linear-algebra

[vapid abyss]

Does orthogonality imply linear independence?

Like, if two or more vectors are orthogonal to each others, can you then say that they are also independent?

[lavish jewel]

you can

[wintry steppe]

no. for example (0,0) and (1,0)

[lavish jewel]

i forgot the 0 vector, oof

[wintry steppe]

never forget the 0 vector. He's the only one who's always there ❤️

[heavy crown]

Help anyone ?
Given: Let A be nxn order, and c is an eigenvalue of A.
Prove: for any integer m ≥ 1 , cᵐ is an eigenvalue of the matrix Aᵐ

I was thinking maybe proving it by showing |cᵐ * I - Aᵐ| = 0 but I got stuck on how to get there 😦

[marble lance]

If v is an eigenvector for c, i.e. Av = cv.

Then how can find an eigenvector for c^m based on v, i.e. A(what) = c^m(what)

[heavy crown]

you mean A^m(what) = c^m(what)

[marble lance]

Oh

I misread

😵‍💫

[heavy crown]

its okay no worries :]]

[marble lance]

But yes

Then I mean that

And that makes it easier

[heavy crown]

is it

[marble lance]

Yes

The eigenvector is just v. Every time you multiply by A, another c comes out

You can use induction to prove that A^m v = c^m v

[heavy crown]

hmm induction

[heavy crown]

I'm sure there must be other way :[

[nocturne jewel]

I don't think you even need induction tbh

[heavy crown]

how?

[nocturne jewel]

just multiply by A m-1 times

[marble lance]

Yes, that's true, but writing it out nicely that it works for all m in N is basically induction

[heavy crown]

Av = cv / *A^(m-1) ?

[nocturne jewel]

$Av=cv \ AAv=Acv=cAv=c^2v \ \implies A^2v=c^2v$ repeat

[stoic python]

Mosh

[nocturne jewel]

But yeah, it basically just builds it up recursively

[heavy crown]

sorry but why cA = c^2?

[marble lance]

No, Av = cv, so then cAv = ccv

[nocturne jewel]

c[Av]

[marble lance]

cA is a matrix, c^2 is a scalar. They are not equal.

[nocturne jewel]

cA is a matrix

[marble lance]

Shh

[heavy crown]

oh right

I understand ))

[marble lance]

The induction isn't particularly hard if you know how to do induction

[heavy crown]

yea induction would make it more organized,
but I believe doing what he said and adding one more sentence to conclude it to m would be sufficient for me

thank you two for your help 🙂

[marble lance]

I mean, the induction is really basically the same thing:

Note that $Av = cv$. Suppose $A^m v = c^m v$, then
$$A^{m+1} v = A (A^m v) = A(c^m v) = c^m (Av) = c^m (cv) = c^{m+1} v.$$

And that's it.

[stoic python]

Lunasong the Supergay

[heavy crown]

yes it's more neat 🙂 looks good thank you

[heavy crown]

I got stuck again 😦

any clue?

how do I even tackle this

so I want to find two eigenspaces of dimension 1 that together create B. I know that Av = cv for existing c eigenvalue.
And also from the given, cv = Pu(v) then idk

[outer goblet]

can someone tell what this is called in english??

is this span?

[wintry steppe]

idk, i dont know that language

[gray dust]

not quite but close

[heavy crown]

lol metoo

[outer goblet]

norwegian lol

[gray dust]

you can say the span of the $u_i$'s is the set of all linear combos of them, ie
$$\Span(u_1,\ldots,u_n)=\brc{\sum_it_iu_i:t_i\in\bR}$$

[stoic python]

RokabeJintaro

[heavy crown]

did you try translating the defintion names?

[gray dust]

we can think of the x_0 term as shifting span(u_i's) from the origin by x_0

[outer goblet]

hm

[heavy crown]

anyone got any clue with this? 😢

[slow scroll]

Put the two vectors on the first line as the first two columns of a matrix. Put the next three vectors as the next three columns. Put the matrix in REF. Each pivot column corresponds to a basis vector in the original matrix

It’s possible to see the pivot columns in REF. But rref works too

[vast iron]

We can split this into two matrices, A = R + i C
Here both R and C are symmetric, real matrices (with only the last row and last column being non zero).
If R and C commute, they are simultaneously diagonalizable and therefore A is also diagonalizable.
Let's rewrite a_k as r_k + i c_k, then, we want RC - CR to be the 0 matrix.
(RC)_ij = (R)_il (C)_lj, i,j = 1,2,3...,n, where I'm using the Einstein summation convention (meaning there's a sum over l from 1 to n).
We know that (R)_il is 0 unless i=n or l=n, similar (C)_lj = 0 unless l=n or j=n.
So,
(RC)_ij = (R)_nl (C)_ln, where summation is only over l.
Similarly,
(CR)_ij = (C)_nl (R)_ln, where summation is over l alone.
Now we want
(R)_nl (C)_ln - (C)_nm (R)_mn = 0
Or,
(r_n1 c_1n + r_n2 c_2n + ... + r_nn c_nn) - (c_n1 r_1n + c_n2 r_2n + ... + c_nn r_nn) = 0
But r_ij = r_ji and c_ij = c_ji, so this is identically 0.
It means that the matrix is diagonalizable for all complex a_k.

I'm not sure if I made an error somewhere, hopefully someone can correct me if I did.

[worn fjord]

I am creating a website for my robot https://eudbriel.wixsite.com/curriculum

[glad acorn]

How to evaluate this det?

[zinc timber]

a circulant matrix

[glad acorn]

I thought it was 9*(-1)^86

[zinc timber]

well yes

and (-1)^86=1 so answer is just 9

[glad acorn]

it's not the answer, I was wrong. And I don't know why it's wrong

[zinc timber]

ok wait, let me check again

No it isn't

you also have the last column full of [10, 10, .., 9]

[glad acorn]

you mean my answer is correct or what?

[zinc timber]

ok here's a stupid way to calculate it

notice that -1 is an eigen value with multiplicity 86

and since sum of all the rows/cols is same = 10*86+9, this is the last eigen value

os det = product of all EVs = (-1)^86*869 = 869

[zinc timber]

no it's not

[glad acorn]

How about this, I have no idea how to evaluate the last det using the others

[zinc timber]

re arrange, transpose, factor out etc you'll get that matrix back

[glad acorn]

can you be more specific? I don't come up any method in my mind that can expand this det

oh i solved

[zinc timber]

ok can you write $\mqty[-4 \ -7 \ -10]$ as $-3\cdot \mqty[1 \ 2 \ 3] -\mqty[1\1\1]$

[stoic python]

Ryu

[glad acorn]

I just cannot see this relation

I do it in this way

[zinc timber]

ok but since you hv solved it, no need to see the relation anyway

[glad acorn]

although it's the same thing

[zinc timber]

[glad acorn]

Ryu are you an undergraduate student?

[zinc timber]

no

[glad acorn]

ok i have confused you with another user

[zinc timber]

[vocal isle]

Hey guys can someone help me get from the 2nd line to the 3rd line? I don't know how to do calculus with vectors and matrices

[lavish jewel]

you can expand the product into y^Ty - 2px^Ty + p^2x^Tx

differentiate w.r.t. p to get -2x^Ty + 2px^Tx

set that equal to 0 to find a stationary point and solve for p

for nonzero y and x, this should be strictly convex wrt p

so that the minimizer is unique

[wintry steppe]

What does this mean in layman's terms?

[lavish jewel]

what exactly?

if none of it seems familiar... i have bad news

[wintry steppe]

Is this just saying that one can define a linear transformation by mapping every basis vector of V to some vector w \in W?

[wintry steppe]

wdym..?

[lavish jewel]

this is right

[wintry steppe]

Ok...what is the significance though and isn't it pretty self-explanatory..?

[lavish jewel]

if you know the vectors and their image under the transformation, that's enough to define the transformation

[wintry steppe]

I didn't quite understand the existence/uniqueness proof in Axler tbh

[wintry steppe]

What exactly is meant by "define" the transformation (sorry for the dumb questions)?

[lavish jewel]

all the information you need to construct it

[zinc timber]

p is the component of the projection of y onto x

[wintry steppe]

Ok, this is again really dumb but how does one construct a linear transformation? Is it like analogous to a function (single-variables, f:R --> R)?

[vocal isle]

Thanks Edd, can I ask where di you learn this? wahat was the course called? I'm just a dumb Engineer and I have't stuidied how to do matrix calculus 😦

[lavish jewel]

hmm

i'm also an engineer

let me think

some optimization course, maybe

after linalg and calc 3

if you need references, wikipedia and the matrix cookbook will go a long way

just wiki matrix calculus

i don't remember anyone ever explicitly teaching it tbh

but if you want to convince yourself, just rewrite everything as sums and compute partial derivatives component-wise

[random axle]

My textbook demonstrates how to show that a set of elements forms a vector space. But it doesn't teach much else about it. What is the consequence of a set forming a vector space? Can we apply vector operations to all sets that form a vector space? Could we find the cross product of two polynomials for example?

[vocal isle]

thanks Edd, I'm surprised you havne't done a direct course about it. I know "normal" calclus really well. But when it comes to matirces and transposes I'm a complete noob. I find out I need to expand the vectors and matrices out every time, rearrange them, and then try and put them back into matrix form

i can see in that line you've done something similar to expanding the brackets out

i've also managed to find this

but it's not super comprehensive

[lavish jewel]

https://www.math.uwaterloo.ca/~hwolkowi/matrixcookbook.pdf

few things have proof in there, but there are plenty of references

or you can do it by hand once

[zinc timber]

you can define angle between 2 polynomials if that surprises you

[wintry steppe]

Can you please elaborate on this?

[lavish jewel]

due to linearity and the definition of a basis, if you know what the transformation does to a basis of its domain, that's all you need

[wintry steppe]

Isn't that...common sense though?

Since any vector can be constructed using the basis

[lavish jewel]

for an arbitary v in V, you can build its image only from T(v_i)

well, they're driving that idea home

[wintry steppe]

Ok....

[lavish jewel]

and it's important to know it's unique

[wintry steppe]

How do you prove such a statement though?

The uniqueness part

Why is it necessarily unique?

[vocal isle]

Thanks this resource is extremely helpful! If you or someone else finds/remembers a course that formally taught you these things, then please let me know

[lavish jewel]

here

you can decompose two linear transformations for which T(v_i) = w_i = S(v_i) into their "action" on the basis, and show that T(v) = S(v) for all v

and then use the def of two maps being equal

[random axle]

so if the elements of a vector space are vectors then what can we do with them? you're saying we can't do vector operations with polynomials? then why does my book say they're vectors? #mindfucked

[zinc timber]

I didn't say we can't

it's just the cross product is not something you can define on arbitrary vector space, u can define dot product without any difficulties

[gray dust]

generally to show smth satisfying a property is unique, we suppose x,y are objects satisfying the property then show x=y

in this case we suppose T,S are maps satisfying the given property then show T=S

[ebon river]

I need some help understanding the intuition behind the steps in this proof. The finite dimensional version was easy to understand since I can start working with a basis and take inner products appropriately. But how does the author in particular think of the 'tau' vector or the expansion above it? In the sense how could one come across it independently?

Sorry the page was cut off

[glossy plank]

what are schur modes? i cant find any info on their definitions online besides this nyu paper thats kinda hard to read for someone at my level

[wintry steppe]

can anyone explain F is not a characteristic 2 this ??

F is a field

[dim epoch]

certain fields have the characteristic 2

for example F_2

and some Theorems don't work as they do for other fields on these

[wintry steppe]

but what is meant by F is characteristic 2 ?

[wintry steppe]

okay

@dim epoch

[silk scroll]

just let τ be k•(α-γ), k in K, so we could get the function of τ, then get the min. btw the ||β-α- kγ|| could let you thought the proof of C-B inequality.

[dusky epoch]

the characteristic of a field is the smallest number of copies of 1 whose sum is zero, if it exists (otherwise the characteristic is 0)

@wintry steppe

[wintry steppe]

actually they have not introduced abstract algebra yet . i believe i should skip that que .

thanks tho Ann n Timo2

[dusky epoch]

can you show the question

[wintry steppe]

@dusky epoch

[dusky epoch]

see Appendix C

you were told exactly where to look, no?

[wintry steppe]

Yep but there is no such thing called as characteristic 2

[dusky epoch]

appendix C contains not even any mention of the word 'characteristic'?

[wintry steppe]

Yepp

[dusky epoch]

in any case, for the purposes of your question, you can restate the condition as: "assume 1+1≠0 in F"

[wintry steppe]

Okaay

[wintry steppe]

Oops my bad i found it

Thanks @dusky epoch

[winged prairie]

yo im having a hard time wrapping my head around the idea of congruence between vectors and subspaces. Like when looking at modular congruence, if u have x =(congruent) to y mod n that makes sense to me (int terms of integers), but when u put vectors in it doesn't

[granite mesa]

Here it is like x=y mod M, meaning that x-y is in M. The equivalence classes are x+M. Note that x+M=M iff x is in M. x+M is afine space. So it is a vector space iff x+M=M iff x is in M.

[winged prairie]

ok ty

[vocal isle]

Can someone help me understand why in this example
$x^T A^T b = b^T A x$

[stoic python]

The Smelly Cherry

[vocal isle]

https://www.youtube.com/watch?v=EnNH3SxyZEI&ab_channel=MathTheBeautiful
Reference at 8:44 of this video

I thought the rule was:
(x^T A^T b)^T = b^T A x
?

[honest socket]

Considering the solution of this problem

I can't quite understand the part where we form the matrix:

As in, I understand the reason why the order doesn't change the determinant, I understand what we're trying to do, I believe I understand why we order them like that, but I'm not sure why this is precisely the resulting determinant

Also not sure if I should put this in competition math or here, since it's a putnam problem but I got it as part of a homework problem set in linear algebra class

I suppose I am also not too familiar with block matrices, besides having a general idea of what they are

[spring pasture]

I haven't watched it but is the resultant here 1×1 matrix

[vocal isle]

oh i see

because the transpose of a 1x1 must be the same

thank you!

[severe heron]

Can someone help with showing R(A^3)<= R(A^2) where R denotes the Rank?

[slow scroll]

show that the null space of A^2 is contained in the null space of A^3 and apply rank nullity.

*im not sure what inequalities and stuff you are allowed to assume, but this would be a pretty barebones way to go about it

and i can elaborate more if needed

[glad acorn]

this is obvious since the degree of freedom has been decreased, if we think in terms of column space.

also the rank would surely remains unchanged in some cases

[slow scroll]

^ ye, and you can use the size of the null space to encode the decrease in "degrees of freedom"

[zinc timber]

has your problem been resolved?

[honest socket]

No, I still can't figure out that particular detail

Haven't been sleeping much recently, but even if I were I doubt I'd fully get it, I feel like there's something missing in my knowledge perhaps?

I know where the first "block" comes from,

And I can see how the second and third case can give rise to the other Mn-1 ones in some way

Well since there's induction, maybe it's something that the induction step does?

also I'm not sure how the other blocks come to be

I can get 1's along the main diagonal since elements with the same indices obviously equal 1

[zinc timber]

can you understand why det should be invariant under any permutation of S_i's?

since Si are all possible subsets if {1,2,..,n} except the empty set, if you interchange any 2 to Si, Sj,, you must also change row i with row j, as well as col i with col j so net effect is +1, so it's invariant under permutations of Si

so we can just look at one permutation that is convenient to us, I would rather choose one that looks like
{1},{2},{3},..,{n}, {1,2},{1,3},... and so on

Now think recursively, construct the matrix M_n+1 by adjoining the (n+1) to all the previously achieved subsets, like
{1, n+1}, {2,n+1}, ... and so on

and take the last subset {n+1} and you have the complete set of the subsets

[honest socket]

yes

[honest socket]

That seems similar to what they do in the second step of the ordering, yes?

yeah i get that part of the ordering

[zinc timber]

Now for the older subsets that we had, their intersection, that is
$S^{(n+1)}_i \cap S^{(n+1)}_j = S^n_i ∩ S^n_j$ for first of the subsets

[stoic python]

Ryu?

[zinc timber]

because all of them now have one element added to them

Ok let me rephrase it properly

[honest socket]

nah i get that

I wrote that exact same thing in the clarifications i was making to the solutions

well very similar

it's the next bit i don't get

how that induces the particular matrix we see I suppose

[zinc timber]

Let $S^N_i$ be a subset of {1,2,..,n}. denote $S^{n+1}_i$ mean the n+1 is unioned with the subset

[stoic python]

Ryu?

[zinc timber]

so the subsets now can be written as $S^n_1, S^n,2, \cdots, S^{n+1}_1, S^{n+1}_2, \cdots, {n+1}$

[stoic python]

Ryu?

[zinc timber]

Now if you try to construct the block matrix given these subsets, you see that for first $2^n-1$ it's basically what you had before with n elements because n+1 doesn't come into picture yet

[stoic python]

Ryu?

[honest socket]

yeah that's as far as I got

[stoic python]

Ryu?

[honest socket]

and those would have the same values as the previous ones

oh wait

[stoic python]

Ryu?

[zinc timber]

so that gives another M_n

[honest socket]

though I think you shuold have n+1 on the S_i too

[stoic python]

Ryu?

[honest socket]

oh wait

now this is where I don't get it

[zinc timber]

do we?

[stoic python]

Ryu?

Ryu?

[honest socket]

also, this is where I kinda stop understanding I suppoe

[zinc timber]

upto this is fine?

[honest socket]

I know we definitely need to have the ones along the main diagonal so that makes sense, and that we got the second M_n, but I'm not sure about the symmetricity thing

[zinc timber]

look and this one and try to understand the construction

hence symmetric

[stoic python]

Ryu?

[honest socket]

oh okay

yeah that part makes sense now

[zinc timber]

upto what? everything I have done by now?

[honest socket]

kinda iffy on that part but i think it's up to minor stuff that i need some more sleep to get since I thought something very similar but kidna wrong

[zinc timber]

what we are actually doing is just adding the element to the Si'th subset and creating a new one

try with some examples

say we have {1,2} then subsets are given by

[honest socket]

oh yeah i think i get that wait lemme translate what I wrote earlier

[zinc timber]

{1}
{2}
{1,2}
Now we add 3 to it, but keep the first 3's.along with them

{1} = S²_1
{2} =S²_2
{12} =S²_3

{13} =S³_1
{23} =S³_2
{123} = S³_3

{3} = S³_4 say doesn't matter

you understood the construction?

[honest socket]

no yeah I understood that but I'm not sure which part I'm not understanding, it's kidna dumb so I'll probably get that on my own tomorrow

[zinc timber]

[honest socket]

so I just need to know where the other parts of the matrix comeee

[honest socket]

assume it's fine up to this point

[zinc timber]

ok tell me this one $S^3_i \cap S^2_j = S^2_i \cap S^2_j$ or not

[stoic python]

Ryu?

[honest socket]

yeah it is

[zinc timber]

because S³ one contains 3 but S² doesn't

so, for the next block, when we take S³ ∩ S² we'll just get S² ∩ S² which was our M²

[signal mulch]

I wrote an answer for a question but im not sure abt it, Can i send it here to confirm?

[zinc timber]

try one thing write down those subsets in sequence on paper and try to construct the matrix

S³ one

see if you get the M² blocks back like I mentioned

[honest socket]

oh wait i get why i was wrong about what i was talking about lmao

ye sure go on

[zinc timber]

can you figure out why there should be a block of 1's,

what they all have in common?

[honest socket]

is that the n+1's intersecting

[zinc timber]

yup

also you can figure out the rest?

[honest socket]

the problem i had was very dumb, I basically couldn't figure out that for one of the matrices we're have indices before 2^n for one and after that for the other and then the reverse for the other one

hmm

I could try

oh yeah

lmao

yeah so for the zeroes we're intersecting the only {n+1} with the S^n's

[zinc timber]

yup

[honest socket]

lmao thanks i was being super dumb xD

[zinc timber]

and for S^(n+1) we have n+1 common with {n+1} so rest is 1

now time for the determinant

[honest socket]

oh that's the part I wasn't getting, after that point it's just doing matrix stuff to the matrix xD

[zinc timber]

try column manipulation and make the block of 1's = block of zeros

and google determinant of a block matrix

I ran out of energy to type latex on my phone

[honest socket]

ye ye ye, I posted only the relevant part of the solution, that's described in the next one but I felt it wasn't pertinent to my question

[ebon river]

Oh I see...thank you very much

[outer goblet]

in my language "range" means rank, and not range, why does everything need to be confusing like that 🙃

[granite mesa]

Somebody has already answered u earlier.

[wintry steppe]

diagonal matrices are funny

[lethal tangle]

Hi could some one help me with this exercise?

[granite mesa]

,tex Take $W_1=\langle (1,0)\rangle$, $W_2=\langle (1,1)\rangle$, $W_3=\langle (0,1)\rangle$.

[stoic python]

RaD0N

[granite mesa]

They're pairwise direct sum. So I guess you could go by this way and see what more conditions are necessary...

[upper badger]

Can anyone help me understand why if two prime numbers divided by two gives an integer yet divided by four does not?
For instance,
(3 + 7)/2 = 5
However,
(3 + 7)/4 = Decimal value

[lethal tangle]

perfect, thank you!

[nocturne jewel]

1. not LinAl
2. not even true to begin with

2+3 is 5, 5/2 isn't an integer

In fact any (2+p)/2 will never be an integer for odd primes p

Also 11+5=16 which is 4^2, so your entire statement is false

[upper badger]

Forgot to add that the prime has to be greater than two

and yeah not everyone knows how to use LinAl m8

[nocturne jewel]

LinAl... the branch of math

[nocturne jewel]

But I still disproved your claim for divisibility by 4

[wintry steppe]

I don't understand how they got (1, -1, -5)

if someone could explain

[nocturne jewel]

Just solve AU=B

[wintry steppe]

do i isolate U? or how do i do it

[nocturne jewel]

If A is invertible then sure

[wintry steppe]

ok thank you!

[granite mesa]

Ahh ok

[shell kindle]

Can someone tell me what happened in this step?

Like, how did they change signs?

I know they factored out the variable, but how come the inside of the parenthesis changed signs but not the variable outside the parenthesis?

[north hedge]

they just multiplied both sides by -1?

[shell kindle]

But the variable outside the parenthesis never changed signs

[north hedge]

it doesnt matter the other side is 0

[shell kindle]

Oh, you right

[north hedge]

i mean for example, if x+3=0 then -x-3=0

[shell kindle]

Right, thanks!

[wintry steppe]

Can anyone explain what is the meaning of eqn in part c??

And i have not understood part c que too

[nocturne jewel]

they want you to show those 2 expressions

[obtuse wraith]

Cosets of subspaces can be represented by more than one vector, generally. So, if W is a subspace, and u-v is in W, then u+W and v+W are the same coset. If you want to define addition of cosets, you have to show that it doesn't matter which representative you take. That's what it means to be 'well defined'.

[nocturne jewel]

aka the operation is independent of the choice of v

[wintry steppe]

Ohh okaay

[wintry steppe]

Okay

Thanks !! Mosh n carthelis

[torn leaf]

Can anyone please help on how to do this

[gray dust]

if $ED=B$ and a certain row operation on $D$ gives $B$ then $E$ is the result of doing the same row op on the identity matrix

[stoic python]

RokabeJintaro

[torn leaf]

Can U explain more

[signal mulch]

Help pls!

Until now I was only able to prove that SUS' spans Rn but thats only 1 part of showing that its an orthogonal basis for Rn

How can i prove the other half? and how can i deduce *?

[torn leaf]

Which matrix do I need to do row operations of?

[gray dust]

what row op on D gives B?

[obtuse wraith]

Well, remember that row operations don't have to use more than one row

[torn leaf]

Do we just multiply bd and fb?

[obtuse wraith]

A row operation on D could be multiplying one of the rows by something.

[gray dust]

i'm not asking you to multiply any matrices

[spring pasture]

For * use dim(W U W') formula

[gray dust]

what elementary row operation, when done on D, will give B?

[obtuse wraith]

To prove the other half, you know that all the u's are orthogonal to each other, and you know that all the v's are orthogonal to each other. What can you say about any given u and any given v (considering which spaces they are in)?

[gray dust]

it helps to recall the types of elementary row operations

[obtuse wraith]

Also how to write row operations as a matrix

[torn leaf]

BD^-1 and FB^-1?

[spring pasture]

It's me

[torn leaf]

?

[spring pasture]

Nou need to find inverses

[torn leaf]

Then what I’m confused

[spring pasture]

They explained you bove

[obtuse wraith]

Forget about trying to find a matrix for a second. Can you see what you would need to change about D in order to make it turn into B? What row has to change, and how does it have to change?

[torn leaf]

Row 2 it’s multiplying by -3

[obtuse wraith]

Well, to go from D to B, it's the other direction, right? So instead of -3 it's....

[torn leaf]

3?

[obtuse wraith]

You're right that you want to change row 2. So, if you start with row 2 of D, what do you multiply by to get row 2 of B?

[torn leaf]

1/3?

[obtuse wraith]

Almost

[torn leaf]

-1/3?

[obtuse wraith]

There you got it! Now, an elementary matrix is a matrix that represents a row operation. Your row operation is multiplying row 2 by -1/3. Do you know how to write the elementary matrix for that row operation?

[torn leaf]

No

[gray dust]

the corresponding elementary matrix is obtained by doing the same row op on the identity matrix

that matrix is the answer to the first question

[torn leaf]

I’m not getting

Sorry

[gray dust]

the row op that we do on D to get B is to scale the 2nd row by -1/3

do the same row op on the identity matrix

[obtuse wraith]

Do you know what the identity matrix is?

[torn leaf]

Inverse identify matrix?

[gray dust]

for this problem we mean the $3\times3$ identity matrix. it has 1 on the diagonal and 0 elsewhere
$$I=\m{1&0&0\0&1&0\0&0&1}$$

[stoic python]

RokabeJintaro

[obtuse wraith]

Whenever you see or hear 'Identity Matrix', it's a matrix like that. With 1 on the diagonal entries and 0 everywhere else.

[torn leaf]

Yes I know this

But how to use in this i dont know

[gray dust]

the row op that we do on D to get B is to scale the 2nd row by -1/3
do the same row op on the identity matrix

[obtuse wraith]

So, you determined that you need to multiply row 2 by -1/3. Do that to the identity matrix. What do you get?

[signal mulch]

To prove that SUS' is an orthogonal basis I need to show this right?

1. SUS' should span Rn
2. the subset itself is orthogonal

[torn leaf]

1 0 0
0 -1/3 0
0 0 1

[obtuse wraith]

That is the elementary matrix for the row operation. If you multiply that matrix and D, you will get B

[torn leaf]

So that’s the answer?

[gray dust]

yes

[torn leaf]

Ahhh ok got it thank u both

Also

[obtuse wraith]

I_2 and I_3 are both identity matrices, I assume?

[signal mulch]

How can I prove that S and S' are orthogonal?

Bcuz that would allow me to prove that the subset is orthogonal

[obtuse wraith]

Take any u from S and any v from S'. Where does S live and where does S' live?

[torn leaf]

Is it correct?

[signal mulch]

S is in the basis of W and S' is in the basis of W^perp

[obtuse wraith]

The first part is a drop-down menu? I think you need a different option for these.

[obtuse wraith]

So, if u is in S, than u is in W. And if v is in S', then v is in W^perp. What does that say about u and v?

[obtuse wraith]

So, for the first one, if you start with the identity matrix, what do you have to do in order to get to E?

[signal mulch]

its orthogonal then right?

[obtuse wraith]

Right, by definition of W^perp

[signal mulch]

because W^perp is the orthogonal complement of to W so S and S' are orthogonal

[wintry steppe]

How do I write (6t-4) as a linear combination of (t+1) and (t-1)?

[torn leaf]

So only the first value gets replaced by -6

[obtuse wraith]

Right, so are you adding any rows together, or interchanging any rows?

[torn leaf]

Adding

So that’s the answer

Add k times…

[obtuse wraith]

I don't think adding rows will help here. Row 2 is 0,1. If you add that to row 1, it won't change the number

[torn leaf]

Oh

So multiplying?

[obtuse wraith]

Exactly. What are you multiplying?

[obtuse wraith]

I would write a(t+1)+b(t-1)=6t-4 and try to solve for a and b

[torn leaf]

First answer ok?

[obtuse wraith]

Beautiful

Your second one is close. Just take a look at the rows you have picked out

[torn leaf]

It’s not row 2?

[obtuse wraith]

Right now you say you are adding -4 times row 2 to row 1. Is that what you want?

[wintry steppe]

What's a and b? :"<

I've been trying to find those values for 30 minutes now.

[torn leaf]

That’s what I think

[obtuse wraith]

I was a bit flippant, I suppose, I'm sorry. So, you want to write 6t-4 as linear combinations of (t+1) and (t-1). If I were to solve this, I would assume it was possible, and the answer looked like a(t+1)+b(t-1)=6t-4, for some a and b. If I expand that, I get at+bt+a-b=6t-4, or (a+b)t+(a-b)=6t-4. Then I'd match up terms to make a system of two equations in a and b.

[obtuse wraith]

That would change row 1 but wouldn't change row 2.

[torn leaf]

So swap and i = 1 j=2?

[obtuse wraith]

Exactly! Do you see how that changes row 2?

[torn leaf]

Yes but I don’t understand what is happening in row 1 like nothing is changing tho

[obtuse wraith]

When you say "add -6 times row 1 to row 2" what you are doing is you're taking each entry of row 2, and adding to it -6 times the entry from row 1 that's above it. You're not doing anything to row 1.

[torn leaf]

Ojhhhh ok got it

Lastly q3

And the last box is 1

[obtuse wraith]

Again, how exactly are you changing row 2? Are you adding rows to it, or are you just multiplying it?

[obtuse wraith]

Which row are you multiplying?

[torn leaf]

2

[torn leaf]

2

[obtuse wraith]

Right now it says 1

[torn leaf]

I= 2, j=1

[obtuse wraith]

Do you need to do anything to row 1?

Or with row 1?

[torn leaf]

No

Do it should be @ (which means blank no answer)?

[obtuse wraith]

Correct!

[torn leaf]

Got it thank youuu so much

[vast iron]

Pretty sure it's diagonalizable for all complex a_k, unless I'm mistaken of course.

[vast iron]

Look at that (not sure if you posted this before or it was someone else).

[barren sentinel]

guys, how would i go about finding eigenvector of a matrix with eigenvalue 1?

in code?

[dusky epoch]

do you have a means of finding nontrivial solutions of Ax = 0?

[barren sentinel]

u mean row reduction?

if so, then yes

i have an algorithm for row reduction + back substitution

[dusky epoch]

no, i mean what i said.

[barren sentinel]

wdym exactly? could u explain a bit more?

also. just fyi, im not trying to find eigenvalues

[dusky epoch]

do you have a black box that can take a matrix A and output a basis for the solution set of Ax=0?

[barren sentinel]

im trying to find eigenvector given eigenvalue 1

[barren sentinel]

i think so?

i mean isnt row reduction + back substitution just that?

[dusky epoch]

row reduction and back substitution fucks up on degenerate matrices does it not

[barren sentinel]

no?

if u swap rows it shouldnt

and if its an inconsistent system it wont work anyways

then u have no Ax=0

[dusky epoch]

Ax=0 is never inconsistent, beastblaze

maybe i should have phrased this as, do you have a way to solve homogeneous systems

[barren sentinel]

yes

[dusky epoch]

okay then solve the homogeneous system (A-I)x = 0

[barren sentinel]

ok then? thats my eigenvector?

[dusky epoch]

yes, the solutions of (A-I)x = 0 are your eigenvectors by definition

[barren sentinel]

ok cool thnx

so i just take matrix A, subtract 1 from diagonals, and then solve the matrix using row reduction + back substitution

[zinc timber]

very specific question, but your matrix happened to be a symmetric matrix?

[barren sentinel]

no

[winged prairie]

anybody know how to do this?

im triyng to define a linear map that is bijective but im not sure how to do it

[vast iron]

@blazing sage did you get it?
It's not diagonalizable if
$$a_n^2 = -4 \sum_{i=1}^{n-1} a_i^2$$
(except when ai = a_{i-1} = 0, where the matrix is already diagonal)

[stoic python]

Dovahkiin

[silk scroll]

maybe just prove the dimension is equal.

[winged prairie]

Ye but we don't know the dimension of W

or V

nvm i did it

ty

[silk scroll]

maybe proju(v)=Σ(＜ui,Σ(＜ui,v＞ui)＞) ui

then the Orthonormal basis can get it

[wintry steppe]

so I could not use it

[outer goblet]

find a and c such that Ax=b is solvable

any tips on how to approach this?

[zinc timber]

A is the augmented matrix?

[outer goblet]

augmented? what does this mean

[zinc timber]

nvm it's not

have you heard of gaussian elemination?

[outer goblet]

yea lol

i just dont get what they mean by = span

[zinc timber]

ok, think you need to find a,c s.t. the space where Ax=b is solvable for b is the span of (1,2,3),(1,1,2)

[outer goblet]

ax=b is solvable when b is in the column space of A

but i cant see how to relate that to span

[zinc timber]

hm hm

[frank blade]

Because the left-hand side is a vector space. Whenever b_1 and b_2 lie in the image of A, so does b_1 + b_2.

And, as you said, the sets are the column space of A. Phrased a bit differently, the span of the columns of A is the span of the two vectors on the right.

[outer goblet]

is what im suspposed to show is that clumns of A, have the same span as the two vectors on the right?

[frank blade]

Depending on how you show that, yes.

[outer goblet]

so i can rephrase it to be "Find a and c such that span(A)=span{(1 2 3), (1 1 2)}, and that b is in the column space of A

[zinc timber]

first part is fine

showing that already means "b is in the column space of A"

[outer goblet]

so i have to find basis for column space

[zinc timber]

or use eye

since (1,2,3) is already a column of A, find another one, ex if u take a=1, you get (1,1,2) which is our 2nd required vector

(2,1,c) just has to lie in the span,

[outer goblet]

its earlier exam question so dont think they would be satisiified if i said "use eye" lol

[zinc timber]

or you can try to find for which $a$, you get
$$\left| \mqty{ 1 & 1 & a\2 & 1 & 1 \ 3 & 2 & 2 } \right| =0$$

[stoic python]

Ryu?

[zinc timber]

if you don't want to use your eye lol

[spring pasture]

There's a fast way bye row operation 2

[granite mesa]

Take a linear operator $f$ from the product to $W$. Note that $g_i=f\circ \iota_j :V_j \to W$ is also a linear operator, where $\iota_j:W_j\to W_1\times ... \times W_m$ is the canonical embedding (wich is also linear operator). Then send $f$ to $(g_1,...,g_m)$. Then check that this is an isomorphism.

[winged prairie]

so it is wrong to check if their dimensions are equal?

[stoic python]

RaD0N

[granite mesa]

I w'd say it's not right, since you don't know if some V_j or W are finite dimensional or not.

[winged prairie]

ok makes sense

thanks!

imma try it out

[granite mesa]

For surjectivity, if you take $(g_1,...,g_m)$ arbitrary, u can put $f=\sum_j g_j\circ \pi_j$ where $\pi_j: V_1\times ... \times V_m\to V_j$ is the projection map, which is linear operator.

[stoic python]

RaD0N

[granite mesa]

Then check that $f\circ \iota_k=g_k$.

[stoic python]

RaD0N

[winged prairie]

👍

[silk scroll]

👍

great answer

[sinful valve]

ay for singular vectors / eigenvectors etc

is there any exact reason why the first singular vector / singular value is the most prevalent feature

or important aspect of the eigenspace

like the basis of PCA for face recongition is saying the singular vector is the average face and its corresponding singular values and then the other faces are like different in some sense as less average

like the ordering is so significant but how does it arise?

like it kinda makes sense given the SVD is taken from the covariance matrix of the flattened images

[wintry steppe]

yea

cuz

removing bigger singular values from the matrix results in a bigger change of the distance between the original matrix and the new one

[sinful valve]

im just tryna make sense of the SVD of this thats all

this being like covariance matrix of our image matrix

where the image matrix has each image in the rows and there are like n rows for n images in our dataset

calling it matrix M

get $C_x = M^TM$

[stoic python]

shriller44

[sinful valve]

so this gives us like the relative similiarity of each image where the image with itself is just gonna be the singular values or something

but im not too sure exactly

i think the concept was he diagonalized it by extracting the diagonal then pulling out the singular vectors to the left

i think the understanding according to this

like first singular vector would be the average A with its singular value etc

and so there would be like for 5 images of same person and different configuration as in smiling/angled computes a different like singular vector and value that becomes more hazy i guess

really weird tbf the natural ordering of singular values in this context it doesnt make sense

why would the singular value of vector 6 be less than the first 5 singular values its just a new dataset image set

[lavish jewel]

because given the entire set of images, the other vectors "explain" the entire set better

there are more different from each other

idk if that's what you meant, now that i think about it

[sinful valve]

uh its kinda true but basically all it does before svd as i mentioned

was do that covariant matrix to do that similarity thing you were on about

[lavish jewel]

well, that's exactly what covariance means, right?

[sinful valve]

yeah so first row is image 1 compared with every other image ?

[lavish jewel]

since one first subtracts the mean from all the images, this operation is related to 1/N expected value{(M - mu)^* (M - mu)}

[sinful valve]

of the covariant

[lavish jewel]

yeah

[sinful valve]

then its image 2 compare with other image

[lavish jewel]

indeed

[sinful valve]

but then we convert to SVD for diagonalization

but why exactly

[lavish jewel]

why what?

why do we svd?

[sinful valve]

something about getting the diagonal values where image 1 correlates with itself

like the diagonal is the variance not covariance

[lavish jewel]

sure

that's before the svd, though

[sinful valve]

but like im confused how this idea is translated into the svd

[lavish jewel]

it kinda isn't, not directly

the important part is that the covariance matrix is symmetric, so it is diagonalizable in an orthogonal basis

what you actually want is the EVD

for real symmetric matrices, the EVD and the SVD are the same

[sinful valve]

wdym EVD

[lavish jewel]

eigenvalue decomposition

[sinful valve]

oh yeah

nah it still gotta be svd

[lavish jewel]

they're literally the same here

[sinful valve]

i cant assume its gonna be a square

[lavish jewel]

the covariance matrix is always square

[sinful valve]

oh shit ig yeah

[lavish jewel]

or you mean SVD of the original data input

[sinful valve]

wait is it always a square

taking any matrix M doing M^TM = square matrix

[lavish jewel]

yes

[sinful valve]

fair that makes sense ye obv

so the SVD sort of breaks the covariance matrix into the vectors we require to see features of our input etc

thats the kinda cryptic part

I guess i could think of it as like the first bit of SVD i view as rotation

[lavish jewel]

it gives you an orthonormal basis for the domain and codomain

that's why i was suggesting you don't think of it with the svd

if you already understand the evd

if you don't understand the evd though, either one is ok 😛

[sinful valve]

eh same thing just eigen got both orthonromal no?

like yeah he broke it inot eigen here

[lavish jewel]

but do you understand what i means

[sinful valve]

yeah relatively

[lavish jewel]

well, it's equivalent to the svd here, so

[sinful valve]

yeah i get that

so he converts it to this diagonal form shown here and like

its just kinda not too clear how its ordered xd i dunno

[wintry steppe]

Is ={(x1, x2...xn) : x(2i)= 0} a subspace of R^n?

[lavish jewel]

how what is ordered?

[lavish jewel]

without further context, i can only say "sure i guess, looks like it"

[wintry steppe]

I edited my question.

[lavish jewel]

it seems like it, yeah? if you take 2 vectors in that set and add them together, all of the entries that were 0 remain 0

[sinful valve]

wait i think i considered it wrong tbf

this is the first n singular vectors shaped into images

[lavish jewel]

and scalar mult works as usual

[sinful valve]

but each image becomes a linaer combination of these i guess

somehow

this generates the unique faces

[lavish jewel]

shriller, you should review how the eigenvalue decomposition works

[wintry steppe]

thanks :) But if we have {(x1, x2... xn): x1 + x2 + ... + xn = 1} - that is not a subspace of R^n, right?

[lavish jewel]

right, think of scalar multiplication and notice it's not closed under that operation

[wintry steppe]

thank you :)

[sinful valve]

i get the decomp just tryna understand the application here ig

[granite mesa]

I think it is natural answer xd

[lavish jewel]

yes, all images in the set will be linear combinations of those eigenvectors

and the hope is that the covariance matrix has full rank, so that ALL images of the same size can be represented with the same vectors

and the hope is that the eigenvalues become small very quickly for other face images

that would give you some compression power

[silk scroll]

just for me 233333

[sinful valve]

yeah acc makes sense and to compare similarity we can just project a new image

and compare its projection

to the desired face

[lavish jewel]

that's exactly what an eigenvalue decomposition does

when you do QDQ^-1, Q^-1 is a change of basis into the eigenbasis

[sinful valve]

what he says is the linaer combination value is in the

[lavish jewel]

more importantly, for real symmetric matrices, Q^-1 = Q^T, and so QDQ^T is the EVD

[sinful valve]

$\Sigma V^T$

[lavish jewel]

right

[stoic python]

shriller44

[lavish jewel]

say you have a new face x

then you can take Q^Tx, which is exactly an orthogonal projection

and then the sigma matrix, which i called D, tells you how much to scale each of the basis vectors

[sinful valve]

okay so yeah that links to the idea of basis changing and scaling / rotation/scaling that i learnt it from originally ig

just in many dimensions ig

[lavish jewel]

rotation/scaling/rotation, sure

[sinful valve]

V tranpose in SVD all i know is it converts it back to the original basis

after scaling

by sigma

[lavish jewel]

hmm?

[sinful valve]

he said the linear combinations are stored in sigma v transpose

not sure precisely what he meant

my understanding of V transpose bit is kinda vague

just a reversion of basis

[lavish jewel]

matrix multiplication by a vector is a linear combination of the columns of the matrix

i guess what the mean is V^T x is an orthogonal projection, then you scale these coordinates by sigma

and finally you take a linear combination of the columns of U

or idk if you took the SVD of the data vs the SVD of the covariance mat

i'm taking about the latter

in the former, then it would be V instead of U

[sinful valve]

yeah wait he expands acc to say sigma v transpose is the mixture of U

oh wait yeah

u said the V and U would differ

this video takes the SVD of the matrix of images originally

but the original person i watched takes the covariance first before svd

so sigma idea is same but we use U as you said i think

[lavish jewel]

that changes what U and sigma are

(not V)

well

also V, depending on if you work with M^TM or MM^T

the basic intuition is the same, but what it means in an applied scenario is completely different

[sinful valve]

bruh just made this more confusing knowing theres 2 ways

[lavish jewel]

it changes the basis and the linear combination

[sinful valve]

whats the point of doing the covariance matrix then

[vocal isle]

i've been told that the minimum of this quadratic polynomial:

occurs when x = A^-1 b

but I can't find a proof for it. Does someone know where I can find this proof?

[winged prairie]

does anybody know a good textbook to aconomy Linear Algebra done right?

like on that has determinant-free proofs

[sinful valve]

oh i guess i need to remember that the svd or technically evd is literall inherently the covariant matrix in itself

just expressed in that form

and the order of A^TA and AA^T determines which vector to get from

[lavish jewel]

the covariance matrix MM^T has as row and column spaces the subspaces where the images are

[sinful valve]

wait no they the same arent they

why would they differ

the V part really confuses my entire understanding of this task

or last part of SVD

so you use U sigma if A^TA and V sigma otherwise basically

and applying V^T * image = projection but how

projection onto U and sigma

image projected on that space ???

what space

why is his projection U^T tf

[lavish jewel]

you need to review projection matrices

and they are probably working with MM^T

and the svd of M

[sinful valve]

i understand them separately just this context is confusing

but yeah the difference of MM^T and M^TM i need to check out

[wintry steppe]

Let M be a nontrivial subspace of V. Show there is a basis {v1, v2... vn} for V : v1, v2... vn not in M.

[sinful valve]

because somehow that changes what value you project by

thats the only thing i need to gauge to understand this task really

[zinc timber]

you sure such basis will exist?

[wintry steppe]

yes

[zinc timber]

[spring pasture]

[zinc timber]

I ask again, are you sure?

[wintry steppe]

Sorry, I didnt mean to be rude. But thats my homework, it says there exists such basis and I need to find one.

[zinc timber]

thing is there isn't such a basis

[spring pasture]

Show a photo perhaps

You may have written something off

[zinc timber]

either "non-trivial" or "show doesn't exist"

[wintry steppe]

I am confused.

[spring pasture]

Just show the question mate

[wintry steppe]

sorry not in English, but thats it, just one sentence

[wintry steppe]

could you elaborate?

[zinc timber]

any basis of V must contain atleast one element from M, otherwise you can't get a basis

there's no 'whole' without the 'part'

[dusky epoch]

??

[wintry steppe]

i am confused

[dusky epoch]

M is a nontrivial subspace of V

does this not mean M != V

also you can have like, V = R^3 and M = {z=0} and you can have a basis without any vectors in M just fine

[wintry steppe]

we defined a nontrivial subspace as M != 0 or V

[dusky epoch]

@zinc timber

[zinc timber]

non trivial also means ≠{0}

[dusky epoch]

yeah and?

you're saying that every basis of V has to contain something from M?

contain directly, not in span.

[zinc timber]

not directly, in span

needed the basis, I see

thanks for correcting

[dusky epoch]

okay so like here's what you can do

take a basis of M, call it B

add some arbitrary nonzero vector in M to it, call the result B'

and then take some vector v not in M and consider {v+x | x in B'}

[wintry steppe]

Thank you, Ann Could you explain how did you come up with this solution? {v + x} is a basis for V? I am a beginner

[dusky epoch]

imagine taking 3 vectors on the xy plane in R^3 and translating them all by (0,0,1)

[wintry steppe]

Is "(0,0,1)" here a nonzero vector in M or some vector v not in M? And how do we know {v + x} is a basis for V?

[dusky epoch]

(0,0,1) is a nonzero vector in M

also i think i may have lied

my method only works for subspaces of codimension 1

oops

[winged prairie]

Yo I’m having a hard time wrapping my head around having a basis of the space of linear maps

Does anybody have an intuitive way of visualising this is r^2 or r^3

Like it implies that every linear map going from let’s say V to W can be written as a linear combination of the basis linear maps but it doesn’t make sense to me in my head

[dusky epoch]

look at matrices rather than linear maps

you can have a basis consisting of matrices where one entry is 1 and the rest 0

[winged prairie]

ah i see

cuz im trying to visualise the dual basis

for F^N

so each would be n*1 matrix with all 0's except 1

sorry a 1*n

?

[sinful valve]

for the projection onto the eigenvectors of evd /svd where do the singular values come in

i dont think the singular values are even cared about in this tbf

im confused so you project to get a linear combination

the scale with eigenvalues?

like the U obtains a basis of principle components to construct a face from ordered by the eigenvalues

to show the importance

so for this face recongition thing it doesnt care about the actual values associated with each feature as long as its building a classification from the most important principle components based on the eigenvalue ordering of the diagonal

[lavish jewel]

what's capital X here? a matrix of face vectors?

[sinful valve]

capital X in this example is each image flattened to a column

so its shape is like (heightxwidth of image set, number of images)

[lavish jewel]

well, since X = USV^T, X and U span the same subspace

and so one wants to do a change of basis

if you put the images in X, then the images from some set, let's call it V, can be obtained by simply taking Xe_i, where e_i are the canonical basis vectors

i.e. they have a single 1 in them, and everything else is 0

on the other hand, U is a unitary matrix that has orthonormal columns, but the first r columns span the same subspace as the columns of X

so one usually takes the so-called "economy size SVD" that keeps only those columns. in this scenario, it makes sense to call this U_s, where the s denotes "signal space"

and since the columns of U_s are orthonormal, U_s^T is its left inverse

so that U_s^T performs a change of basis into the basis of singular vectors

[sinful valve]

yeah U^T times x i just see as the basis of which we want to translate some image into , projection to some space

space defined on the most important values

but like the actual singular values arent really used right?

they just exist

for this case at least

[lavish jewel]

they are used in determining U_s

[sinful valve]

are u talking about the projection part

[lavish jewel]

either when the matrix X is rank deficient, or if it isn't, you throw away the columns with small singular values on purpose

yeah

you might wanna project onto a lower dimensional space than the original if the error is small

[sinful valve]

well we presume the new image has the same dimensionality always in this case

as in its not deficient

[lavish jewel]

what are you calling image

like columns of X?

[sinful valve]

yes

[lavish jewel]

that is 1-dimensional and is not what i am referring to

[sinful valve]

x here is on the right is a new image that we have not seen

[lavish jewel]

mhm

[sinful valve]

projected into the basis to just a small sample space

which is deemed enough to correctly identify a face

but yeah thats U right but whats the purpose of V^T

[lavish jewel]

i think you're using projection wrong, as well as dimension

[sinful valve]

wdym using it wrong

[lavish jewel]

you won't need V in this setting

by dimension i'm talking about the dimension of a vector space

[sinful valve]

V the guy said was like a combination of the vectors present in U

[lavish jewel]

setting singular values to 0 is the same as saying you want a subspace of images spanned by fewer basis vectors

[sinful valve]

subspace of images? nah its just one image calculation huh

[lavish jewel]

you are doing the SVD because you want to talk of the subspace spanned by the images you put in X

the whole point of the SVD is to set up subspaces that are orthogonal complements of each other

all of the discussion here of linear combinations and the like is to that end

[sinful valve]

oh yeah so subspace you are saying a subset of another vector space

this vector space being U ?

yes that is what this is ig

but whats interesting is in this other video

[lavish jewel]

the images, i.e. the columns of X, span a subspace

[sinful valve]

he uses V to project

[lavish jewel]

and the columns of U span the same subspace, it's just another basis

one that is orthonormal

if you have a wide variety of images, any new image can be either approximately or exactly represented as a linear combination of the ones you have in X

and using the SVD let's you make this representation easily

[zinc timber]

$$|AB-BA| = \frac{\tr{(AB-BA)^3}}{3}$$

[stoic python]

Ryu?

[sinful valve]

and this approximation is simply just U^T x where x is new column image

but in this altnerative thing they used V output of SVD

and proj = x * V but the image was horizontally stored

is the fact its horizontal as in each row just make sense for V to be used to project ig

[lavish jewel]

if you put the images as rows in X, sure

[sinful valve]

so i assume when he extracts V

[lavish jewel]

then the change of basis would be x^T V

[sinful valve]

thats like transposed already

in his matlab code he just calls its V

[lavish jewel]

notice they're using an EVD, too, not an SVD

since the two are the same in this scenario

[sinful valve]

yeah cause its the same as mentioned

wait the change of basis is x^TV

as apposed to U^Tx

[lavish jewel]

as i said

it isn't

[sinful valve]

if they were columns

[lavish jewel]

sure

if they were columns, sure

PLEASE do yourself a favor, this is the third time i'm telling you

please review EVD, SVD, and change of basis matrices

you keep saying you know and understand them, but

[sinful valve]

yeah i mean i do grasp them but i wouldnty say fully understand in entirerty

[lavish jewel]

exactly

[sinful valve]

i have done lots of looking into them but i never got taught much of the mathematical basis

just the intuition i get

[lavish jewel]

i would said it's the intuition that you lack

you only know they exist

that's the whole struggle you're having here

also with things like vector spaces, spanning sets, and bases

[sinful valve]

for svd my intuition is just changing the basis > scaling in that basis > reverting back , based on some lecture

rotation / scaling/ rotation^T for transfromation matrices was what i learnt it for

PCA just weird

like rotation to new basis > new axis > scaling on each axis is the singular value etc

PCA case just has lots of different rotations with various scalings based on the variance over this axis direction or sumn

i literally only briefly looked over this ill defo review the linaer algebra and formalise the understanding tbf before acc applying

i kinda get it but its still a bit unintuitive

its really annoying as i have spent a long time trying to understand

svd that is

[lavish jewel]

this is what the EVD does, but not the SVD in general

and part of the issue here is that you're not dealing with transformation matrices, but rather looking only to do a change of basis for vectors in some subspace

[sinful valve]

nah the SVD does that wdym? like taking the example

i was shown

trasnforming a unit circle to an ellipse

that was svd

[lavish jewel]

not, really, no

as soon as you have a matrix that isn't symmetric, you don't "revert back"

say your matrix is 2 x 4

[sinful valve]

wait let me show u this diagram

[lavish jewel]

U and V aren't even the same size

[sinful valve]

i may have minterpreted

[lavish jewel]

see, there is no "reverting back"

you start and end at different places

[sinful valve]

yeahhh huh i always thought of V* as like go back to og

thats kinda changed it a bit

it was more to do with it as a transformation i think

[lavish jewel]

no

[lavish jewel]

V* doesn't "go back"

you end up somewhere different

[sinful valve]

i mean go back to original basis

[lavish jewel]

it doesn't do that either

[sinful valve]

okay

[lavish jewel]

EVD does that

[sinful valve]

yeah so that recent image is EVD

i think

yeah i gues it aint acc its just saying that weird ellipse is the same as M huh

which was the idea obv

[lavish jewel]

no, it's saying that it transforms the circle into the ellipse

M is only the arrow in the diagram

[sinful valve]

oh yeah shit i assumed M and that ellipse was RHS and LHS bruh

yeah did seem odd

so the diagram is just demonstrating some generic M transformation as being teh same as its SVD then

so linking back to my thing its like my covariance matrix is a transformation in this case and we are finding the axis it varies on following that diagram , with that being the principle components or sumn

my PCA idea is kinda new so i will look over that way more

[lavish jewel]

so that's where the issue comes in, technically your matrix isn't doing any transformation. you can still talk about the column and row spaces though

[sinful valve]

its not transforming but its being taken apart to analyse its components right thats the idea

[lavish jewel]

yeah

[sinful valve]

where principle componets according to this video is just sigma times U

[lavish jewel]

sounds about right

[sinful valve]

you know how its not a reversion or whatever bs i was on about

whats the signifance of the V part then conjugate transpose to anything

[lavish jewel]

it's a change of basis

[sinful valve]

oh yeah it applies it opposite the matrices bruh

so it does basis change * singular value scale * singular vector

[lavish jewel]

the matrix does a transformation M: R^n -> R^m

V's columns are an orthonormal basis for R^n

[sinful valve]

like after being in the new basis the 2 other matrices are just representing the vectors of the axis and the length

[lavish jewel]

U's columns are an orthonormal basis for R^m

sigma scales the vectors in the V basis and then either projects onto a lower dimensional space or embeds into a larger one

[sinful valve]

sigma scales the V basis ay?

[lavish jewel]

and also either "deletes" some components by multiplying them by 0, or appends 0s

[sinful valve]

yeah, i thought sigma scaled U tbh

[lavish jewel]

it could do either

matrix mult is associative

[sinful valve]

yeah ofc

[bold sun]

hey i need some guidance on this question

this is what ive tried

[sinful valve]

ill do some more research to actually formalise my understanding into something coherent rather than random linear algebra and diagram intuition

[bold sun]

basically im not sure on how to go about the show that aspect of it .....

[lavish jewel]

can you think of the relationship between the determinant and a matrix being invertible?

[bold sun]

determainant is non zero

[lavish jewel]

mhm

though i guess you don't know what the char poly of a diagonal matrix looks like

[bold sun]

what?

[lavish jewel]

that aside

recall that if a matrix A has an inverse B, then AB = I

and notice also that your matrix A is already very close to I

[bold sun]

yeah

[lavish jewel]

the entries of A only need to be divided, right?

[bold sun]

the inverse is 1/a11 1/a22

yeah

[lavish jewel]

all right, so you have that

have you seen how to determine whether the columns of a matrix are linearly independent?

[bold sun]

nope not yet

im going in order like so ive done till just b4 determinants so im doing qs then yeah gnna look at that section

[lavish jewel]

do you know how to find the determinant of a diagonal matrix?

[bold sun]

if that makes sense

no i havent looked at that yet ive done only 2x2 matricies det

[lavish jewel]

i'm not sure how they want you to show it's invertible then

[bold sun]

ooh okk

its basically im revising but im doing it all in order so i dont think im not meant to use other stuff yet if that makes sense

but thank you for the try 🙂

[lavish jewel]

i understand. there must be some clever trick that i can't see

[bold sun]

yeah i thnk so maybe

[glad acorn]

how to evaluate det U?

[lethal tangle]

Hi could someone help me with this exercise? I have no idea how to start.

[still turret]

clearly my calcs are wrong but I don't know why

[wintry steppe]

im not sure the steps u took but the last value should be 44

[dusky epoch]

check the last coord

44? why 44

[wintry steppe]

thats what u get if u do B | x

and rref it

[dusky epoch]

,rccw

[dusky epoch]

you made a typo

the last equation is -4=4a**-**2b+c

[still turret]

ohhh

thanks 🙏

[still turret]

did you do it my method?

[wintry steppe]

no, i did something else but ur method seems like it works as well

i set up a matrix that looks like this:

[1, 0, 0, -3; 5, 1, 0, 3; 4, -2, 1, -4]

where we basically have the matrix for B augmented with the vector x

and then just rref the matrix and your solutions will be lined up on the right side

it's just a more organized way to do what u did basically, dont worry about it too much if your class hasn't covered it yet

[still turret]

what does rref mean?

oh

reduced row echelon form