linear-algebra

[winter harbor]

And in a way, you did that tho.

Try doing some more examples and you get all of it right.

https://yutsumura.com/category/linear-algebra/

If you want some problems in linear algebra to try out, take a look at the problems at yutsumura.

there's a bunch of them related to trying to find the matrix of a linear map wrt to given basis.

https://yutsumura.com/range-null-space-rank-and-nullity-of-a-linear-transformation-from-r2-to-r3/

https://yutsumura.com/a-basis-for-the-vector-space-of-polynomials-of-degree-two-or-less-and-coordinate-vectors/

https://yutsumura.com/basis-with-respect-to-which-the-matrix-for-linear-transformation-is-diagonal/

https://yutsumura.com/matrix-representation-of-a-linear-transformation-of-subspace-of-sequences-satisfying-recurrence-relation/

https://yutsumura.com/matrix-of-linear-transformation-with-respect-to-a-basis-consisting-of-eigenvectors/

Some problems you may try out.

[violet pecan]

-2 4
1 -2
so i have this 2x2 matrix above and i found the basis vector for the column space of it which i found was
1
-1/2
but the TA marked it wrong and told me the answer is supposed to be
-2
1

aren't they both right?

[cosmic ingot]

Do you happen to have a pic of the work you did? Ill see if i see any errors. I got -2, 1 too.

[violet pecan]

yess

my book shows this method

[cosmic ingot]

There are two methods but if you choose to take the transpose route then the column space of A is equivalent to the row space of the transpose of A so regardless of transpose or not the answer would still be -2,1. The column Space has to be found by using the ORIGINAL columns of A that correspond to the pivot positions of the row echelon form of A. And that would be the equivalent of the transpose of A thenn reducing it and as you can see when you reduced the transpose of A then the first row would be the only valid one. So then you go back to the ORIGINAL transpose of A and select the first row which is still -2,1. Im not sure if I explained that well but tell me if you need clarification. The main thing is that it needs to be from the original matrix when dealing with column spaces. If the question was only asking from row spaces from the start of A^T then ur answer would be correct since the basis for row spaces correspond to the rows starting with 1’s aka the pivots. I do see sometimes when doing row spaces after the reduced form that you still have to use the original matrix for the row space but personally i never ever did that in my class, it was always the reduced rows for a row space. Anyways, since the objective was a basis for a column space and you were finding the row space for A^T then it for sure still has to be for the original row of A^T

[violet pecan]

hmm

but if it's a basis vector wouldn't [1 -1/2]^T span the column space just as well as [-2, 1]^T

after all, they're just a constant multiple of the other

but yes i do understand what you mean about using the pivots to go back to the original matrix, that is the second method listed in my book

[violet pecan]

thanks for your answer though

[vital shoal]

which matrix to do with (0,1) and (1,0) exactly? and why we need to do something with (0, 1) and (1,0)?

[cosmic ingot]

Technically yes. The way i learned it is that if you start with column and do the transpose then it will be the original row of the transpose in order to preserve the original operation of columns. But if you start with row spaces then you can do either the original row or the reduced row. But yes how i said above that some professors do it differently and even books with contradicting rules and since thats how the book did it, (especially if the book is assigned by the class) you can dispute it with the TA or prof because sometimes they dont know it or any of the technicalities and they’re simply just following the answer key. Even if they did know it and the actual prof wrote -2,1 then they still have to follow what the answer key says so talk to them about it if it was a test or something to not hurt your grade. My profs had made mistakes on my exams and id always dispute it while backing up why my answer is correct. Worked every time so def talk to them 👌🏼

[lavish jewel]

it is indeed the same thing, nick

and as you noted, you can prove that you can get those vectors from the one you chose by simply scaling

in R^n, subspaces other than the 0 vector have infinitely many bases, and the one you chose is just fine

you can speak to the prof if the TA won't listen to reason

[unkempt fiber]

Hi community help please!
I have a overdetermined system with full column rank of linear quations of the form Ax=b ( A is long matrix)
I have A_1x = b1 which is composed of rows of Ax=b but height is smaller than A.( A_1 is Wide matrix)
I want a least squares solution of Ax=b but such that A_1x = b1 strictly holds.

[lavish jewel]

i'm assuming this means the overdetermined system is inconsistent

you can try lagrange multipliers

you'd want to do $J = \Vert Ax - b \Vert_2^2 + \lambda^\text{T}(A_1x - b_1)$

[stoic python]

Edd

[lavish jewel]

and then find the stationary point of that (i say point because if A has full column rank, the 2-norm squared term is strictly convex and should have a unique minimizer, if it exists)

the gradient w.r.t. x, assuming $J: \mathbb{R}^n \to \mathbb{R}$, should be $\nabla J = 2 A^\text{T}(Ax - b) + A_1^\text{T} \lambda$

[stoic python]

Edd

[lavish jewel]

set that equal to 0 and see if you can come up with something

the reasonable thing would be to substitute the constraint in. if A_1 is wide though, using the constraint gets annoying

[unkempt fiber]

@lavish jewel no it is not inconsistent

[lavish jewel]

then you can ignore the constraint

it's regular least squares

[unkempt fiber]

@lavish jewel solving Ax= b as a least square will ensure A1x = b1?

[lavish jewel]

if it is consistent, sure

projecting onto the row space of A will give an exact solution

not only for A_1, but all of A

(i was aski g whether Ax=b is consistent btw, not A_1x=b_1)

[unkempt fiber]

@lavish jewel wait may be I am confusing b/w inconsistent and indeterminate?
Are they the same thing?

[lavish jewel]

inconsistent is that it has no solution

indeterminate is that it has infinitely many

for example, since you have a tall matrix A, it can be that satisfying all of the equations is impossible

but satisfying a small subset A_1x = b_1 can be done exactly

then since A_1 is wide, it has a nontrivial kernel, and so A_1x = b_1 is indeterminate, and Ax = b can be inconsistent

that's the scenario i thought you were describing, but i could be wrong

[vital shoal]

please help me with this one

so do I need to multiply the matrix a b c d with the matrix 1 0 0 1?

[unkempt fiber]

@lavish jewel ;okay I messed up with the definitions
Yeah you were correct A_1x = b_1 is indeterminate but Ax = b is inconsistent for my case
For this i shall use langrange multipliers right?

@lavish jewel one more thing so a indeterminate system is always consistent?

[vital shoal]

can anybody help>

[unkempt matrix]

do I need to check if the first vector is a linear combination of these three?

[torpid portal]

yes.

specifically, a linear combination with scalars in [0,1]

[unkempt matrix]

why exactly 0,1

ok i solved it and all scalars in [0,1] but i dont understand why

wait

no

one of them negative

[torpid portal]

The box formed by those is kinda defined as the set of linear combinations with scalars in [0,1]. I'm having trouble giving a technical motivation as to why but for me its all intuitive in terms of visualising it.

https://www.geogebra.org/3d/qbbb7ssa shows the situation. I could try give an explanation as to why its defined that way in terms of those vectors forming a basis, but I don't know enough linalg to be able to give that confidently sorry.

[quasi vale]

@urban magnet Try to prove that ker A is a subspace of ker A^2

and then use the fact that both subspaces have same dimensions

Yes.

If U is a subspace of V and dimU=dimV, then U=V.

(true for finite dimensional vector spaces, not sure about infinite-dimensional)

any other info given?

other question can be done but im having trouble showing dim kerA^3 = 4

Okay so have you seen the fundamental theorem of linear maps?

Or that yeah

sec i messed something up

[lucid glacier]

Kind of a weird question, but is there a way to show that 2 norms in R^d are topologically equivalent without showing they're lipschitz?

[quasi vale]

yeah im stuck as well

but if we somehow prove dim(kerA^3) = dim(kerA), then n=13(the second question you said) can be shown easily by the rank-nullity theorem

because n = 4 + 9

[wintry steppe]

can someone explain me linear transformations and change of basis vectors. having trouble understanding it

Hi! Could somebody help me solve this problem? Find the span of a set {(a,b,c): a=2b; a, b and c are real numbers} Thanks

[zinc timber]

so, can you tell what will be dimension if your resulting subspace?

[tranquil steeple]

you can for example view C as either a circulant matrix (with scalar generating symbol) or as a block Toeplitz (with 2x2 blocks, and matrix valued generating symbol) So either A is a diagonal matrix or a block diagonal matrix

(and the block Toeplitz view is also circulant by the way)

(and the matrix can be diagonalized by for example https://en.wikipedia.org/wiki/DFT_matrix )

[wintry steppe]

Sorry, I am not sure.

[quasi vale]

@wintry steppe You can write all vectors of the form (a,b,c) = (2b,b,c) as a linear combination of two vectors.

Since we have two free parameters here, namely b and c.

Hint: (2b,b,c) = (2b + 0c, 1b + 0c, 0b + 1c). Recall how we add two vectors.

[wintry steppe]

@quasi vale mate can you explain matrix transformation and change of basis vectors. its hard to visualize

[quasi vale]

@wintry steppe Do you mean change of basis matrices

[wintry steppe]

yes yes

[quasi vale]

But yeah anyways no I can't I'm not an expert on this topic. But I can advise you to read the book "No bullshit guide to linear algebra", it explains it well

Or atleast for me

[wintry steppe]

ok thanks

[quasi vale]

or prolly watch videos

[wintry steppe]

didnt help 😦

i have doubts

cuz im confused

(2b, 1b, 0b) + (0c, 0c, 1c) = (2b + 0c, 1b + 0c, 0b +1c)

Thank you, @quasi vale

I am still confused.

Our set consists of infinite number of elements. For example, (6, 3, 7), (62, 31,4), (12, 6, 0), (-2pi, -pi, e)... I do not understand how to tell span when we do not have a finite number of elements. It seems that I can write any vector in R^3 as a linear combination of elements of our set.

[quasi vale]

@wintry steppe No you can't write any vector in R^3 as a linear combination of (2,1,0) and (0,0,1). We need 3 linearly independent vectors to span R^3.

The linear combination of the vectors (2,1,0) and (0,0,1) forms a plane and so only the vectors in that plane can be written as a linear combination of (2,1,0) and (0,0,1).

And yes this plane consists of infinitely many vectors.

[wintry steppe]

@quasi vale I understand. All vectors from our set are in the same plane. Thank you very much! You are great. :)

[lean spoke]

First image is The question and seconds is how Im trying to solve it

But Im stuck

Any ideas?

[tranquil steeple]

you are done, the matrix polynomial p(x)=c where c is a scalar just means p(A)=c*I where I is the identity of same size as A

[lean spoke]

Oh okay then thanks

[grave oasis]

some problems from the MIT OCW courses keep emphasizing the computational time (in terms of # of operations / big O notation) for solving various situations. i.e. problem 1 here: https://nbviewer.org/url/web.mit.edu/18.06/www/Fall17/1806/psets/pset4.ipynb ... but none of the examples I've seen or read about have really involved matrices THAT big where optimization would matter. what's the motivation for this? what field has huge matrices? economics / financial modeling?

[winged prairie]

anybody know how to get started on 4c

[tranquil steeple]

Discretizations of PDE/FDE/SDE, unless you use residual based or matrix-free solvers. And what do you mean by huge? I regularly work with matrices over 100kx100k

[grave oasis]

thanks. I guess I just mean larger than the 5x5 examples we keep working with by hand. it feels silly to talk about computation time mattering without giving actual examples. I assume time complexity matters with 100kx100k

[tranquil steeple]

yes, a full double precision matrix of that size takes about 80GB just to store it in memory

so complexity of the algorithms you use is important

[grave oasis]

I agree that it must be symmetric, but doesn't this contain a typo? why did the B^2 turn into BA after the first equals sign?

A and B are symmetric (but not related) in this problem. (from https://nbviewer.org/url/web.mit.edu/18.06/www/Fall17/1806/psets/pset4-sol.ipynb#Solution)

[raven badger]

How would I go about finding the linear trans-matrix of a double transformation?

h(x) = g(f(x))?

Is it just multiplying F * G

Where

xF = f(x)
and
xG = g(x)?

[limber sierra]

Matrix multiplication corresponds to linear function composition

But I'm not sure why you're writing xF = f(x)

That's... Very nonstandard

Typically we write matrix representations of linear transformations via left-multiplication

f(x) = Fx

And then composition corresponds to multiplication as g(f(x)) = GFx

If you do xF instead (which maybe your class is doing, but is very very uncommon), yeah you'll have to swap the order

[raven badger]

When I say

xF, I mean vector matrix multiplication for x*F

[lavish jewel]

that works and all, but you'll find that considering vectors as column vectors is much more widespread

so that you would rather find that in books as x^T F^T

[raven badger]

Idk about that, but x*F is fine for me and my class

I will clarify a notation now for what I am about to say then;

iFe means, that a vector with base in e multiplied with iFe will result a linear transformation into a space with the basis of i.

[lavish jewel]

your notation seems to imply the opposite since you put your vectors on the right

[limber sierra]

Okay that's unusual but fine

[lavish jewel]

but sure, all good

[limber sierra]

Then f(g(x)) = xGF

[raven badger]

Say that the first linear transformation matrix is on the form of

wFe

and the second is eGw.

I am asked to find the linear transformation matrix for h(x) with a basis in e.

I.e eHe

Do I have to convert both of the transition matrices to e(Matrix)e?

Everybody else says so, but I do not understand why.

Surely I go:

e -> w ->e?

[limber sierra]

The entire point of the way we define matrix multiplication is to make it agree with function composition

Wait I don't understand what you're doing

[raven badger]

Then we're two 🙂

[limber sierra]

Are you saying that your matrices are the wrong size to multiply?

[raven badger]

No

[limber sierra]

Then you can't compose them as functions either

[raven badger]

Wrong basis

[limber sierra]

Basis shouldn't matter

[raven badger]

?

I have to find the transformation matrix for H

with regards to standard basis e in R^4

f: R^4 (basis in e) -> R^3 (basis in w)
g: R^3 (basis in w) -> R^4 (basis in e)

I am asked to find a function such that

h:R^4 (Basis in e) -> R^4 (basis in e).

If it were a matrix it would be

eHe.

What I am wondering about, is whether or not

eHe = eGw * wFe

Or if I would have to do

eHe = eGe * eFe?

If so, why?

[lavish jewel]

eGw * wFe works fine

[raven badger]

Righ?!

I thought so too

[lavish jewel]

if you were to change the basis in between, you would need to put two matrices that are inverse to each other

you can if you want to, but if you associate the matrix products, you get an identity matrix

say some matrix wBe

you'd get eGw wBe (wBe)^-1 wFe

[raven badger]

Ah

But we've actually been given

eFe also, but I felt like it was an uneccary step

unessecary

not needed

[lavish jewel]

that's what wBe^-1 does

[raven badger]

Oh, but they give the identity matrix in your example?

[lavish jewel]

that's exactly the point

[raven badger]

Oh

It would give my first equation regardless

[lavish jewel]

changing basis from w to e is achieved with the inverse of the matrix that changes from e to w

(i'm using inverse loosely here, since your case is rank deficient)

[raven badger]

Right, and that would cancel each other out?

So that it is the same as just not doing it?

[lavish jewel]

precisely

[raven badger]

Perfect!

Now I can also back up my claim with a proof >:)

[lavish jewel]

should be doable in your case because the gramian of the change of basis matrix should yield a 3x3 identity

[raven badger]

(Do not mind me replying)

But, then my next question is;

Since I am going from standard basis to standard basis, I would be using both vector coordinates and just the vector itself on the matrix, right?

Gramian is not a word, I have heard before 🙂

[lavish jewel]

i'm not sure i understood what you meant

i do have to get going tho, so i leave it to nami or someone else to follow up :x

[raven badger]

Uhhh

this si the matrix I get

Suppossedly this is the one I should've gotten

[wise oriole]

can someone help me with this one

[winter harbor]

Use Laplace expansion...

twice

[wise oriole]

my friend is telling me you first need to make a under triangle matrix and then do it is that true ?

and laplace expansion i get that but in this example the last row has a zero which number do i need to take then

because you need a row with zeros right? and then take that number in the row for the laplace

[wintry steppe]

you should learn latex

[cosmic ingot]

I would do it that way too. Do row operations till you get an upper triangular matrix and multiply diagonal elements

[wintry steppe]

matrix-vector product = application of linear transformation to a vector in coordinates

mistersystem will answer you

for i cannot do so satisfactorily

[winter harbor]

Basically, the correspondence between n x m and linear maps f : F^m -> F^n is given by the ''matrix-vector produt somehow''.

To make this more precise, yesterday we did the direction where given a linear map f : F^m -> F^n and basis, we define a matrix. Now, what we can do is, given an n x m matrix A; is to define a linear map:
f_A : F^m - > F^n and takes a collumn vector x = (x_1, ..., x_m) and sends it to Ax.

So that, as TTerra pointed out before, the matrix-vector product is basically applying a linear map to a vector, but wrt to coordinates.

[winter harbor]

[winter harbor]

👉 👈

For the dot product part, I'd recommend you reading this.

https://www.lesswrong.com/posts/rs2focaRymwvkW2jS/inversion-of-theorems-into-definitions-when-generalizing

[wintry steppe]

determinant
is it time to start exterior algebra posting?

[winter harbor]

Ofc that the dot product also shorthands notation

But historically, it exists because people started playing around with coordinate geometry.

And they noticed that the dot product basically encodes the geometry of the plane R^2 and of the space R^3

In the sense that angles and length are all encoded in the dot product

Which were notions taken as primitives in say, classical euclidean geometry.

[spice storm]

LEARN LATEX

[winter harbor]

What is funny is that nowadays, we don't take the notion of angles and lenght to be primitives

But the notion of inner product instead

Which is a generalization of the dot product

For other vector spaces

And we define length and angles in terms of the inner product

And not the other way around

[wintry steppe]

do away with angles and you get psuedo inner products

this pleases the relativist

[winter harbor]

The determinant is somewhat more difficult to explain, because you can interpret it in a bunch of ways.

Basically, you can think about the determinant as a certain monoid homomorphism det : M_n(F) -> F which restricts to a group homomorphism det : Gl_n(F) -> F^x which satisfies certain universal properties and basically encodes ''multiplicative information about matrices''.

Or

You can think about it as a certain alternating multilinear form

And that somehow represents ''signed volume'' (this interpretation makes more sense if we think about it over reals).

[noble swan]

I'm having a hard time understanding what I'm being asked to do here. Can I get some pointers, please?

[winter harbor]

Yeah, that's a good way to view it.

And this basically comes from the fact that the determinant is alternating.

Btw

If you want to derive the formula for the matrix product from composition of linear maps.

[noble swan]

Wait, so the change of coordinates matrix is just the B given to me?

[winter harbor]

Try thinking about the matrix associated to a linear map wrt to basis like this.

It makes things easier I guess.

[winter harbor]

You need to find how to write (1,0) and (0,1) in terms of the basis vectors in B, which is basically trying to find the matrix (wrt to the standard basis) associated to the linear map dank just described.

[noble swan]

Sorry, I can't seem to wrap my head around this

So it'd be something like 1[x_1 x_2] + 1[y_1 y_2]?

And what is x?

[winter harbor]

You need to find a matrix that takes a vector written in the old basis and sends it to this same vector written in the new basis, right?

[noble swan]

Okay, I'm following

[winter harbor]

Nice

So basically

We need to see how this matrix

acts on {(1,0),(0,1)}

Because basically

the collumns of this matrix

will be given by the values associated to how it acts on (1,0) and (0,1)

that's alright tho

anyways

we need to find

how we can write the vectors (1,0) and (0,1) in the new basis

i.e

find numbers a,b

for which (1,0) = a*(-2,-9) + b* (1,8)

this amounts to solving a system of linear equations

can you see?

we have the system
\
\
\begin{cases}
1 = -2 a + b \
0 = -9 a + 8b
\end{cases}

[stoic python]

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

[noble swan]

So I row reduce the system twice

Once for (1,0)

And another time for (0,1)

[winter harbor]

yup

and notice that

the solutions of this linear system

will be the collumns of your matrix

[noble swan]

Wha

How'd you see that

[winter harbor]

Well, I could say ''ah, that's how the change of basis matrix is defined '' but I will try to give a more intuitive view.

(It is defined like this tho)

anyways

Intuitively

what we have is a linear map T : (old) R^2 -> (new) R^2

which takes a vector in the old basis

which in this case is {(1,0),(0,1)}

and sends it to its coordinates in the new basis

[noble swan]

Ohhhh

[winter harbor]

which is {(2,-9),(1,8)}

[noble swan]

But since it's to the same dimension

[winter harbor]

yeah

and this map is linear

so can be represented by a matrix

[noble swan]

So then it's the same thing for this one, right?

Because it doesn't give me a new dimension

[winter harbor]

And this is the change of basis matrix.

[noble swan]

As long as it's in the same dimension using standard basis, the change-of-coordinates matrix will just be the column vectors of the matrix given, right?

[winter harbor]

Idk exactly what you mean by that

But in any case

notice that the change of basis linear map T : (old) R^2 -> (old) R^2 is completely determined on how it acts in the old basis

meaning that if we have another vector x* (1,0) + y * (0,1) we have that T(x * (1,0) + y *(0,1)) = x * T(1,0) + y * T(0,1)

And that's what I said before by ''the change of basis matrix is completely determied on how it acts on (1,0) and (0,1)''

we basically just need to know what are the coordinates of (1,0) and (0,1) in the new basis

to write any other vector in the new basis

[noble swan]

Damn, I was doing fine in this class until we started to have to understand the theories 🥲

[winter harbor]

That's alright tho

Just do some more examples

[noble swan]

Do the new & old basis' change if you're still in the same dimension?

[winter harbor]

wdym by that tho?

[noble swan]

Well, standard basis doesn't change if you don't move dimensions, right?

Standard basis for R^2 is always (1,0) & (0,1), yeah?

[winter harbor]

oh

[noble swan]

Same for R^3, standard basis is always (1,0,0),(0,1,0),(0,0,1), right?

[winter harbor]

Yeah, that's correct

Anyways

[noble swan]

So using the explanation you gave me

The answer for 10 would also just be the same columns, right?

[winter harbor]

No

It won't

[noble swan]

[winter harbor]

Unless I am interpreting this incorrectly

if we are taking as ''old basis'' B and finding how to write it in terms of the standard basis

then yeah

that is correct

But that would be a weird exercise lol

[noble swan]

Ye, that's what the question is asking, right?

[winter harbor]

There's nothing to do

[winter harbor]

If that's the question, then sure.

[noble swan]

That's why I was confused lmfao

Aww, I was overcomplicating it

I actually understand it now, though

[winter harbor]

In any case, try to find how you would do the opposite tho

[noble swan]

If it wasn't to the standard basis, I'd have to set up the given matrix equal to each column of the basis I need to find the change-of-coordinates matrix of

And row reduce

[winter harbor]

find the change of basis matrix from the standard basis to B.

yeah

[noble swan]

Sick, tysm!

[winter harbor]

[noble swan]

It's okay XD

Your thing looks like you can solve the same problem via matrix multiplication?

[winter harbor]

Dank, do you have background in another area of mathematics/physics or?

Good thing you are learning LA, probably one of the most useful things to learn as a computer scientist.

[valid geyser]

I'm not quite sure what linear algebra actually is, though I guess I've been learning it. So what's that? What do you learn in it

[vale yoke]

Can anyone help me?

[noble swan]

Linear algebra w/ computer science yeesh

There was something that we went over in class that was the only way you could program a computer to find

I think it was determinants or inverses?

Was uber nasty

Took forever to do by hand

[vale yoke]

I'm in vc if you wanna join

sure lol

you got a deal

I am also a CS student lol

[winter harbor]

what does the L((2,0,1)^t, (-1,1,0)^t) notation mean?

is L like, a notation for ''span''?

Well, you basically want to find another subspace, say $W \subseteq \mathbb{R}^{3}$, of $\mathbb{R}^{3}$ for which:
$$
\mathbb{R}^{3} = \text{ker}(A^{2}) \oplus W
$$

[stoic python]

MISTERSYSTEM

[winter harbor]

That's what they mean by complementary subspace

One for which R^3 is a direct sum of ker(A^2) and this, so called, complementary subspace

Are you following up until now?

No

Because direct sum

also means that ker(A^2) and the span of (1,0,0)^t have trivial intersection

i.e

the intersection of (1,0,0) and ker(A^2) is only the 0 vector

and here's why they chose (1,0,0)^t

first

they needed to check that (1,0,0) is not in the kernel of A^2

which I guess you can check yourself (idk what A is)

and then

they also needed to check that the complementary subspace of ker(A^2) has dimension 1

and this is easy

because the dimension of the kernel of A^2 is 2

and the dimension of R^3 is 3

and the dimension of the direct sum of subspaces is exactly the sum of their dimensions

so 3 = 2 + 1

and the complementary subspace of ker(A^2) is spanned by only one vector

and there you have it

with this information

we know that all we have to do in order to find the complementary of ker(A^2)

is just to find a vector in R^3

which is not in the kernel of A^2

and take the span of this vector

they chose (1,0,0) because it was an easy choice I guess

[winter harbor]

Suppose that you have linear maps f : V - > W and g : W -> U between finite dimensional vector spaces

and you take a basis B = {e_1, ..., e_n} for V, a basis B' ={b_1, ..., b_m} for W and a basis B'' = {u_1, ..., u_l} for U.

so the thing is that we want to write the linear map g ° f wrt to the basis B and B''

First

Let's remember how to find the matrix of a linear map

Here

Basically

To find the matrix of a linear transformation wrt to given basis

in this case B and B''

We need to find how it acts wrt to these basis

more specifically

let's suppose we want to find the matrix of f : V -> W wrt B and B'

what we want to do

is find how to write f(e_j) in the basis B'

For each e_j in the basis B

so that for each $j \in {1, \cdots, n}$ we have that $\exists \alpha_{1,j}, \cdots, \alpha_{m,j} \in \mathbb{F}$ for which:
$$
f(e_{j}) = \sum\limits_{i=1}^{n} \alpha_{i,j} b_{i}
$$
and we define the $(i,j)$-th entry of $[f]^{\mathcal{B}}{\mathcal{B'}}$ to be $\alpha{i,j}$.

[stoic python]

MISTERSYSTEM

[winter harbor]

is that good so far?

notice that the matrix of f wrt to B and B' is an mxn matrix

now

let's try to find the matrix of g wrt to B' and B''

Is that a notation for the vector f(e_j) wrt B'?

if so, that's the correct idea

nice!

so ok

now we need to find [g(b_k)]_B''

we will denote it by:

We will write
$$
g(b_{k}) = \sum\limits_{j=1}^{l} \beta_{j,k} u_{j}
$$
for each $k \in {1, \cdots, m}$

[stoic python]

MISTERSYSTEM

[winter harbor]

where the b_j,k will be the coefficients of the matrix of g wrt to B' and B''

i.e, the entry (j,k) of this matrix is given by b_(j,k)

notice that this matrix has size (should be) l x m

wait I'm trying to make myself sure of the sizes and indices

yeah

it should be right

the matrix of g wrt B' and B'' has size l x m

and the matrix of f wrt B and B'' has size m x n

[winter harbor]

There should be an ''m'' at the top of the summand, not an ''n'' btw.

this is because W has dimension m

thus f(e_j) which is an element of W is written as a combination of m vectors

this is what was bugging me

but in any case

so far, so good

now

Idk what you mean by this

but the thing is

oh

I guess I know what you mean

the values for b_(j,k) and a_(i,j) are generally different

so that's why I am denoting these differently

anyways

now we want to find the matrix of g ° f wrt B and B''

You might already know how to do that, right?

how would you do that?

we need to know how g ° f acts on the basis B

exactly!

but notice

[noble swan]

Am I going crazy? How come I have opposite signs of what I need?

[winter harbor]

we have already f(e_j) wrtten as a certain linear combination

of the basis vectors in B'

let's use that

[winter harbor]

.

maybe you mean this?

yeah

b_k is not f(e_j)

b_k is a basis vector in B'

and f(e_j) is the image of e_j under f

which is not generally a basis vector in B'

but a linear combination of those

which is what the first equation gives us

So,
$$
g(f(e_{j})) = g \left(\sum\limits_{i=1}^{m} \alpha_{i,j} b_{i} \right)
$$
But g is a linear map, thus:
$$
g \left(\sum\limits_{i=1}^{m} \alpha_{i,j} b_{i} \right) = \sum\limits_{i=1}^{m} \alpha_{i,j} g(b_{i})
$$

[stoic python]

MISTERSYSTEM

[winter harbor]

can we see what we can do now?

g(b_i) is also written in terms of vectors in B''

yup, in the end this should give us this:

We should get these as coefficients in the end

Which is how we define the product of two matrices

Oh, that's correct.

OH you were referring to this?

I don't exactly understand your question tbh

Where did I denote something differently?

[winter harbor]

Yeah, in any case, you wrtie g(b_i) = b_j,k u_j and then you reorder the sums

you will get exactly a sum of the form a_i,j b_j,k

you only need to get the indices right lol

Yeah, I am also using sloppy notation now lol

I need to get these indices right

Oh yeah

right

yeah, I am a bit pissed off I am getting the damn transpose for the product

it should be u_i there (or whatever index lol)

Anyways

Now I get what I was doing

It was ok

We will just get a different notation than that of Serge Lang's book

notice that we can write
$$
g(b_{i}) = \sum\limits_{k=1}^{l} \beta_{k,i} u_{k}
$$
Then,
$$
\sum\limits_{i=1}^{m} \alpha_{i,j} g(b_{i}) = \sum\limits_{i=1}^{m} \alpha_{i,j} \left(\sum\limits_{k=1}^{l} \beta_{k,i} u_{k} \right)
$$
and so, by rearranging the summation, we have that:
$$
\sum\limits_{i=1}^{m} \alpha_{i,j} \left(\sum\limits_{k=1}^{l} \beta_{k,i} u_{k} \right) = \sum\limits_{k=1}^{l} \left(\sum\limits_{i=1}^{m} \beta_{k,i} \alpha_{i,j} \right) u_{k}
$$

[stoic python]

MISTERSYSTEM

[winter harbor]

and we are done lol

You see what we just did?

notice that we wrote:

$$
(g \circ f)(e_{j}) = \sum\limits_{k=1}^{l} \left(\sum\limits_{i=1}^{m} \beta_{k,i} \alpha_{i,j} \right) u_{k}
$$

[stoic python]

MISTERSYSTEM

[winter harbor]

This means that

we actually wrote for each e_j in the basis B

how g o f acts on e_j

wrt the basis B''

it

we found a linear combination of (g o f)(e_j)

in terms of the u_k

i.e, the basis vectors in B''

this actually tells us

that the (k,j)-th entry of the matrix of g o f wrt to B and B''

is given by this sum:

[wintry steppe]

[winter harbor]

$$
\sum\limits_{i=1}^{m} \beta_{k,i} \alpha_{i,j}
$$

[stoic python]

MISTERSYSTEM

[winter harbor]

So yeah

That is precisely what we would get for the (k,j)-th entry

if we did the matrix product of [g] with [f]

and that's the formula for the matrix product

of an l x m matrix with an m x n matrix

I actually wanted a formula for m x n and l x m, but I didn't plan ahead and messed up the indices

It doesn't really matter anyway

I'd recommend you doing these computations again

You should get this

in the end

up to like, ordering of the indices lol

and I know what I messed up with the indices now

you should actually work with g : U -> V and f : V -> W where U has dimension l, V has dimension dimension n and W has dimension m.

and compute (i,k)-th entry of f o g

This should be enough

In fact, I will do this now

dim(V) = n

dim(U) = l

and dim(W) = m

Write
\begin{equation}
g(u_{k}) = \sum\limits_{j=1}^{n} \beta_{j,k} e_{j}
\end{equation}
and
\begin{equation}
f(e_{j}) = \sum\limits_{i=1}^{m} \alpha_{i,j} b_{i}
\end{equation}
Then, from equation (1)
$$
f\left(g(u_{k})\right) = f\left(\sum\limits_{j=1}^{n} \beta_{j,k} e_{j} \right)
$$
Then, by linearity of f
$$
f\left(\sum\limits_{j=1}^{n} \beta_{j,k} e_{j} \right) = \sum\limits_{j=1}^{n} \beta_{j,k} f(e_{j})
$$
and from equation (2), we get (after rearranging the terms):
$$
\sum\limits_{j=1}^{n} \beta_{j,k} f(e_{j}) = \sum\limits_{i=1}^{m} \left(\sum\limits_{j=1}^{n} \alpha_{i,j} \beta_{j,k}\right) b_{i}
$$
and this tells us precisely that the $(i,k)$-th entry of the matrix product
$$
[f] \cdot [g]
$$
wrt the appropriate basis is precisely given by:
$$
\sum\limits_{j=1}^{n} \alpha_{i,j} \beta_{j,k}
$$

Here

[stoic python]

MISTERSYSTEM

[winter harbor]

oof

anyways, just made the right index choices now

in order to get the formula from Lang's textbook

the basis are still the same

{e_1, ..., e_n} is still the basis for V

notice that I basically just changed the order of the arrows

and that's it

I just wanted to get the corrected formula with the indices at the end lol

but the idea is still the same

Yeah, I am all this time avoiding talking about ''coordinate maps''

and instead talking about the basis explicitly

by a coordinate map I mean an isomorphism from a vector space V to F^n for some n.

Lang's goes through this

he proves that there exists a unique determinant map det : M_n(F) -> F with these properties

Yeah, nobody bothers to go around computing determinants

[safe jay]

hey guys why a set of divergent sequence is not a vector space? Because I think that it satisfy all the axiom, while my teachers said that it does not contain 0

[winter harbor]

consider the sequence f = (1,2,...) given by f(n) = n

if this sequence were to be closed under addition

then f - f = (0,0,...) were to be an element of V

but (0,0,...) is a convergent sequence

[safe jay]

ohhhhhhhhhhhhhhh

[winter harbor]

Your prof is correct, (0,0,...) is convergent.

[safe jay]

I just thought f is R, i dont think f is R^n

Thanks, you enlight me :>

[winter harbor]

I should sleep now

[golden kindle]

Hey guys,

I'm taking a course on Khan Academy for Linear Algebra for fun (Yes Ik that's weird) and I am confused how I would find the angle between the vector and the x-axis I was never taught this concept.

I would assume I can find the magnitude using the Pythagorean theorem, right?

Cheers,
Tom

[quasi vale]

@golden kindle

Can you find theta?

For the magnitude, $||-4v|| = |-4| ||v|| = 4 ||v||$

[stoic python]

Sup?

[lavish jewel]

@keen flame

do something similar for $\langle u+v, u+v\rangle$ and $\langle u-v, u-v\rangle$

[stoic python]

Edd

[lavish jewel]

then see if you can notice a way to get rid of unneeded terms

[crystal hound]

Does linear programming come under this channel or another?

[dusky epoch]

linprog is fine here i think?

not sure where else it'd fit

[lavish jewel]

hmm somewhere here, in multivar calc, or applied computational math. the line is blurry

you can try here and we see

[crystal hound]

Ok so I’ve just started the linear programming half of a linear algebra and linear programming module and I’m just wondering what is meant by this proposition because if we have a finite optimal solution, How can we have infinite solutions? Is the wording of this proposition implying there’s either possibly 1 unique optimal solution or infinite sub optimal solutions? The theorem is below

Proposition If the linear programming problem has a finite optimal
solution, then it has either a unique solution or infinitely many solutions.

I’m just a bit confused because the wording is implying to me even though there’s a finite optimal solution, there’s also infinite optimal solutions? Which doesn’t make sense

[lavish jewel]

it means there is a possibility there are many solutions, and all of them are equally good

if you're familiar with convexity, you can use that to show that if a minimizer exists, it is global, but this doesn't mean it is unique

[crystal hound]

So what does the finite optimal solution bit mean, if the prop is then saying there could be infinitely many such solns

[lavish jewel]

the thing is that, without constraints, a linear program deals with an expression of the form $c^\text{T}x$, yeah?

[stoic python]

Edd

[lavish jewel]

and in general you can make the result of such an expression be arbitrarily large

[crystal hound]

Yep I’m with you so far

[lavish jewel]

after adding in the constraints, it might be that the feasible set does not allow that anymore

and so the feasible values of x can only yield results that are so large

so there could exist some optimal x* such that c^T x* yields the smallest result possible, considering the constraints

the constraints don't always enforce this, though. it could be that one could still just set all the entries of x to -inf

aside from that, c^T is a rank 1 linear transformation, and so it has a nontrivial null space

so it could be there exist values y such that c^T y = 0

and so if c^T x* yields the minimum value, then c^T (x* + y) = c^T x* + 0 also yields the same optimal result

the constraints may or may not restrict this

[crystal hound]

Ahhhhh that kinda makes sense now, thank you 😁

[wintry steppe]

this might be a stupid question but
why does the determinant not readily extend to quaternion matrices? Diudonné has generalised the concept of the determinant but I don't see why the sum_sigma permutation sign(sigma) a_sigma(1..n) doesnt immediately work with quaternions
also theres the nth exterior algebra which is one dimensional for which you get that the determinant is equal to the scalar factor but this shouldnt readily generalise as at least to my knowledge you need your associative algebras to be over a commutative ring

[dusky epoch]

quaternion multiplication is noncommutative

[wintry steppe]

yeah that's kind of my whole question: why doesn't the definition of the determinant readily extend to noncommutative rings

[dusky epoch]

you lose some nice properties

im not even sure if the determinant ends up being multilinear in the noncommutative case

[wintry steppe]

very heuristic but if we require the det to be a homomorphism into the quaternions then wed need to have det(AB) = det(A)det(B) but on the LHS the terms are products of terms ab whereas on the RHS the terms are products of products of terms of A and products of terms of B and there is no immediate way to mix the terms on the RHS

also you can apparently use homotopy theory to show that no nice determinant function exists

[livid idol]

Does anyone mind giving me a hint on how to approach this? I can't think of any clever way to prove this.

I know that rank(A) ≤ min(m, n), but I am not sure if that is helpful here.

[stable kindle]

rank-nullity surely?

i mean this is basically equivalent

[lavish jewel]

as a hint, the vectors are orthonormal

[hollow void]

how did you figure that out.

and if it is ortho. i just have to find adj?

[lavish jewel]

the third column is ortho to the other 2. for the others, i guess cuz i've seen them before and i recognize it's a rotation mat

[hollow void]

i mean ik that det of ortho 1

[hollow void]

ohh ok

@lavish jewel sir i have one que

to find adjoint, you have to multiply (-1)^(i =j) with its minor?

[lavish jewel]

you're thinking too hard

if two vectors are orthogonal, u^T v = 0

[granite kraken]

If I'm really struggling with this class do any of u tutor? I'm willing to pay?

[lavish jewel]

you have so many basic questions, i think you should go back and review the last few sections/chapters

[wise oriole]

is this correct or am i wrong

[lavish jewel]

looks ok

[wise oriole]

yeah?

[hollow void]

😩

im self learning

[tranquil steeple]

you are spamming.

[hollow void]

sorry from now i wont

[winter harbor]

If one of its eigenvalues of P is 0, then P is non invertible.

Which implies what about its determinant?

[hollow void]

cant understand this msg

[wise oriole]

To return to my previous question, is this the same because if I calculate all my matrices, I do not get the right one?

[tranquil steeple]

that is correct, then X=A'B'-I

[wise oriole]

I cannot follow you there

[tranquil steeple]

X=(BA-I)'
X=(BA)'-I'
X=A'B'-I

[wise oriole]

i follow you there but what is '

[tranquil steeple]

transpose

(or conjugate transpose if you have complex matrices)

[wise oriole]

then i just need to transpose A and B and - I

right?

[tranquil steeple]

yes

[wise oriole]

and that's it

you have a matrix x

[tranquil steeple]

you need to multiply A' and B' and then subtract I, then you get X

[wise oriole]

but then you don't get a matrix w fractions bcs my friend has a lot of fractions in his matrix

[tribal moss]

I'm bit confused on how to prove the following:
For c a scalar and vec(a), prove that if
c.vec(a) = 0 then c =0 or vec(a) = vec(0). I have proven it for if the scalar is different fromzero and used the inverse element following from the field axioms but idk how to do it if the vector is different from zero to prove that the scalar is zero

[tranquil steeple]

Think of it like this ```
X'*inv(A)+inv(A)=B
(X'+I)*inv(A)=B
X'+I=BA
X'=BA-I

Compute BA-I in the last line and then transpose it (since BA-I is equal to X'). Do you see any fractions in the elements of X? no.

[wise oriole]

then my friend is wrong right? and thank you very much for the help I appreciate that

[wise oriole]

My last question and then I’m done but do I need to transpose matrix B and A first and then multiple or do I need to multiple first and then transpose?

[tranquil steeple]

either (BA)' or A'B' it is the same

(order matters usually)

[wise oriole]

and that is the same outcome ok ty

[tranquil steeple]

see property 3 https://en.wikipedia.org/wiki/Transpose#Properties

[muted loom]

Can anyone walk me through this?

my prof doesnt do a great job of explaining dynamical systems, any advice is appreciated.

[winter harbor]

Hint

I suppose they meant x_(k+1) = Mx_k, right?

They didn't even define A.

[muted loom]

yes, the notation is weird

they do not define A, no

[winter harbor]

If that is the case, the idea is to diagonalize M.

And in this way

You can easily compute the powers M^k

And study the convergence.

[muted loom]

why does diagonalizing allow this?

[winter harbor]

If M = P D P^(-1) for some diagonal matrix D and P invertible, then M^k = P D^k P^(-1).

For every k ≥ 1

[muted loom]

oh i see

that makes sense

[winter harbor]

But computing powers of diagonal matrices is really easy

And like

Notice that the k-th term of this sequence is given by M^k (x_0)

[muted loom]

i wish the professor had actually just walked through an example for this

correct

@winter harbor so after i diagonalize, do i just find D^inf?

[winter harbor]

You take the limit, yes.

[muted loom]

so lim of k-> inf for (3/10) ^k and 1^k

[winter harbor]

Yes, you need to find the matrix P of the eigenvectors of M too.

[muted loom]

I guess my biggest issue after that is putting everything together

I have P and D, and i did the limit of each value in D, but then what

[winter harbor]

I mean...

You did everything by now

Like, matrix product is a continuous function

So if you want to know the limit of M^k

And you know M = P D P^(-1)

All you need to do is calculate the limit of D^k

And then compute P (lim D^k) P^(-1)

This gives you lim M^k

[muted loom]

okay, that makes sense. So i just need to multiply everything together at the end

[winter harbor]

Yeah, and apply it to the initial vector.

[quaint sage]

When describing a plane in planar equation form (Ax + By + Cz + D = 0) are A, B and C the unit normal vector?

Or does it hold true for a normal of any magnitude so long as it is in that direction? I.E does it really matter?

[wintry steppe]

Could maybe someone help me with this problem?

[quaint sage]

you do math in greek? god help you

[wintry steppe]

hahaha yh cause a greek but i wrote it in english

if someone has the time I would appreciate it

[quaint sage]

Props to you, I can't even imagine what it'd be like to do math if all the symbols were in my spoken alphabet... Maybe it would kill some of the mystique and fun of it... Maybe it would just converge with normal words and make alphabet soup in my brain.

Anyhow, I am completely unqualified to answer your question so I'll just bow out. Good luck.

[jagged tinsel]

^

[marsh trail]

I'll try to be of help, but I'm not sure how much I'll be.
(Also, this was assuming that the numbers are 9s, if not then just replace all the 9s in this with the number that it is (2?))
So the first one, A⋂B is asking for the numbers in both A and B, so any numbers that are both multiples of 9 and divisors of 18 in the range 1 <= x <= 19.
The second asks for everything that is in both, so any numbers in the range that are either multiples of 9 or divisors of 18, so they only have to satisfy one of them.
The third part is asking for everything that is not in the second part, and so that should be everything that's in omega but not in your answer for the second part.
The fourth is asking for everything that isn't in A, but is in B. So all divisors of 18 that are not multiples of 9

I hope that may have been of some help

[magic venture]

I can't figure out if it's a 9 or a 2
In both cases, listing the elements of A and B will make it much simpler and less abstract

[marsh trail]

Yeah, they are both significantly smaller subsets of omega, which only has 19(?) elements

[wintry steppe]

i solved it but thenks anyway for the intrest

[marsh trail]

No problem. Just out of curiosity, was it a 9 or a 2?

[wintry steppe]

It is the first one

@marsh trail *

[marsh trail]

Ah nice, OK. It was a 2 then

[near crescent]

How can i show that the rank of the matrix A doesnt change, when its multiplied by elementary matrix (S, T)

rk SAT = rk A

[worldly bear]

When determining the basis for an eigenspace, if i have an eigenvalue with multiplicity 2 for example, I should find exactly 2 free variable when looking for the basis vector corresponding with that specific eigenvalue right?

[winter harbor]

By multiplicity you mean algebraic multiplicity, right?

If that's the case, then no, because the dimension of the eigenspace associated to an eigenvalue (which is called its geometric multiplicity) is less than or equal to the algebraic multiplicity.

So like

If you know that the algebraic multiplicity of an eigenvalue is 2.

You know that the dimension of the eigenspace associated to it is at least 2 (could be 1 or 2).

So you are looking for either 1 free variable or exactly 2.

[zinc timber]

elementary matrix always have full rank. Say E be our elementary matrix and S be our original matrix. Now we know that rank(MS) <= min{rank(M), rank(S)} = rank(S) as M is of full rank.

to show that rank(MS) = rank(S), assume they are not equal, i.e. there is a vector v st MS(v) = 0 but S(v) =/=0. this cannot happen as M(S(v)) cannot be zero as M is of full rank. So we must have rank(ES) = rank(S)

also I wrote M instead of E, correct that

[worldly bear]

okay, so I can't have more free variables than the algebraic multiplicity associated with the eigenvalue im working with.

[winter harbor]

Yup

[tribal fossil]

hey someone help me withthis

how can I find all 3 * 3 elementary matrices?

[zinc timber]

google it

[tribal fossil]

not there bro

pls help

[odd kite]

try an example of each type, that may give you an idea

[teal grotto]

there are only a couple types of them. do you know what they are

[tribal fossil]

nope

hey bro u mean this?

[teal grotto]

yup

do you know the form these matrices?

[tribal fossil]

no 🥲

[teal grotto]

🥲

okay.

[tribal fossil]

im very new to LA

[teal grotto]

there are three matrices that swap rows. the one that swaps rows one and two is
$$E_{12}=\begin{bmatrix}0 & 1 & 0\ 1 & 0 & 0 \ 0 & 0 & 1\end{bmatrix}$$
the one that swaps rows two and three is
$$E_{23}=\begin{bmatrix} 1 & 0 & 0\0 & 0 & 1\ 0 & 1 & 0\end{bmatrix}$$

[stoic python]

c squared

[teal grotto]

try and recognize the pattern here

[tribal fossil]

the other is 1 & 3 swap?

[teal grotto]

yea. can you figure out what swapping 1 and 3 is?

[tribal fossil]

yea

[teal grotto]

okay, great. these are the only ones in the space of three by three elementary matrices that you need to know. the one that doesnt swap any rows is the identity matrix, but, its not really a swapper. these are all invertible because they have non-zero determinant

do you have access to or are aware of the properties of the determinant?

[tribal fossil]

not fully

we know how to calculate it

[teal grotto]

oh nvm

i missed the question

you need to find inverses for them

whats the inverse for the matrix that swaps rows 1 and 2?

[tribal fossil]

swap rows 2 and 1 👀

[teal grotto]

yea

but thats just the same matrix

[tribal fossil]

its the same..

[teal grotto]

so it is its own inverse

[tribal fossil]

nice

[teal grotto]

thats true for any elementary matrix that swaps rows

so E_ij is its own inverse, if E_ij denotes the elementary matrix that swaps rows i and j

that all make sense?

[tribal fossil]

yea

[teal grotto]

okay. the matrix that multiplies row 2 by a non-zero scalar $c$ is given by
$$E_{2,c}=\begin{bmatrix}1 & 0 & 0\ 0 & c & 0\ 0 & 0 & 1\end{bmatrix}$$

[stoic python]

c squared

[tribal fossil]

ok...

[teal grotto]

what is the matrix that multiplies row three by a non-zero scalar?

[tribal fossil]

E3c

{{1,0,0}, {0,1,0},{0,0,c}}

[teal grotto]

awsome

what do you think the inverse of these guys should be?

[tribal fossil]

mutliply the row with c by 1/c?

[teal grotto]

yea. so $(E_{k,c})^{-1}=E_{k,1/c}$

[tribal fossil]

or replace c with 1/c? 👀

[stoic python]

c squared

[tribal fossil]

but the inverse again gives identity matrix right?

[teal grotto]

wdym by that?

[tribal fossil]

{{1,0,0}, {0,1,0},{0,0,c}}

inverse of this...?

{{1,0,0}, {0,1,0},{0,0,c*(1/c)}}

?

[teal grotto]

get rid of the first c

[tribal fossil]

{{1,0,0}, {0,1,0},{0,0,1/c}}

?

ok nice

[teal grotto]

you can check this by multiplying E_{k,c} with E_{k, 1/c} to make sure you get the identity matrix back

[tribal fossil]

yea now im gettting it nice

so the final one is Ri - k*Rj ?

i mean inverse of the final type elementary matrix 👀

[teal grotto]

yea

[tribal fossil]

yo man thanks for the help

[teal grotto]

yw

[tribal fossil]

😁

[teal grotto]

the inverse of R_i - kR_j should be R_i + kR_j (please check tho)

better notation might be L_{ij}(k), add k times row j to row i

then the inverse of this is L_{ij}(-k), add negative k times row j to row i

[tribal fossil]

😮

[cloud fox]

hi

im year 1

what is linear algbre

what is f(x)

[limber sierra]

see pins

[limber sierra]

a function f applied to some number x.

this isnt the place to ask about function notation; try a questions channel (see #❓how-to-get-help )

[dark brook]

If I want to look at det(A_n) where A_n = (a_(ij)) and a_(ij) = (i - 1)n + j, why is it n needs to be n > 2?

[limber sierra]

what do you mean by "needs to be"?

you can take that determinant for any positive integer n

[dark brook]

It says I need to explain det(A_n) = 0 for n>2

[limber sierra]

well, its trivial for n = 1 and n = 2

so presumably they want you to check the nonobvious cases

and not worry about the obvious ones

[dark brook]

I know if n=3 I get a bottom row of zeros, but if I didn't do any calculations, I wouldn't know

So my TA wrote back to me that there is another reason for we assume n>2

This is what I've got for A_n

[limber sierra]

oh, i see what youre saying

if you have at least 3 rows, then:

• row3 = row3 - 2 * row2
• row3 = row3 + row1

note what row3 is now.

n > 2 is necessary so you have a third row to work with

so that you can apply these operations to it

(for n = 1 and n = 2 you can check manually)

[dark brook]

I did this

[limber sierra]

that seems fine, you just need to do it in general

you only gave a 3x3 matrix

[dark brook]

Oh

So I would do row operations with the generalisation I showed earlier?

[lavish jewel]

you can show that the rows from 3 on are linearly dependent on the first 2 rows

[dark brook]

Hmm how would you do that?

[lavish jewel]

well, nami already showed you how to do this for n = 3

you should be able to see it from there

e.g. the 3rd row is the 2nd row times 2 minus the 1st row

try a couple more rows and see if you pick up on the general pattern

[dark brook]

You guys are really sweet. I don't see the pattern, but I'll try and fiddle with it a bit and see if I'll pick it up, thanks a lot 🙂

[dark brook]

How / what would be a general way of setting it up?

[lavish jewel]

i'd probably do something like this. the nth row r_n, with n >= 3, is given by a linear combination of rows 1 and 2, which we will call x and y, as follows. r_n = (n-1)y - (n-2)x

[pallid rampart]

I wonder if this is true in general, that is is d^2/dx^2 a self adjoint operator on C^2([-π,π])

[dusky epoch]

is it true that <f'', g> = <f, g''>?

[pallid rampart]

Of course

Not

[dusky epoch]

well there you have it

[sinful valve]

Can anyone help with understanding eigenvalue decomposition / SVD ?

[lavish jewel]

what about it

[sinful valve]

So i was reading this right

then on like maths exchange i found this question

I was confused as to what this even means i cant lie

[lavish jewel]

the notation in the last image, or?

[sinful valve]

How would you read the first line Ive heard its an inner product space or sumn but not sure what that means

[lavish jewel]

inner product indeed, but since you're working with matrices, the inner product is the dot product

so $\langle Ax, y \rangle = y^\text{T}Ax$

[stoic python]

Edd

[sinful valve]

oh so its just y dotted with Ax

what about rhs though

why are they equivalent

[lavish jewel]

do you know what it means for a matrix to be symmetric?

[sinful valve]

where its tranpose is the same matrix

[lavish jewel]

precisely

in general it is true that $\langle Ax, y \rangle = \langle x, A^\text{T} y \rangle$ because $\langle x, A^\text{T} y \rangle = (A^\text{T} y)^\text{T} x = y^\text{T} A x$

[stoic python]

Edd

[lavish jewel]

later on in the proof they also go on to use that A^T = A for other stuff

[sinful valve]

so wait for the first line they have used the idea that A is symettric basically

from the original question being asked

[lavish jewel]

no, not yet

i misguided you by asking that too early. it's used later on

in the first line it's not needed yet

[lavish jewel]

as written here

cuz you get A transpose transpose

[sinful valve]

wait so you have shown the statement of the first line works

because the dot of x and A^Ty = y^TAx

how does that show anything though

[lavish jewel]

it doesn't, it's just the first step. more stuff is written below

ngl, i think you might be too early for this :x

especially for the svd

[sinful valve]

wdym too early

[lavish jewel]

you're struggling with taking transposes

[sinful valve]

uh i can take a tranpose wdym im confused with how you have notated it out

them angle brackets are confusing

acc wait lookin a bit more its calm

so you have just worked on the rhs using teh dot product to get to the original lhs which is y^TAx

i understand that and then what about the part after

[lavish jewel]

they use the definition of a symmetric matrix and of eigenvalues

[sinful valve]

yeah why does the lambda / mu eigenvalues apply only to x / y in the product space thing

[lavish jewel]

write out the dot product as a sum and see for yourself

or maybe it's easier if you start from <Ax,y>

then if x is an eigenvector of A, Ax = lambda x

so <Ax,y> = <lambda x, y>

then write the dot product as a sum $\langle \lambda x, y \rangle= \sum_{n_1}^N \lambda x_n y_n = \lambda\sum_{n_1}^N x_n y_n = \lambda \langle x,y \rangle$

[stoic python]

Edd

[sinful valve]

right yeah

thats clear but what exactly is the relation between the eigenvalue and the eigenvector to only apply to x / y in product space

like i have seen how eigenvectors and values are calculated as such

[lavish jewel]

what?

[sinful valve]

so they say x and y are eigenvectors of A

found via the eigenvalues lambda and mu

why is it not $\lambda \langle x, y \rangle = \langle x, \lambda y \rangle$

[stoic python]

shriller44

[lavish jewel]

that's also valid, but useless for the proof

because lambda is not the eigenvalue associated to y

so lambda y is not A y

whereas lambda x is also A x

[sinful valve]

oh yeah i guess that makes sense

fair

so after the what is this deducing

so it ends up being the dot of x and y is 0 so they are orthogonal eigenvectors

but what about the first bit

(lambda -mu)<x,y> =0

[lavish jewel]

because <Ax,y> = <x, A^T y> = <x, Ay>

and Ax = lambda x, and Ay = mu y

so lambda <x,y> = mu <x,y>

[sinful valve]

oh yeah okay

[lavish jewel]

it's on the line just above 😛

[sinful valve]

wait so lambda = mu basically the idea

so its just 0 <x,y> = 0

[lavish jewel]

exactly the opposite of that

[sinful valve]

acc not equal like

[lavish jewel]

lambda is not mu

[sinful valve]

yeah ofc ofc

[sinful valve]

so the lambda - mu bit they just brush as non zero

but the important part is that dot of eigenvectors is orthornormal

he doesnt seem to fully explain it though tbf

[lavish jewel]

the explanation is pretty clear

[sinful valve]

okay fist off

what is an eigenspace

i guess its just a load of the same eigenvector with different scales on them

acc no it says the same eigenvalue acc nvm

[lavish jewel]

the subspace spanned by the eigenvectors associated to an eigenvalue

[sinful valve]

so these 2 eigenspaces are mutually orthogonal as shown from the last line above i guess

[lavish jewel]

mhm

[sinful valve]

im confused as to what

they say these vectors together give an orthonormal subset of R^n

what does this mean exactly

could you explain this a bit

[lavish jewel]

they have unit norm and are mutually orthogonal

idk what else to say, that's a definition

[sinful valve]

yeah

just new terminology is confusing to understand tbf

[lavish jewel]

none of these should be new if you're studying eigen-stuff, which is what i meant when i said you're too early

you skipped something in the middle

[sinful valve]

its not new new its like

a bit different

so they make an orthogonal set which contains the vectors of eich eigenspace from which we know are not equal, therefore the inner product space of evaluates to 0

and diagonalizable part relating to basis I think that makes sense ill have to check on that

like a basis only affects one dimension so logically i guess it does

each part of the basis i mean column

[lavish jewel]

i have no idea what you mean

[sinful valve]

yeah ignore that bit

so yeah if its diagonalizable we have got the basis n shit

so back to my thing

how does this basis we showed show this then

[lavish jewel]

i strongly recommend you review old topics

[sinful valve]

what topics you on about

this book simply isnt a maths book tbf its for computer graphics

its not necessarily important to understand it all really but it seems random atm

like i know orthonal matrices/ transposes/ symettric matrices n stuff

jsut trying to think how they combined it

[lavish jewel]

so, assume you know for a fact your matrix of size n x n has n linearly independent eigenvectors

[zinc timber]

https://youtube.com/playlist?list=PL221E2BBF13BECF6C

watch this

[lavish jewel]

in your case where you were looking at a symmetric matrix with distinct eigenvalues, you also just now showed that the eigenvectors are orthogonal to each other

you also know that the definition of an eigenvector with its associated eigenvalue is that Ax = lambda x

what you can do is put all of the eigenvectors as columns of a matrix, call it Q

and put the eigenvalues on the main diagonal of a matrix D

then AQ = QD

it's the equivalent of Ax = lambda x, but done for all of the eigenvectors at the same time

now, if the columns of Q are linearly independent, you can take its inverse

so that A = QDQ^-1

or also Q^-1 A Q = D

the latter is known as diagonalization, because there exists a matrix Q that makes A into a diagonal matrix D

the only extra step is that, in the case of a symmetric matrix with distinct eigenvalues, you just showed that the eigenvectors in Q are orthonormal

this means that Q^-1 = Q^T

so A = QDQ^T, and Q^T A Q = D

[lavish jewel]

this can be done for any matrix for which the geometric multiplicity of each eigenspace matches the algebraic multiplicity of each eigenvector

otherwise, the matrix is a so-called "defective matrix" that cannot be diagonalized

[sinful valve]

yeah i was looking into that tbf with the null space span thing

[lavish jewel]

hmm i haven't mentioned the null space

but sure, 0 is a perfectly good eigenvalue

[sinful valve]

yeah i was watching a video on diagnoalization that mentioned it

for geometric multiplicity

[lavish jewel]

ah looking at A - lambda I, sure

[sinful valve]

yeah anyway that reasoning does make it clearer tbf

for singular values though this is weird part xd

i think ill look into them more before questioning here again though if i get confused

[lavish jewel]

you should follow ryuzaki's suggestion and watch that video series. it does seem like youe course is skipping a lot of important stuff

[sinful valve]

its not necessarily a course

its like base linear algebra for CG

the applications being more important

[zinc timber]

that's why I specifically recommended it

[sinful valve]

ill watch them definitely then for sure

this book is fine but it does as you said not contain every single detail in math you need

[noble swan]

I am to find the basis of this system. I've noticed that it's the null space w/ pivots in each row & column, but how does that lead to it having no basis?

[zinc timber]

show what have u written.

also, you'll always have a basis, could be empty but there will be a basis

[noble swan]

The book claims the answer is that there is no basis

[zinc timber]

ok what are we finding? basis of the null space of basis of the range space?

[zinc timber]

that was a dumb question.

yeah so you see that the matrix is of full rank

so only possible point is (0,0,0)

i.e. the subset will contain only one element, 0

it means that the basis is the empty set ∅

[noble swan]

Ahhh, gotcha

[zinc timber]

u can't/don't say that the basis doesn't exist

[noble swan]

I wasn't comprehending that it's literally just 0's lmao

[zinc timber]

every VS has a basis, be it empty or non empty