalars without looking at a+ib decomposition?

for C over C

[native rampart]

Yes

[fringe burrow]

nice, thank you

[native rampart]

We don't need to decompose it in C

[fringe burrow]

But is there a theorem or something to calculate explicitely the dimension of a vector space over C?

I had an exercice in exam Schur's lemma but I didnt succeed it 😦

and my professor never mentione this kind of things

[native rampart]

I guess it will always be double?

dim over R=2 dim over C

[fringe burrow]

alright, thank you!

[wintry steppe]

Infinite dimensional.

[native rampart]

Not sure if it does in that case

[wintry steppe]

It doesn't determine the operator?

[native rampart]

Yes

[wintry steppe]

But the notes I'm reading are telling me that it does.

[native rampart]

Does the space have any more structure?

I guess they might be implicitly assuming choice

[stoic python]

SK

[wintry steppe]

isn't this immediate from non degeneracy

Sure, how do I get the operator then?

give me a second im on my phone

How can I determine how the operator will act on a given vector?

i can tell you that if A and B are both linear operators satisfying your condition, then A = B, but i can't give you a detailed description of how A looks

if $A, B$ are linear operators on $V$ such that $$\langle v, Aw \rangle = \langle v, Bw \rangle$$ for all $v, w \in V$, then, if $w$ is fixed, $$\langle v, Aw - Bw \rangle = 0, \quad \forall v \in V.$$ by non degeneracy, $Aw = Bw$

[native rampart]

AKA <x,x>=0 implies x=0

[stoic python]

R2T2

[wintry steppe]

and since w was arbitrary, A = B

[teal grotto]

my internet went out

[wintry steppe]

Not what I want.

okay then

what i wrote is what "uniquely determines the operator" means, but ok, what do you want?

What we know is
[\langle \psi, A \phi\rangle \forall \psi, \phi \in V]

[stoic python]

SK

[wintry steppe]

yes?

Let's take a random $\sigma \in V$. What is $A\sigma$, knowing the above?

[stoic python]

SK

[wintry steppe]

in full generality, i don't know. if you were able to choose an orthonormal basis you'd be able to find it, but your space is infinite dimensional

It's separable; now?

[native rampart]

Choice moment

[wintry steppe]

just do the same thing with a countable orthonormal basis xd

[crude falcon]

if C = {x,y,z} is a basis of V, how can I prove that B = {(x+z),(y+z),(x+y)} is also a basis?

[lavish jewel]

you know ax + by + cz = 0 only has a trivial solution w.r.t. the combination parameters a,b,c

show that you can do something similar for B

i.e. a'(x+z) + b'(y+z) + c'(x+y) = 0 needs to only have trivial solutions

you can rewrite that as (a' + c')x + (b' + c')y + (a' + b')z = 0

[crude falcon]

yep I see now

[wintry steppe]

can anyone help with some linear algebra terms?

[wintry steppe]

just ask

[atomic hound]

Hey guys what's wrong with my logic in this proof (from Ex 19 Sec 5B of Axler's LADR)?

Exercise: Suppose $V$ is a finite-dimensional with dim $V > 1$ and $T \in \mathcal{L}(V)$. Prove that ${p(T) : p \in \mathcal{P}(F)} \neq \mathcal{L}(V)$.

My Solution: Can I just use the idea that there exists $p(z) = a_0 + 0z + 0z^2 + \dots + 0z^m$, then $p(T) = a_0$. If $a_0 \neq 0$, then $a_0 \not\in \mathcal{L}(V)$ since that implies $T(0) \neq 0$?

[wintry steppe]

what is the big difference between rref and ref

im very confused on the differences

[wintry steppe]

backslash your set brackets

[atomic hound]

thank you

[wintry steppe]

ok so

gaussian elimination right, thats the same as forward elimination

[stoic python]

HimmyHow

[wintry steppe]

and gauss jordan is when its forward elimination and then backward elimination

[teal grotto]

quick question, what is P(F)?

[wintry steppe]

i see some people do gauss elimination to bring a matrix into row echelon form, then they bring the matrix back into a systems of equations and back-substitute the equations to get the solutions, what would this process be called? Its not gauss jordan right? Cause gauss jordan is finding the solutions using only matrix form by making the REF into RREF by backward elimination

[atomic hound]

It's the set of all polynomials. In 2.13 of Axler he defines $\mathcal{P}_m(F)$ as the set of all polynomials with coefficients in $F$ and degree at most $m$ where $m$ is a nonnegative integer.

[stoic python]

HimmyHow

[atomic hound]

And he defines $p(T)$ as follows. Suppose $T \in \mathcal{L}(V)$ and $p \in \mathcal{P}(F)$ is a polynomial given by $p(z) = a_0 + a_1z + a_2z^2 + \dots + a_mz^m$ for $z \in F$. Then $p(T)$ is the operator defined by $p(T) = a_0I + a_1T + a_2T^2 + \dots + a_mT^m$.

[stoic python]

HimmyHow

[teal grotto]

ok thank you himmy

your idea works himmy. i like it

[atomic hound]

thanks @teal grotto , it seems like my answer took a very different approach than some of the other ones I've seen so I wasn't sure if I was missing something.

[teal grotto]

what were the other approaches

[atomic hound]

and another one here:

[teal grotto]

why are their's so complicated?

[wintry steppe]

If a whole column for a variable is 0, is it a free variable?

[atomic hound]

yeah that's what i'm wondering too... was worried that i was missing something.

[teal grotto]

is it because P(F) is infinite dimensional? there is no subscript here, unless its implied somewhere

[atomic hound]

It's possible, I interpreted P(F) without the subscript to mean it can be either finite or infinite (not necessarily infinite).

[teal grotto]

it cant be infinite dimensional in the first proof if F is a finite field...

[wintry steppe]

p(T) will be a_0 times the identity operator, not the scalar a_0

[teal grotto]

ah. thats what it was

[atomic hound]

If it helps, in Axler's book he restricts F to mean either R or C.

[atomic hound]

ahh

that's dumb of me

[teal grotto]

it got me too 😦

[atomic hound]

i led you astray

but thank you all i'll revisit the proof

[teal grotto]

nothing gets past @wintry steppe tho

[wintry steppe]

quite a bit gets past me

the other day i mixed up the rows and columns in working with the basis rep of a linear map

[teal grotto]

[lavish jewel]

considerable tteppa

[teal grotto]

god these emojis are so good haha

[wintry steppe]

i once (wrongly, of course) proved that every riemannian manifold is flat

[teal grotto]

how did that go

[wintry steppe]

it was stupid

[teal grotto]

lol i’m curious now

[atomic hound]

hey guys i'll ask another question here. I posted it on math exchange cause it seemed pretty involved but so far no response. No worries if you can't every question. https://math.stackexchange.com/questions/4170069/the-linear-map-that-sends-p-in-mathcalp-n2-mathbbc-to-pt-in-ma

(if it's bad practice to post on both here and math exchange lmk).

[wintry steppe]

What are the possible echelon forms of a nonzero 4x2 matrix?

[patent wigeon]

this is in the teacher's notes, is this right?

why is it 0 to 180?

2d is 0 to 360 so why is it changed to 0 to 180 in 3D?

[wintry steppe]

@atomic hound to answer your second point, yes, p(T) will be identically zero on V. this doesn't contradict "4.7" since p(T) isn't a polynomial, it's a linear operator

your suspicions are correct

Hw problem btw not exam for clarification

I think it’s A C D but idk

The examples r inconsistent

[wintry steppe]

for the first point, i think the definition of S you gave is enough to tell us what it is. it's the linear map $P_{n^2}(\bC)$ that takes a polynomial $p(z)$ to the linear operator on $V$ given by $p(T)$. if i wrote $$S(a_0+a_1z\cdots+a_kz^k)v = a_0v + a_1Tv + \cdots + a_kT^kv,$$ would that tell you what it looks like?

[stoic python]

R2T2

[wintry steppe]

for the third, i think it's pretty clear that S is well-defined and linear. the only times you need to worry about a map being well-defined is if you're making some kind of choice to define what it is on each element

[teal grotto]

@wintry steppe do you have a math stackexchange?

[wintry steppe]

yes

[teal grotto]

if you do, you should just answer it on there

[wintry steppe]

does anyone know the q I posted above? I’m very confused

[teal grotto]

or at least i would haha

[atomic hound]

@R2T2 wow thank you that helps a lot. and yes you are right for the first point. don't know how i missed that. that makes it much clearer.

[wintry steppe]

i might post an answer on mse

i'm not hungry for upvotes, and people on MSE will shred me alive if i make even a tiny error

[teal grotto]

only reason i say that is because it is a more permanent record than trying to scroll through discord threads

[teal grotto]

true true. but isn’t that what makes ppl better at writing proofs?

[wintry steppe]

people on MSE are ultra-pedants

there's a difference between being nitpicked to the bone and making a small error

[teal grotto]

they can be really irritating sometimes

i’m on there quite frequently but i try not to act like that. they are really condescending at times.

[atomic hound]

On this point, if p(T) is identically zero on V, does the proof imply that every v in V is an eigenvector for T? Or am I misinterpreting what it means for p(T) = 0?

[wintry steppe]

also, the proof you provided is wrong

p(T) will be zero, so c(T - lambda_1 I) ... (T - \lambda_m I) is the zero operator. this implies that one of the T - lambda_k I is not injective, i.e., lambda_k is an eigenvalue

it's not the case that one of the T - lambda_k I is zero (another error here, you didn't write the I)

it's only the case that one of them is not injective

[tropic turret]

Let $V=M_{n}(\mathbb{C})$ with the inner product $\left\langle A,B\right\rangle =tr\left(AB^{*}\right)$ Let $A\in V$ unitary diagonalizable. Show that $T:V\rightarrow V$ s.t $T\left(B\right)=AB$ is also unitary diagonalizable.

[stoic python]

go

[tropic turret]

Help?

[wintry steppe]

no, it just implies that c(T - lambda_1 I)...(T - lambda_n I)v = 0 for all v in V

the point is that injectivity of one of the T - lambda_k I fails (for, if they were all injective, their composition, the zero operator, would be injective)

i am going to make a post

[atomic hound]

So for all v in V, injectivity for at least one of the (T-lambda_k I)'s will fail, but not necessarily the same lambda_k?

hold on ignore this statement. let me think about it for a bit.

i guess what i'm confused about now is say one of the (T-lambda-k I)'s fail injectivity. So say it maps from dim n^2 to dim n^2-1. How does that eventually get you to the zero operator?

[wintry steppe]

are you confused about why one of them will fail to be injective

i don't understand what you mean

anyways someone already sniped me and posted a very similar answer

lmao

haha

[atomic hound]

oh okay maybe i was thinking about it the wrong way. i was thinking that starting from the assumption that one of the (T-lambda-k I)'s fail injectivity, it follows that you get the zero operator. but actually it's the other way around: starting with the fact that you get the zero operator, it follows that one of the (T-lambda-k I)'s fail injectivity?

[atomic hound]

loool

[wintry steppe]

right, since the composition $$(T - \lambda_1I)\cdots(T-\lambda_mI)$$ equals the zero operator, one of the $(T-\lambda_kI)$'s is not injective

[stoic python]

R2T2

[wintry steppe]

and if you write out what it means for T - lambda_k I to be non-injective, you'll see that it's equivalent to lambda_k being an eigenvalue

qed

[atomic hound]

amazing, thank you so much

appreciate you taking the time

[wintry steppe]

no problem

[tropic turret]

<@&286206848099549185>

[wintry steppe]

so i just finished a lin alg class but im still kinda struggling on linear transformations

does anyone have a good reference ?

OH and geometry of vector spaces

[signal current]

Try Sheldon Axler’s Linear Algebra Done Right. It’s a more theoretical approach but it is considered one of the best textbooks for undergrad LA

[wintry steppe]

tysm 💗

it also has a backwards treatment of determinants

so you'd want a supplementary book in that regard

i see some people do gauss elimination to bring a matrix into row echelon form, then they bring the matrix back into a systems of equations and back-substitute the equations to get the solutions, what would this process be called? Its not gauss jordan right? Cause gauss jordan is finding the solutions using only matrix form by making the REF into RREF by backward elimination(my instructor says backward elimination is the equavelent of back-substitution but in matrix form)

[limber sierra]

i dont think theres a name for the process

just "row reduce and back-substitute"

[wintry steppe]

why would you do that instead of just doing it through gauss jordan by keeping it in matrix form?

[torpid venture]

how can we use Vector Spaces and Linear Transformation in real life ?

can anyone give examples please?

[wintry steppe]

by taking a linear algebra class you use vector spaces and linear transformations in real life

[nocturne jewel]

refer to class examples \s

[tropic turret]

I'M LOST 😢

[wintry steppe]

one example is optimization (linear programming)

that's got quite a few "real world" applications

[torpid venture]

what do you mean ? i learned how to get null space, column space and all that but we didn't take how we use that in real life

[nocturne jewel]

Least Squares

[wintry steppe]

you used it in real life when you took your linear algebra class

[nocturne jewel]

The Schrodinger Equation is an eigenvalue problem

[wintry steppe]

a very large amount of modern mathematics depends on linear algebra in one way or another, that's another real life application

[torn hornet]

you will be very hardpressed to find something where linear algebra doesnt atleast make it easier

[primal valve]

I was wondering if someone can double check one of my proofs

im second guessing

it

[brisk fractal]

the A is not necessarily the same for W_1 and W_2

[torn hornet]

I mean you have the right idea, but yeah what bacono said

[torpid venture]

sorry it's my first time taking linear algebra that's why i'm a bit lost

[torn hornet]

try showing that they are subsets of each other or something like that

[torpid venture]

but anyway thanks guys

[primal valve]

damn

What if i show it like this?

does that work instead

[torn hornet]

this shows M_n \subset W1+ W2

and the other side is obvious (but making sure just bc)

so it would be sufficient

just make sure you understand everything thats going on

[primal valve]

bet thanks, so ig i ll rewrite it to reflect the screenshot

[atomic hound]

i can only speak for myself, but one thing (of many) that inspired me to study linear algebra was coming across a recommender system that used singular value decomposition. i believe a lot of machine learning stuff relies heavily on linear algebra. computer graphics too.

[deep warren]

Does anyone know if given any matrix A one can find an isometry S such that S^{-1}A is inferior triangular?

[torpid venture]

ohh i see, thank you so much i will do some research

[hollow finch]

if a matrix A has gaussian integer entries and a nontrivial null space, does it follow that there exists a basis for the null space such that all the basis vectors have gaussian integer entries?

[torn hornet]

actually disregard that (deleted somethign i said that was wrong)

I believe so, what did you try @hollow finch

[hollow finch]

not sure how i would prove it rigorously, especially for gaussian integers, but when finding a basis for null space of an integer matrix you seem to always get expressions for the vector in terms of the entries (at least of the REF or RREF form). sometimes they end up rational instead of integer, but rationals can always be scaled to integers. i imagine that the logic isnt too much different for gaussian integers.

sorry thats a bit rambly, im not quite sure how to articulate it

[torn hornet]

yeah thats basically the though i have

you do some elementery operations to reduce the matrix (each time the solution staying in Q[i]) and then you scale it up in the end

[hollow finch]

right

not sure if id have to prove it stays in Q[i] for all the operations (or at least there is a set of operations such that it does) as a lemma

[torn hornet]

probably how id go about it, describe the reduction as a sequence of certain type of operations and then claim all of these modify the coefficients of the matrix to stay in Q[i] and etc etc

[hollow finch]

the common thread between both ideas seems to be that its always possible to find integers x1 and x2 such that ax1+bx2=0 (or instead of 2, going up to n). first in just finding a vector in the null space, and second for reducing the matrix (like getting zero entries in the right places with row operations). seems basically like homogeneous linear diophantine equations. maybe using that i can prove the null space thing without needing to consider reducing the matrix?

[wintry steppe]

why does this have a unique solution when it has a row with full zeros? I can see theres no free variables but why does so many people say that if theres a row of full zeros then its infinite solutions?

am i misunderstanding the concept?

[dire thunder]

@wintry steppe in that case row of zero is fourth row

you had initially 4 equations in 3 variables

[wintry steppe]

ah yeah im dumb,

i get it now

[dire thunder]

good

[wintry steppe]

what if the third column in the picture was only zeros? What would that be consided as ?

[dire thunder]

@wintry steppe then assuming third row would be also zero row you would have free variable

[lavish jewel]

#real-complex-analysis

[sweet ridge]

Help pls

[crude falcon]

I have a linear map V->V f with that transformation, the vectors of the space are a linear combination of those elements, and C is the basis, how can I find the matrix Mcc(f)?

I guess I can take the transformation and reduce it to get the individual images, eg f(train), f(star)

and then take a vector with basis in C and apply that transformation?

[lavish jewel]

sounds about right

use what is given to find the effect of f on the canonical basis vectors, and put those as columns of a matrix

[dreamy iron]

Given vector space $\mathbb{V} $ What’s the group theory notation for this vector space: $\mathcal{L}(\mathbb{V} , \mathbb{V} )$

[stoic python]

a rainbow powered ninny

[dreamy iron]

$\mathcal{L}(\mathbb{V} , \mathbb{V} ) \stackrel{?}{=} End(\mathbb{V})$

[stoic python]

a rainbow powered ninny

[dreamy iron]

Or is it Endo?

Like how do I write T is in the endo-morphisms of V?

Or do I say endomorphism of V to itself?

[dire thunder]

@dreamy iron but not all endomorphisms are linear

trivial example f(x)=x^2

[heavy crown]

ey it should be simple but I'm having trouble ah
Given: Let A, B be nxn matrices. AB=0, BA=A
Prove: (A+B)²=A+B²

[teal grotto]

can you expand (A+B)^2 ?

[heavy crown]

Yes

A²+AB+BA+B² = A²+A+B² and then I need to show somehow A²=0

[teal grotto]

A=BA so A^2=ABA=(0)A=0

[heavy crown]

oh I see now

I should have thought about it

[dreamy iron]

But the set of all linear transformations from V to itself is an Endomorphism?

[dire thunder]

it is subset of set of all endomorphisms

[teal grotto]

in what context? is V an arbitrary vector space here?

[dreamy iron]

Yes

Arbitrary vector space over a field

[native rampart]

I guess your morphisms are linear transforms

[dreamy iron]

This is very light group theory being introduced….for which my skill points are lacking.
Context is light group theory for a second pass at linear algebra.

[dreamy iron]

I don’t even know what a morphism is

The most ive gathered is that it’s a generalization of a function.

And there are several prefixes one can affix to the word morphisms:
epi, endo, auto, iso, homo

[native rampart]

This is actually category theory

A category has morphisms

[dreamy iron]

Why did you delete it?

[heavy crown]

because nobody replied and i felt awkward haha

[dreamy iron]

Just wait for someone. Lol.

someone will get to it

No need to delete.

i was actually messing with it myself.

[heavy crown]

oh I'm sorry

I'll get it back hm

can someone help me understand what I'm doing wrong? I tried to find the basis of U∩W

[dreamy iron]

So you set the system to zero because youre trying to find a non-zero element of the null space

??

[heavy crown]

nope, because

Let v ∈ U∩W
v = M1(u1) + M2(u2) = M3(w1) + M4(w2)
0 = M1(u1) + M2(u2) + M3(-w1) + M4(-w2)
M1(x²+x) + M2(2x-1) + M3(-x²-1) + M4(-x-2) = 0
x²(M1 - M3) + x(M1 + 2M2 - M4) + 1(-M2 - M3 -2M4) = 0

[dreamy iron]

My next step was to attack it with this

[marble lance]

To show the dimension is >= 1, you just have to find any nonzero element of the space.

So just find any element in both W and U

[heavy crown]

after RREF:

[heavy crown]

yes my problem was telling the basis

what's the basis?

[lavish jewel]

you could try something geometric. get normals using cross products, solve for the line of intersection

then take 2 points on the line

[dreamy iron]

I’m going to try and add U and W next, using a subspace sum.

[heavy crown]

I don't think it's appliable in my exam

[heavy crown]

what's the next step after this?

[lavish jewel]

do they specify which techniques to use?

[heavy crown]

yea

unfortunately

[lavish jewel]

aight

[dire thunder]

a(x^2+1)+b(x+2)=ax^2+bx+(a+2b)

and

c(x^2+x)+d(2x-1)=cx^2+(c+2d)x-d

a = c

c+2d=b

-d = a+2b

c+2d=b
-d=c+2b

c+2d=b
c+2b=-d
2d=b
2b=-d

d = b = 0

wtf

[lavish jewel]

(-5,3,-5,1) is in the null space, yeah?

so that gives you the linear combination weights

[heavy crown]

yea (-5, 3, -5, 1) is the basis to null

[marble lance]

@dire thunder wtf are you smoking? 😂

[lavish jewel]

would mean (5,-3) in one basis is equal to (-5,1) in the other basis

[dire thunder]

vodka

[lavish jewel]

the vector you get is the basis of the intersection

[heavy crown]

you mean (-5, 3) would equal to (-5, 1) yes

so you mean I need to do

[lavish jewel]

i flipped the sign on purpose, cuz we want av1 bv2 = cu1 + du2

[dire thunder]

oh

i lost sign

[marble lance]

How did you go from c + 2d = b and c + 2b = -d to 2d = b and 2b = -d? Is c = 0? Then, yes, b and d have to be 0 too.

[dire thunder]

[lavish jewel]

but you have v1 v2 u1 u2 times (a b -c -d) = 0

[heavy crown]

oh I see

[lavish jewel]

(do double check, but i'm pretty sure this should work)

[dire thunder]

a=c
b=-d-2c gives parametrization for linear combination

[lavish jewel]

oh wait you flipped the sign of the other 2 vectors

nvm, no sign change needed, then

[heavy crown]

so if we take the vectors that didn't flip (the first two) which are -5 and 3 and do
{ (-5t) * (x²+x)+(3t) * (2x-1)| t eR} = {t(-5x²-5x+6x-3)| t eR} = span(-5x²+x-3)
does it mean the basis is (-5x²+x-3) to U∩W?

[lavish jewel]

looks ok

[heavy crown]

yea :] theres only one thing that confused me

[heavy crown]

they said we could get rid of the arbitrary vectors to find the basis so according to that logic it would be without the fourth vector,
But it is inequal to (-5x²+x-3)

I mean like there's another method when we do C(A) and get rid of arbitrary, I guess you can't use it here because it's not C(A)

[lavish jewel]

i'm not familiar with the notation

can't comment

[heavy crown]

when you do column space of a matrix you get the span of the matrix, but it isnt lineraly independant
So to make sure it's linearly independent you do RREF to C(A) and only those vectors that aren't arbitrary are together linearly independent

[lavish jewel]

by C(A) do you mean column space?

[heavy crown]

yes

sorry for not mentioning

[lavish jewel]

yeah you could gram schmidt everything if you wanted

[heavy crown]

isn't it column space?

[heavy crown]

okay I guess I'll work by this method for safety

thankyou edd :]

[heavy crown]

Anyone ?
Given: span{v1, v2} ∩ span{v3, v4} ≠ {0} , v1 ∉ span{v2, v3, v4}
__Prove: __ {v1, v2} are linearly independent

I know that it means if v1 isn't a part of their span, so adding it to their span won't ruin its linear independancy, but I don't know that {v2, v3, v4} are linearly independent?

I'm not sure what this tells me span{v1, v2} ∩ span{v3, v4} ≠ {0}

that the dim of their intersection is >= 1?

[dusky epoch]

yes

i think this may be best done by just applying the definition tho

take $c_1 v_1 + c_2 v_2 = 0$ and attempt to prove $c_1 = c_2 = 0$

[stoic python]

Ann

[dusky epoch]

i think you may not even need the first condition? just saying v1 not in span(v2, v3, v4) should suffice.

[teal grotto]

c_1v_1=-c_2v_2 soo...

[dusky epoch]

-c_2v_2

but yes

what i was going to say is that assuming $c_1 \neq 0$ leads to a contradiction with $v_1 \notin \mathrm{span}{v_2, v_3, v_4}$

[stoic python]

Ann

[dusky epoch]

thus c1 = 0, thus c2v2 = 0, thus c2 = 0 or v2 = 0

okay so you DO need that first condition

to rule out the possibility of v2 = 0

[heavy crown]

the first condition tells me that v2 != 0?

[dusky epoch]

not directly it doesn't

but v2 = 0 leads to a contradiction

[heavy crown]

yes I see

youre right

[dusky epoch]

you would have dim(span(v1) cap span(v3, v4)) = 1, putting v1 in span(v3, v4)

[heavy crown]

yes good point

that was helpful thank you

[wintry steppe]

is it possible to determine number of solutions from just REF? or does it need to be in RREF?

[teal grotto]

it’s possible but i think in some cases it just might be more difficult to determine

[heavy crown]

Help?

I found that the determinant of AB is (2x-1)(x+1)

I'm not sure what to do next

do I need to do |AB| = 0? |A| |B| = 0 ?

like how I'm supposed to get to rank(B)

I know that rank(B) < 3 because not invertible

[lavish jewel]

is A known? or only AB?

[heavy crown]

that's all we know (AB)

weird question

[dusky epoch]

what values can x take

,w determinant {{1, 0, 5}, {6x-3, 2x-1, 30x-15}, {2x+2, 6x+6, 11x+11}}

[stoic python]

Results provided by WolframAlpha

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

Requested by Ann#0413

[dusky epoch]

2x^2 + x - 1

this needs to be zero in order for it to be possible that B is not invertible

it looks like 2x^2 + x - 1 factors as (2x-1)(x+1)

[heavy crown]

yes

so we know the x for B to be non invertible

[dusky epoch]

so x = 1/2 or x = -1

[heavy crown]

what does it tell us about rank(B)

[dusky epoch]

what's the rank of AB in both cases?

[heavy crown]

rank(AB) = 0 too

[dusky epoch]

0?

impossible

the only matrix of rank 0 is the zero matrix

[heavy crown]

right

so it can't be 0

[dusky epoch]

no

[heavy crown]

then to find rank(AB) i need to RREF based on specific x?

[dusky epoch]

i'm saying you definitely went wrong in calculating rank(AB)

no way you could've gotten zero

[heavy crown]

I see what you mean, it can't be 0 because otherwise it'd be zero matrix

and it's not the zero matrix

so rank(AB) != 0

so let's say for example x = -1

,w rref {{1, 0, 5}, {6(-1)-3, 2(-1)-1, 30(-1)-15}, {2(-1)+2, 6(-1)+6, 11(-1)+11}}

[stoic python]

Results provided by WolframAlpha

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

Requested by username#1754

[heavy crown]

it means rank(AB) = 2 in this case?

and rank(AB) <= rank(B) <= rank(Matrix of order 3x3)
so 2 <= rank(B) <= 3

how do I tell if its 2 or 3

[dusky epoch]

B is non-invertible.

[heavy crown]

oh

right

so < 3

then 2

[dusky epoch]

yes

[heavy crown]

I said it earlier but forgot

[heavy crown]

good point

and for x= 1/2

,w rref {{1, 0, 5}, {6(1/2)-3, 2(1/2)-1, 30(1/2)-15}, {2(1/2)+2, 6(1/2)+6, 11(1/2)+11}}

[stoic python]

Results provided by WolframAlpha

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

Requested by username#1754

[heavy crown]

oh it's the same yea 2

nice 😄

[wintry steppe]

in $\mathbb{R}^3$ with usual cross product $\times$, is $$ (Tv)\times (Tw) = (\det T) T(v\times w)$$?

[stoic python]

Carla_

[dusky epoch]

don't think so?

[sudden narwhal]

what is v and w ? is T a matrix ?

[wintry steppe]

T is linear transformation, v and w vectors

Let's say T has determinant 1 or -1 also

[dusky epoch]

i have a feeling you will need T to be orthogonal

[wintry steppe]

hmm sure, i don't really get why would this be true in that case tho

intuitively makes sense but how to prove it

[sudden narwhal]

looks weird

Or I dont get the question

[wintry steppe]

let me ask again the previous question was wrong

Let $T$ be a orthogonal linear operator in $\mathbb{R}^3$. Does it follow that for any $v,w \in \mathbb{R}^3$ that $Tv \times Tw = (\det T)T(v\times w)$ where $\times$ is the usual cross product?

[stoic python]

Carla_

[dusky epoch]

i think it does now?

Ti × Tj = det(T) Tk

and likewise for the rest of the basis vector pairings

[wintry steppe]

why does it follow that Tv x Tw = det(t) T(vxw) from that?

[dusky epoch]

something something (bi)linearity

[wintry steppe]

oh yeah ofc i'm super stupid

thanks Ann you are amazing and everyone loves you

[heavy crown]

how do I show two linear transformations are equal?

I understand logically I need to show S(v)=T(v) for all vectors v∈V but I'm not sure if this is the right

[wintry steppe]

yea, but there can be easier ways sometimes

it's not difficult to show that if S(w)=T(w) for all vectors w of some basis of V then S and T are equal

it depends how the linear transformations are presented tho

[heavy crown]

that's the question I had

and need to show T=S

i mean literally you can see the span vectors are in S and you can see theyre equal but

[wintry steppe]

Since those are a basis, the value the linear trans takes on them uniquely defines the trans

it's not hard to prove it

[heavy crown]

am I supposed to do this?

(i never had a question of proving two equal linear transformations so thats why i'm clueless sorry)

[dusky epoch]

wish that was actually true

[wintry steppe]

Try this. Write a generic element of V. Then compute S of it using linear properties. Maybe you'll get something cool

[dusky epoch]

@heavy crown if two transformations agree on a basis of their domain then they agree everywhere

[heavy crown]

here's a generic element of V. how can I compute it into S if S only accepting two specific matrices ?

[heavy crown]

You mean because {first matrix in V, second matrix in V} form a basis to V then W = ImT = ImS = V ?

[dusky epoch]

no

[dusky epoch]

S accepts as input any matrix from V, not just the two for which the values are given explicitly

remember S is linear

[heavy crown]

oh right so because S(a(v1)+b(v2)) = a * S(v1) + b * S(v2)
which is exactly the general element of V

yea

and their sum is exactly T(v)

okay I got it 😄

thank you ann!

[wintry steppe]

How many solutions does this have , anyone know?

That's a matrix

yea its in RREF

im confused cause im not sure if its no solution or infinite. The bottom row suggest its no solution but isnt column 2 and 4 free variables? Which would mean infinite

You mean the homogenuous system of linear equations associated with that matrix?

yes

so the system of linear equations is Av = 0 where A is your matrix

the bottom row does not suggest no solution then

it just says that the last coordinate of the vector is 0

over R or C this has indeed infinitely many solutions

[lavish jewel]

this seems to have no sol

even tho it has free variables

[dusky epoch]

so this is the augmented matrix

[wintry steppe]

if its augmented matrix then it's not the homogenuous system of linear equations associated with that matrix

oh, why are they talking about number of solutions if they havent performed any row operations on it

but in any case, is it possible for a matrix in RREF to have free variables and only one solution? Or is that contradicting, cause free variables automatically mean there is infinite solutions right

[clever karma]

the last row in the augmented matrix has a contradiction (0 cannot equal 1)

[wintry steppe]

but what if a matrix in rref looks like this, then my book says its no solution cause the last row , but isnt column 2 and 4 free variables?

or is there something going on because its not a square matrix

[clever karma]

if one of the rows has a contradiction then the entire system has no solution. for a linear system, the values of the variables has to simultaneously satisfy all the equations for it to be a solution.

[twilit anvil]

maybe the positive definite propety of the inner product? that is, <x,x> = 0 if and only if x = 0 ?

@heavy crown

[heavy crown]

but it's not true for matrices to say that

[twilit anvil]

Ax is a vector

[heavy crown]

hmm I guess it's fine then 😅 I'll just take it as is

[twilit anvil]

that is how i see it, hopefully someone can step in if i am missing something

[wintry steppe]

x^T x is equal to <x, x>

take x to be Ax, then (dot product properties) implies that Ax = 0

[wintry steppe]

you're fine

[heavy crown]

alright thank you both 🙏

[pure thistle]

Hello, can anyone give me a real life example of vector spaces?

[dark wasp]

the space around you is a 3d vector space, just say "this point here is the origin"

bam you got a vector space

[heavy crown]

dats some real life real life example

[crude falcon]

if dim(Im(f)) = 2, how many cartesian equations the image has? (the space is in R^4)

[wintry steppe]

prevalent in modern mathematics

[twilit anvil]

write R^n on your paper

[teal grotto]

many sounds you hear can be described by vector spaces

[wintry steppe]

is matrix dimension and rank the same?

[pure thistle]

Dyou have an example?

[pure thistle]

No ma man, matrix dimension is like 2*2 matrix but rank is the number of column with a leading entry in its ref

[peak walrus]

1. is the first equation the only way to rotate around some generic vector?

2. is the second equation used if a vector v is changing direction to a new vector since after I apply the equation below to rotate v, I can use the rotation matrices for the normal x, y and z axis since it is the same matrices as the ones to rotate along the new x, y, and z axises?

[wintry steppe]

i guess i meant vector space dimensiomn

[lucid glacier]

A matrix is a representation of a linear transformation. The rank of a matrix is the dimension of the image of the linear transformation corresponding to it (Equivalently, it is the dimension of the column space and row space (It can be proven that they always coincide)). In this case, if your matrix is say, m rows by n columns, then the column space would be a subspace of F^m (Where F is your base field)

In this sense the rank of a matrix is the dimension of some vector space if that's what you meant

Note that the row space lies in F^n, and even when m=n, the row and column space might be different, but they will always have the same dimension

[hoary flicker]

can anybody explain part b for me?

[wintry steppe]

a) has infinite solutions and b) has no solution

idk if thats what they meant @hoary flicker

[hoary flicker]

I wrote this for a

But b i don’t understand

[nocturne jewel]

b has no solution, a wants the (hyper)plane of solutions

[hoary flicker]

alright thanks

[ruby basin]

<@&286206848099549185>

[soft burrow]

this is definitely a #prealg-and-algebra problem, also wait 15 min before pinging helpers

[covert canopy]

Just cube it for the 2nd problem, you will see that it comes pretty easily

For the 3rd one just find 1/x,add it with x ,and square and subtract by 2

And yea,this isn't linear algebra

[stable kindle]

gimme a sec

[stable kindle]

$\text{let span}(v_1+w, ..., v_m + w) = W\
\text{dim(V) = m; if dim(W)}\geq\text{dim(V), it is trivial, so assume not}\
\
\text{then w is not in W, as, if it were, }\
v_1+w, ..., v_m + w\text{ would be in W and an obvious contradiction}\

\text{w is in V, as:}\
v_1+w, ..., v_m + w\text{ is linearly dependent, as if not it would be a basis of W and W would have dimension m, contradiction}\
\text{but if it is linearly dependent, w is in span}(v_1+w, ..., v_m + w)\
\text{(by a previous result)\
\text{so W is a subspace of V}

\text{let X = span(w)}\
V=W\oplus X\text{, as:}\
\exists Y\leq Z\text{ such that }V = W \oplus Y\
\text{X is a subspace of Y, as w is in V but not W, so w is in Y, so span(W) is in Y, so X is a subspace of Y}\
\text{Y is a subspace of X, as V is a subspace of }W\oplus X\
\text{ie. }W\oplus Y\text{ is a subspace of }W\oplus X\text{ so Y is a subspace of X}\
\text{so by mutual inclusion X = Y, and so }V=W\oplus X\
\
\text{then dim(W) = dim(V) - dim(X) = m - 1, so we are done}$

[dusky epoch]

bad tex! bad tex!!!!

[stable kindle]

yeah that's not worked

[dusky epoch]

so much \text

you can do plain text outside of dollars $x^2$ like this

[stoic python]

Ann

Kaisheng21
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[stable kindle]

oh right

well that's a lot easier

i'll do that next time

anyway i'm fairly sure this works but can someone check it please and also it feels quite complicated, or at least it took me waaay too long so if there's an easier way can someone tell me it

[dusky epoch]

nobody said dim(V) = m

[stable kindle]

fuck

no, yes

[dusky epoch]

V is not necessarily equal to span(v1, v2, ..., vm)

[stable kindle]

gimme a sec, i had...

a different version...

ok no

rename the entire vector space to idk Z

and let span(v1, ..., vm) be V

and the argument works, i think

[stable kindle]

let $v_1, ..., v_m$ be linearly independent in an arbitrary vector space Z, and let w be in Z\
let span($v_1, ..., v_m$) be V, and let span($v_1+w, ..., v_m+w$) = W\
we wish to prove that dim(W) $\geq$ m-1\
\
dim(V) = m; if dim(W) $\geq$ dim(V), it is trivial, so assume not\
\
then w is not in W, as, if it were:\
$v_1+w, ..., v_m+w$ would be in W and an obvious contradiction\
\
w is in V, as:\
$v_1+w, ..., v_m+w$ is linearly dependent, as if not it would be a basis of W and W would have dimension m, contradiction\
but if it is linearly dependent, w is in span($v_1+w, ..., v_m+w$)\
(by a previous result)\
and therefore W is a subspace of V\
\
let X = span(w)\
V=W$\oplus X$, as:\
$\exists Y\leq Z$ such that $V = W \oplus Y$\
X is a subspace of Y, as w is in V but not W, so w is in Y, so span(W) is in Y, so X is a subspace of Y\
Y is a subspace of X, as V is a subspace of $W\oplus X$\
ie. $W\oplus Y$ is a subspace of $W\oplus X$ so Y is a subspace of X\
so by mutual inclusion X = Y, and so $V=W\oplus X$\
\
then dim(W) = dim(V) - dim(X) = m-1, so we are done

[stoic python]

Kaisheng21

[stable kindle]

alright final draft

i'm fairly sure this works but can someone check it please and also it feels quite complicated, or at least it took me waaay too long so if there's an easier way can someone tell me it

but i really really properly need sleep now so if anyone says anything i'll come back and look at it later

[crude falcon]

if I have a linear map R^3->R^4, with this transformation:

and I want to the find matrices of the canonicals basis associated to the linear map in each space, the obvious one would be the Mr4r3(f) which is composed from the vectors in the image in column form

then Mr3r4(f) would be the inverse of that matrix right?

and then how could I find Mr3r3(f)?

[lucid glacier]

You can't invert a non-square matrix

[dusky epoch]

what even is "Mr3r4(f)"

or Mr4r3(f)

[lucid glacier]

I think he means the transformation matrix of f according to the standard bases for R4 and R3?

[crude falcon]

yeah I called that way cause I thought it was easier to read

[lucid glacier]

Also you can't invert the bases and take the transformation matrix if your LT isn't an endomorphism

So Mr3r4(f) wouldn't even be defined

And even when it is, by laws of transformation matrices it would only give you the inverse if f is inverse to itself

Since you'd have
$[f]^B_C[f]_C^B = I \Rightarrow [f^2]_C=I \Rightarrow f^2=id_V$

[stoic python]

ShiN

[crude falcon]

is a square one tho? 4x4

[lucid glacier]

Your transformation is from R3 to R4

Your matrix will be 4 x 3

[crude falcon]

oh yeah

[lucid glacier]

You can look at it composed eith the inclusion $\mathbb R^3 \hookrightarrow \mathbb R^4$ but since you're going from a lower dimension to a higher dimension, it won't be surjective

And therefore not invertible

[stoic python]

ShiN

[crude falcon]

I'm trying to find the eigenvalues and eigenvectors of this matrix, I got lambda = a^4 as it unique eigenvalue, but when computing the eigen vectors, E(a) = (A-aId)x = 0, which basically gives a matriz full of 0s and the b, so the only eigevector I get is the (0,0,0,0)

but seeing the solution for this problem the eigenvectors are: (1,0,0,0), (0,1,0,0) and (0,0,0,1)

how that possible

[half ice]

What is E(a)?

[crude falcon]

the eigenspace of a

[lavish jewel]

most of what you wrote was right, but you looked at (A - aI)x = 0 wrong

because it is 0 almost everywhere, you can almost free choose nonzero vectors and still get 0 as a result

(A - aI)x, with x = (x1,x2,x3,x4), is simply (0,0,x3b,0)

any values of x1 x2 and x4 will get multiplied by 0, so that already gives you 3 orthogonal vectors that are eigenvectors

and then if b is nonzero, then x3 must be 0

since x3 * b = 0

[crystal oracle]

In the following we don't assume the axiom of choice.

Suppose 𝔽 is a field, V is an 𝔽-vector-space, 𝕂 is a field extension of 𝔽, A is a Hamel basis of V, 𝕂 as a vector space over 𝔽 has no Hamel basis.

Consider V ⊗_𝔽 𝕂 as a 𝕂-vector-space. (remark: this is the construction used to build the complexification of a real vector space, except here we do it for an arbitrary field and its field extension) Is {a⊗1 | a∈A} a Hamel basis of it?

[dusky epoch]

concrete counterparts?

... are these questions vague on purpose

[last egret]

I guess so lol

[dusky epoch]

bruh

[silk fjord]

uhm hi, is the channel occupied?

[nocturne jewel]

no

[last egret]

Can someone give me an example of where random matrices describe a complex system?

[wintry steppe]

what is a complex system?

[last egret]

Complex systems are networks made of a number of components that interact with each other, typically in a nonlinear fashion.

[silk fjord]

Can I get help on how to prove the converse of this? I really don't have an idea on how to start the proof for q->p.

[twilit anvil]

what is adj(adj(A))?

[nocturne jewel]

are you asking or prompting chobu?

[twilit anvil]

prompting

[nocturne jewel]

👍

though I feel like det(adj(A)) would be more useful / direct

[twilit anvil]

^true that is probably simpler

[wintry steppe]

Why do we express costheta in this format?

[lavish jewel]

it allows you to associate an angle to objects that don't natively have that notion

[wintry steppe]

you are defining theta to be something that satisfies that. they should specified domain tho cuz cos is not invertible on every interval

[lone arrow]

let $p$ be a prime, and $\mathbb{F}p$ the field with $p$ elements. it's easy to show that the size of $\text{GL}n(\mathbb{F}p)$ is $\prod{i = 0}^{n-1} (p^n - 2p^i)$. now if I want to find the proportion of $n \times n$ matrices that are invertible, the expression would be $ p^{-n^2}\prod{i = 0}^{n-1} (p^n - p^i) = \prod{i = 0}^{n-1} (1 - p^{i-n})$. anyone know what these products evaluate to? (in particular, some computation `shows' that for each $p$, this product has a limit as $n \to \infty$. what is it?)

[stoic python]

∧res

[jagged granite]

I had following question

and here is my solution:

First we observe that set of the form $x_0 + W$ satisfies the property of $A$. \
Now let $x_0 \in A$ be a nonzero vector. Then any vector in $A$ can be written as $x_0 + y$. Let $W={y \ | \ x_0+y \in A } $. If we prove that $W$ is a subspace then we are done. For $y\in W $, $cx_0 +cy+(1-c)x_0 \in A$ so $cy\in W$. For $y,z\in W$, $2(x_0+\frac{1}{2}y)-(x_0-y_2) \in A$ so $y+z\in W$. And thus we are done.

[stoic python]

bert

[jagged granite]

Does it look okay?

[wintry steppe]

does two block matrices have to be square matrices to be able to multiply them?

[shy atlas]

no

[wintry steppe]

Block matrix A and block matrix B , i dont understand how these can be multiplied

i can only see how the A_11 and B_1 can be multiplied

but what happens to the rest?

A_21 ,A_22 cant be multiplied with either B_1 or B_2 correct?

[shy atlas]

why not

[wintry steppe]

because A_21 is a 1x3 matrix and B_2 is a 2x2 matrix?

[nocturne jewel]

block of A is 2x2, block of B is 2x1

so you get a block of 2x1

[shy atlas]

just multiply it like how u would multiply that

[frosty vapor]

left multiplication by a row vector

is the same as right multiplication by a column

like soap has shown

[grizzled flicker]

anyone?

[lucid glacier]

Excuse me

The *Schwartz *inequality?

So we be taking credit not only from bunyaskovsky, but from cauchy aswell??

[torpid venture]

i have a question, as we know we can use linear algebra to get the colors of a picture using pixels so we can get a matrix from that and then we can calculate column space, row space and null space

but why do we actually calculate column space and null space in the first place? like what does it tell us about the colors

how do we elaborate our answer ?

[lavish jewel]

it can reveal structure, e.g. being spanned by a low dimensional basis

[torpid venture]

umm can you explain more please

[lavish jewel]

idk what exactly you want, this is very open ended

[torpid venture]

for example this is my column space for a matrix of white and black picture (0 totally white, 255 totally black and in between shades of grey)

so what does this explains about the black and white picture ? why do we have to calculate column space ?

https://docs.google.com/document/d/1ojs_u_nMX7TWEEk2pdH9wv8Unq_VTZEXLRkHu9IeAQc/edit?usp=sharing

maybe if you can check this, don't go through the steps just get the idea of what i'm saying hehe

[lavish jewel]

i would have to know what you plan to do with the image to say more, really. all i could say is that the image is a 3 x 4 mat, and it happens to be rank 3. this means some of the image's content* (had a typo) is redundant in some sense

[torpid venture]

so basically i have an assignment that i have to use either vector spaces or linear transformation in real world application and elaborate on that

i don't actually know how to use vector spaces or linear transformation in real world applications i don't know what the use of that i only know how to calculate and get the values

i decided to talk about how we use linear algebra in pictures and all that

but i have to elaborate on why do we actually calculate it

[lavish jewel]

this would make sense if you were trying to do something with the image

but if you just have it... there isn't really a point

[torpid venture]

what can i do with the image?

[lavish jewel]

idk, compress, transform it somehow, express it in a different basis

[torpid venture]

combining two pictures will that work?

like this by using matrices and then calculate column space? or still no point of that

[lavish jewel]

you could dry doing a naive fusion, sure

like taking a low rank approx of each image and adding them together

[torpid venture]

i did take rank of a matrix and nullity but i don't know what low rank approx is, i only took the basics i can say

this is my first time taking linear algebra

and it is killing me for now 🙂

[crystal oracle]

@torpid venture Suppose your grayscale image is of size n times m. You represent it as a matrix and calculate its rank r. If it turns out that rm+rn<m*n, then you can losslessly compress the image by storing two matrices of size n by r and r by m instead of storing one matrix of size n by m.

[merry orchid]

help ?

[wintry steppe]

are all homogenous linear equations a subspace?

because = 0

[merry orchid]

yes

[merry orchid]

I ticked but I'm not sure

[merry orchid]

Do you know which of these alternatives might be?

[pure thistle]

Hello, does anyone know how I can relate rank of a matrix to control theory?

[nocturne jewel]

yes, since in the general case it's a hyperplane which goes through the origin

hyperplane of the co-ordinate space

[wintry steppe]

ah yes

a linear equation is a subspace

[nocturne jewel]

y=x^2 is a subspace of R2 if you freshman's dream hard enough \s

[teal grotto]

this is really vague. im not sure what control theory is, but i know you need to expand on what you mean by 'rank of a matrix'. rank of what matrix? what is said matrix supposed to represent?

[wintry steppe]

when determining linear independence is it sufficient to just see if the RREF of the vectors is a unique solution or infinite solutions to find out?

[copper heath]

when it comes to notation, is there a way to differentiate dot product from normal vector multiplication?

since x is used for cross products, and can be used as a variable name

[wintry steppe]

...just use a dot instead of a cross?

you kinda answered your own question lol

[zenith olive]

https://media.discordapp.net/attachments/853821798687703101/853822735247343636/unknown.png
https://media.discordapp.net/attachments/853821798687703101/853822536865415188/unknown.png
https://media.discordapp.net/attachments/853821798687703101/853822343612727316/image0.png?width=407&height=533 does anyone know this?

i don't know which ones correct

ive seen both done but idk where one of them is messing up

[subtle path]

help??

[copper heath]

but dot can be used as a general multiplication sign in most contexts
think multiplying the corresponding entries of two vectors to produce a third vector
does that operation even have a name?

[wintry steppe]

it's called the hadamard product

at least on wikipedia

not sure how it's used but it's a thing

you can also use $\langle v, w \rangle$ to mean the inner/dot product of the vectors $v$ and $w$

[stoic python]

R2T2

[copper heath]

ah..hadamard product

yes that's what i'm looking for

[peak walrus]

I am confused about the use of the two equations in the picture:

https://share.icloud.com/photos/0-EgsFk-HItL_Qkl2j2WS4Qjg

My understanding is that

1. first equation (top one) is used to rotate a vector around some generic vector

2. the second equation is used if a vector v is changing direction to a new vector w and I want to rotate along w or vectors perpendicular to w. This works well since after I apply the second matrix to rotate v, I can use the rotation matrix transformations for the standard x, y and z axises since those are the same equations as the ones to rotate along the new w, u, and v axises

Is my understanding for both of these correct, I had to think to fill in some of the blanks, so I am unsure?

[wintry steppe]

hey can someone tell me how to find a vector equation of a line given one point and the slope?

[dire thunder]

@wintry steppe R^2?

[sand apex]

baby rudin is to analysis as ???? is to linear algebra? that is, which book is the gold standard for linear algebra?

[half ice]

Hah baby rudin is a very unique book, and linear algebra is not at all similar to analysis. I don't think an equivalent exists.

Still linear algebra done right is considered to be "very good"

@sand apex

[rose umbra]

Is there a quick way to find if A and B have has the same eigenvalues, and characteristic polynomial?

[torn hornet]

yeah there is @rose umbra notice that the matrices are off by labeling of the rows/columns, so they are similar

and hence same char poly

i.e if A has the rows/cols ordered like 1,2,3 then B rearranges that to be 3,2,1

easy to see A= PBP^{-1} for a permutation matrix

[lavish jewel]

it works for this one because the reordering was the same for both rows and columns, leading to a similarity transformation

if the rows and columns had been shuffled differently from each other, a bit more care is needed

for completeness, P is an antidiagonal 3x3 matrix (like a left-right mirrored identity). this is, as johnDS said, a permutation matrix that reverses the order of rows when you put it to the left of another matrix, or reverses the order of columns if you put it to the right of another matrix. it is also an involution, i.e. P = P^-1

B can be obtained from A by flipping rows and columns, doesn't matter which one you do first. that means B = PAP

but P = P^-1

so B = PAP^-1, and the matrices are similar

[rose umbra]

i see thanks

[nocturne jewel]

If they're both invertible, then its RREF is I_n

[vast iron]

In general, can we claim that the set L of all linear operators that act on some vector space V (by that I mean linear operators that take some element of V and give another element of same V), form a group themselves?

[native rampart]

No

You always have the zero operator

[vast iron]

Which has no inverse

Thanks

What if we removed that? Is that the only problem?

[native rampart]

You also have maps which map something to 0

If you remove those you are done

This is assuming you are working in fin dim Vector spave

I don't know how it works for inf dimensional spaces in general

[rose umbra]

ain't that C2->C1 ?

[native rampart]

Yes

[vast iron]

Okay thank you, time to correct my work

[coarse sandal]

if you have a vector <1,2,3,4>, would parallel vectors just to k *<1,2,3,4>?

[nocturne jewel]

yes, and define k to be one of the scalars

[north anvil]

A vector lives in $\mathbb{R}^n$. What is $\mathbb{R}^n$ called? The space that the vector lives in? Or the dimension that the vector lives in?

[stoic python]

Shen

[wintry steppe]

how do i find the projection matrix when i know the orthonomal basis?

[rare scroll]

is there an intuitive explanation to why the constant polynomial would lie in the left null space of A?

[nocturne jewel]

the space, yes

[north anvil]

Thanks, Mosh

👍

Appreciate that

[nocturne jewel]

specifically vector space but

actually Rn is inner product space cause dot product but yeah

dimension means the "size" of the space, so dim(Rn)=n

[wintry steppe]

this is my orthonomal basis

how would i use this to find the projection matrix ?

i guess this ? $P=A(A^tA)^{−1}A^t$?

[stoic python]

zeffs

[lavish jewel]

that's true in general, but overkill here

if your basis is orthonormal, notice that A^T A is an identity

[wintry steppe]

so how would i reformulate that formula above?

$P=A(A^tA)^{−1}A^t$

[lavish jewel]

A A^T

[stoic python]

zeffs

[lavish jewel]

the middle term, inverse included, is just I

[wintry steppe]

alright, but since theres only one column does that make a difference?

[lavish jewel]

no, doesn't matter

you just get the 1x1 identity mat

so, a scalar 1

whose inverse is 1/1 = 1

so you still get A A^T

you would've gotten the same result by doing vector projections and grouping everything nicely

[lavish jewel]

i doubt you've seen projections onto the row space if you're asking this, so idk where you even got that expression

[wintry steppe]

the one above?

which expression

[lavish jewel]

the one above indeed

[wintry steppe]

the one above is from someone elses answer, i just dont know how he got that

[lavish jewel]

yeah, you don't need that

[wintry steppe]

isnt that the projection matrix though?

[lavish jewel]

yes, but do you know why?

if you don't, better not use it until you learn why that works

whereas arriving at A A^T is just a vector projection, which is one of the first topics in linalg