linear-algebra

[subtle gust]

thought it was 5X4 times 4X2

nvm

ty y'all

appreciate the help

[umbral pike]

If I read my D^n correctly, because x_1 and x_2 never converge, plus with 3^n only going towards infinity, there is no limit to the solutions.

Is that a correct assessment of my equation?

[lavish jewel]

assuming M is diagonalizable and has those eigenvalues, sure

[dusky epoch]

(-1)^n

not -1^n

[lavish jewel]

ah true

[zinc timber]

[abstract vale]

Yeah it’s Cramer’s rule

The part above I proved is his rule I think

[limber oracle]

hi guys, need some help here

[dusky epoch]

have you made any progress so far?

also @limber oracle please do not post in multiple channels at once

[exotic horizon]

hi! I'm struggling with an exercise of Markov chains. I unfortunately don't have any notes or lesson, just the exercises, so I don't even know where to look...

"Check that the matrices below are stochastic, then describe, without doing any calculations, what must be the asymptotic behavior of a particle evolving on χ = {1, 2, 3}, according to a stochastic dynamics defined by T. In particular , describe (approximately) what must be the stationary probability vector p∗"

I've already checked that they are stochastic, but I don't know anything about the second part, and lessons on markov chains i find online don't cover this

so if anyone can tell me where to start looking I'd be very grateful

[zinc timber]

If you have no idea what markov chains are, watch a lesson on it rather than posting it on discord and asking someone to solve them for you

[exotic horizon]

hey i'm sorry, maybe i misworded it. But as I mentioned in my message above, I'm not asking for anyone to solve it - I'm asking what should I look for online to understand how to solve this exercise. I have already looked up lessons on markov chains and they don't cover how to answer this specific question.

[zinc timber]

markov chains, the stationary vector you are mentioning is a part of it

it means a state that remains fixed jf you apply the matrix on it

[exotic horizon]

when they ask me about the asymptotic behavior, does it have something to do with the stationary vector?

[zinc timber]

it does

[exotic horizon]

okay thank you!

[granite mesa]

[A-xI_{m+n}=\begin{bmatrix}B-xI_m&C\0_{n\times m}&D-xI_n\end{bmatrix},]
hence
[|A-xI_{m+n}|=|B-xI_m||D-xI_n|.]

[stoic python]

RaD0N

[magic hawk]

Hey guys can I get a hint for part 6a

I’ve tried induction using definition of determinant

i.e. Expansion along column or row i

But doesn’t seem to work out

[abstract vale]

I need help with 5b

I’ve done 5a but I feel like I’m blind and can’t see what to do on 5b

[wicked palm]

Use the fact that the determinant is some signed sum of all the determinants of minor matrices. That’s my first thought

Might lead somewhere swag we will see

[clear iron]

wait isn't |Bn| the size of the matrix?

or am I getting my notations wrong

[magic hawk]

det

[clear iron]

ohhhhhhhhh

I didn't see that notation over here

[magic hawk]

i tried doing that but don't see how it related to B_n-3

like A_12 minor is weird, assuming we are expanding along the first row

definitely looks like an induction thing

but not too sure

[clear iron]

hmm

did you try looking at the determinant for B4x4?

remember that determinant is a recursive computation,

[magic hawk]

I did but don’t see any connection to B1

Whichever minor I take there’s always gonna be a weird matrix as a leftover

[clear iron]

can you show me waht you did?

also, try to take the determinant starting off with the 1st row
and see if you can see something interesting

[magic hawk]

Ok my working is really messy

This circled part is where I expand along first row

I get a huge matrix that doesn’t seem to relate to anything

I’ve tried row simplifying

But that doesn’t seem to help

It’s fine now guys I got it

This is it if anyone was interested

I made a small error in my working which didn’t allow me to see an important step

[hearty rapids]

i have a question

when you are trying to find the inverse of a 2X2 matrix, why not just multiply it by the identity matrix like you would with a 3X3?

instead of the A-1 = (1)/(ad-bc) [[d, -b], [-c, a]] method?

am i making sense

[fringe fjord]

Not really.

What does "just multiply it by the identity matrix like you would with a 3X3" mean?

[fringe fjord]

Yes, I know what an idenity matrix is.

[hearty rapids]

sorry

i meant i saw a video

oh

i am being dumb

[fringe fjord]

I'm confused that you seem to have a method for inverting a 3×3 matrix that consists of multiplying it with the identity matrix. That doesn't sound like it would make any progress.

[hearty rapids]

you are not multiplying the 3x3 matrix with an identity matrix

at all

sorry

my brain just fizzed out

what you are doing is setting the 3X3 matrix equal to an identity matrix and making the 3x3 matrix look like the identity matrix with row operations

[fringe fjord]

One method for inverting matrices involves gluing an identity matrix to the right side of the original matrix and doing Gaussian elimination on the whole thing. That works for 2×2 as well as for 3×3.

[hearty rapids]

ah i see

[fringe fjord]

Yes, what you described.

[hearty rapids]

so this must be another method

[fringe fjord]

Yes, that is Cramer's rule. For 2×2 matrices it boils down to something fairly simple, but it quickly gets too complex to be of practical use for larger side lengths. It is theoretically important, though, because it says that the entries of the inverse are rational functions of the entries in the original matrix.

[hearty rapids]

i see yeah my head was spinning when i was trying to think of how that rule works in a 3X3

it seemed more efficient to attach the corresponding identity matrix

ty so much for your help

oh one extra question

are there situations where you find the inverse, you multiply the inverse by the original matrix and you find that it is not equal to the identity matrix?

or does finding the inverse means that the step after will always prove you correct

[fringe fjord]

The definition of inverse is that you get the identity matrix when you multiply with the original matrix. If you don't then what you have found was not the inverse after all.

[hearty rapids]

got it

so that part where they multiply the inverse matrix by the original matrix to get the identity matrix is like a sanity check

[fringe fjord]

Yes.

[hearty rapids]

gotcha

[subtle gust]

question d

am i messing something.... i got [-30 \n 6 \n 21]

but the book says it is

what i did was B times [3 \n 6\n 0]

[dusky epoch]

,w {{6,-2,4},{0,1,3},{7,7,5}}*{{3,-2,7},{6,5,4},{0,4,9}}

[stoic python]

Results provided by WolframAlpha

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

Requested by Ann#0413

[dusky epoch]

can you show your work for that? maybe youve messed up some arithmetic somewhere

[subtle gust]

Sure thing

Just a sec

[subtle gust]

Nvm 🥲🥲

I found the mistake....

Messed up a sign

Any advice on how to avoid making such stupid mistakes on my test? 🥲

[dusky epoch]

double- and triple-check everything ig

[subtle gust]

Especially that i barely get enough time for all the questions...

[subtle gust]

Probably wouldn't have the time to do that on a test with a question about solving a 6d system of equations+ finding eigenvalues and eigen vectors+ finding determinants of a 3×3 matrix

[dusky epoch]

6d system of equations?

[subtle gust]

Ikr

Saw one of the past exams

6 vars

[subtle gust]

Wait isn't that what they call it 💀

4 equations in 6 variables*

[dusky epoch]

i wasnt objecting to the name

i was just in shock at what theyre making you do

its so pointless

[subtle gust]

agree tbh

if i could do 2 equations in 3 variables

i can probably do anything

there's just more room for stupid arithmetic mistakes

[wraith monolith]

Hi, guys, suppose T is an n by n invertible matrix, and x_1,...,x_n is a basis, then T(x_1),...,T(x_n) is a basis. So is it true that the change of basis matrix from {x_1,...,x_n} to{T(x_1),...,T(x_n)} is T?

[clear iron]

basis for which spaces? what does T map ?

@wraith monolith

[wraith monolith]

for example C^n, T\in GL_n(C)

[wide mortar]

heyyyy guyyss please im struggling whti this question i mean isn't normally i apply this formula to determine the orthogonal projection

why we need this innerproduct expression ?

[nocturne jewel]

cause that defines orthogonality

you can't determine if vectors are orthogonal w/o an IP present

[wide mortar]

@nocturne jewel yeahh but i m already using the dot product here as innerproduct

[wide mortar]

and the given another innerproduct where i can apply this expression ?

[wide mortar]

please someone could help me to solve this question

[nocturne jewel]

yeah you cant do that

cause you're not given the canonical IP

you use the IP provided

u and v are orthogonal iff <u,v>=0

[wide mortar]

thank you s much okay so i can't apply the dot product then how i can solve this question with the provided ip

[nocturne jewel]

Just apply the projection formula

You just need to make sure {e1,e3} is an orthogonal basis of the space.

[lean spoke]

hello

Im kinda brainfreeze you guys have any idea of how to do this

[wintry steppe]

try computing it

[nocturne jewel]

what is span{f,g} by definition?

[lean spoke]

span{cos^2(x), sin^2(x)} | from my understanding it is asking me to find out if the followin a,b,c... can be represented as the linear combinations of sin^2x cos^2x

[nocturne jewel]

Yes, hint think about trig identities

[lean spoke]

I found it out myself too, thanks btw When i shared this I wasnt able to understand what i read

[lime fulcrum]

The least square solution to A’Ax=A’b simplifies to xhat=A’b when A has orthonormal column vectors. Then bhat= AA’b=b. For linear regression this would imply that the estimated response equals the actual response. I think there is something wrong here. Does anybody what it is?. Thanks

[zinc timber]

it's because your A is not always (almost never in regression) orthonormal

and A'A may or may not be identity depending on the size of A, i.e. if size of A is 2x3 then shape of A'A = 3x3, and A'A = diag[I_2, 0] and not I_3

[lime fulcrum]

Thanks for responding. I think that’s where the issue

Is. I was almost treating A as orthogonal when it isn’t necessarily

[molten hill]

This is more of an applied problem, I have two linearly independent sets of vectors $A$ and $B$ in $\bZ^n$. I know that $\mathrm{span},B\subseteq\mathrm{span},A$. I want to find the canonical group decomposition/isomorphism class of the space $\mathrm{span},A/\mathrm{span},B$ (as in the quotient), I don't care what the generators are or any other specifics. Note $n$ can be really big, like in the hundreds. Is there a way that given these sets of vectors in a language like python that I can find the canonical isomorphism class of $\mathrm{span},A/\mathrm{span},B$? Thanks

[stoic python]

@molten hill

[opaque glen]

I have no idea if this is a linear algebra problem (I know barely anything about Lin algebra) but here we go

So

Let’s define M is a 2x2 real matrix with a nonzero determinant,

and set S(r) to be a 2-ball of radius r

[zinc timber]

Z is not a field so Z^n is not a vector space

[opaque glen]

Let’s define a function K(M,r) which does the following

Let’s say K finds the minimum of a set R(M,r)

[zinc timber]

ahem

[opaque glen]

Which basically for every vector in the 2-ball of radius r, the function finds the magnitude of the vector transformed by M, and then divides it by the vector’s original magnitude

Is there a way to essentially find the minimum value here

[molten hill]

If you're responding to me I'm treating Z^n as a module and referring to its elements as vectors (erroneously, I know)

[lavish jewel]

this is the same as the induced 2-norm, is it not, mizalign?

[molten hill]

Although I really only care about its group structure

[opaque glen]

The what

[zinc timber]

yeah, it's a different guy who was typing, thought it was you ignoring what I just said

[lavish jewel]

operator norm or induced 2-norm

[opaque glen]

Google time

[opaque glen]

I’ve never heard of this

[opaque glen]

That’s the supremum

I’m effectively asking for the infimum

though the supremum is also useful

[lavish jewel]

yeah, just replace that with the infimum

[opaque glen]

I’m asking how the fuck I’d find it

[lavish jewel]

your google result should have shown you that the supremum is given by the largest singular value of your matrix

and the infimum, by the smallest

[opaque glen]

I’ve never heard of these terms

[zinc timber]

infimum will be zero?

[lavish jewel]

you can also look up rayleigh quotient

[opaque glen]

I want it to be greater than 0

[lavish jewel]

it would be, ryu, but these exclude the 0 vec

[zinc timber]

like unless you also enforce |v| = 1, then you might get some interesting result like smallest singular value

[lavish jewel]

so only if the matrix is rank def

[zinc timber]

even if it does, like saying v neq 0, then taking inf will again give 0

[lavish jewel]

they did say they want | | Ax || / | | x | |

[opaque glen]

Use a backslash before typing things

*like this*

[lavish jewel]

it's 5 am, gimme a break lol

[opaque glen]

aka determinant is nonzero

[zinc timber]

hm ig not, yeah

no idea what I was thinking

[lavish jewel]

rank deficient, i mean. if it is rank deficient, the mat will have det 0

[opaque glen]

Once again I know only the very basics of linear alg

[lavish jewel]

i see.

[opaque glen]

But idk how I’d find this minimum from a given matrix

and radius

Having both the inf and sup gives me bounds

(Even better)

[zinc timber]

ok so Z is a PID, so these are noetherian modules and finitely generated (given n is finite) then you'll just get the Z^(rank(B)-rank(A))

[lavish jewel]

i suggest you read into the singular value decomposition, then

unless ryu knows another way

[zinc timber]

SVD is the best bet honestly

[opaque glen]

I essentially want to prove that it is nonzero

[zinc timber]

it's not practical to find the inf over a unc set

[molten hill]

No? Suppose I'm working in Z^1, I can set A = {1} and B = {2}. spanA = Z, spanB = 2Z. Z/2Z is not equal to Z^(rank(A)-rank(B))=Z^0.

[opaque glen]

For not rank def matrices

[zinc timber]

say if that is zero means |Ax| = 0 => x is in null space but x is not zero implies that A is singular

[opaque glen]

The whole thing behind this is

I got bored and wondered if

Essentially

[zinc timber]

hmm right, you might need to find the primary factors to get the answer

I haven't done much modules so ask in #groups-rings-fields

[opaque glen]

You define a square lattice from <-N,-N> to <N,N> excluding the zero vector

And basically sum over 1/||Mv||^2k where v are vectors over this lattice and k is some natural number >2

and basically use the circumradius sqrt(2)N

to take the limit as N approaches infinity to essentially and see if it’s convergent

I want to use the possibility that there exists some infimum

Denoted h I guess

[lavish jewel]

since you have ^2k, you can group (||Mv||^2)^N and note that this is the same as having v^TWv, where W is a symmetric positive semidefinite matrix

[zinc timber]

anyway this is a very general question, like you are essentially asking the quotient of every abelian group with a abelian subgroup, which you know is a very broad case

[opaque glen]

That makes 1/||Mv||^2k >= 1/||v||^2k * 1/h^2k

[lavish jewel]

all you have to do is show W = M^TM is full rank

[molten hill]

yes pretty much lol, I'll ask there

[opaque glen]

Do you mean k instead of N

[lavish jewel]

yes, oops

[opaque glen]

“Symmetric positive semidefinite matrix”

I need to take linear alg. I do have a linear alg book I want to read but I’ve been busy

[lavish jewel]

also, ||Mv||^2 / ||v||^2 is what's called a "Rayleigh quotient"

[opaque glen]

OH

[lavish jewel]

for which the min and max values are given by the smallest and largest singular values of M

[opaque glen]

I thought it would be funny to define a sort of “matrix zeta”

This is just to prove convergence

[zinc timber]

"operator theory"

[opaque glen]

Bruh I’m a 17 year old high school student who just reads about math sometimes

Many theories to collect

[lavish jewel]

sorry about that, chrome made my computer crash entirely

but yes, you'll need to read up on the singular value decomp

[opaque glen]

Wait all I need to prove is that it’s greater than 0

Absolute value is a metric so it’s always positive, and it’s a quotient

Therefore it’s only 0 if either it’s numerator is 0 (aka the transformed radius is 0, which is impossible in this case unless the vector is 0 which is excluded)

or

[lavish jewel]

and we're also taking v nonzero. this just leaves M having a trivial kernel

[opaque glen]

If the denominator explodes to infinity

[opaque glen]

Kernel is the set of vectors that get mapped to 0 right?

[lavish jewel]

yep

[opaque glen]

Ah

Problem is

I want to essentially define a limit here

As r -> infinity

and the denominator blows up but

So does the numerator

[lavish jewel]

what happens depends on the matrix

[zinc timber]

$\text{your_norm}(A) = \norm{A^{-1}}_2$

[opaque glen]

Y e a h

[opaque glen]

True

[zinc timber]

(probably)

[opaque glen]

Just need to find out in which cases does it converge to a number not equal to 0

[zinc timber]

should be 1/||A'||

[opaque glen]

I only did eigendecomposition like once to find the exponential of any n^2 real matrix

[lavish jewel]

you're gonna get a geometric series

[opaque glen]

What

k is an input, not an index

[zinc timber]

on the note, there's a pretty neat formula for finding $f(J_k(\lambda))$ where $f$ is analytic and J is the jordan block of order k with ev lambda

[lavish jewel]

i'd have to see the whole thing then, it's too early for me to keep track of all the bits and pieces you've given

[zinc timber]

it's handy

[opaque glen]

I posted it in discussion

Yesterday I think

But basically

For some non rank degenerative 2x2 real matrix M

I want to know the infimum of ||Mv||/||v|| if v is any R^2 vector if it exists

Not sure if said non-zero infimum exists for all R^2 vectors

[zinc timber]

already told you, the smallest singular value of M

[opaque glen]

What exactly is a singular value

[zinc timber]

that you have to learn for yourself

[lavish jewel]

ok, the square root of the eigenvalues of M^TM

that'll suffice, since you said you know eigendecomp

[opaque glen]

Ah k

[lavish jewel]

the answer would be sqrt(smallest eig val of M^TM)

so you need to show that is nonzero for your M

but honestly it doesn't seem you have the needed linalg knowledge yet

[opaque glen]

True

Actually

I wonder if there is some sort of relationship between the Eigenvalues of a matrix’s transpose

i think the eigenvalue formula through the quadratic formula is

((a+d) + ((a-d)^2+4bc)^1/2)/2 I think but the 1/2 refers to both branches because I’m too fucked to get the plus-minus sign

it’s down to just rearranging the vars

Wait I’m fucking stupid

just b and c swap

And it’s commutative

[lavish jewel]

at this point i'd just tell you to pick up a book

[opaque glen]

I’ll start reading it next weekend

Thank you guys for the help

what is the importance of a transpose?

Like I get it flips it but

It’s distributive-commutativity (just what I’m gonna call it because idk what else to call it) seems rather important

I’m going to make a big fucking assumption here and assume that (due to same dim square matrices) determinants are distributive over matrix multiplication, so are Eigenvalues (if you “define” a multivalued eigenvalue operator I guess)

If matrix K = M^T * M

The Eigenvalues are just the Eigenvalues of M squared

And the transpose of K is just M^T * M^T ^ T (which is M) which is itself K

I’m going to stop babbling to myself here and start jotting down notes of this stuff

WAIT A SECOND

Okay now it makes sense lol

[zinc copper]

my linear algebra teacher defined a minimal polynomial of a morphism $\phi : K[X] \to A$ to be monic polynomial that generates $\ker\phi$ as an ideal. When you look at the definition of a minimal polynomial of an $n\times n$ matrix $M$ over $K$, it is the monic polynomial of least degree $P\in K[X]$ such that its evaluation at $M$ in the algebra $M_n(K)$ is $0$.

[stoic python]

𝓛ittle ℕarwhal ✓

[zinc copper]

So to reconcile these two definitions, would you use the universal property of polynomial rings as such:

The $n\times n$ matrix $M$ represents an endomorphism $\phi$ from $K^n \to K^n$. Since $\operatorname{End}(K^n)$ contains $K$ by mapping $K$ to the dilations, we induce a morphism $\psi : K[X] \to \operatorname{End}(K^n)$ that fixes $K$ and sends $X$ to $M$. Then the minimal polynomial of $M$ is just the minimal polynomial of $\psi$?

[stoic python]

𝓛ittle ℕarwhal ✓

[exotic horizon]

do the eigenvalues of a matrix give us any information on its determinant ?

[lavish jewel]

the product of the eigenvalues equals the determinant

[exotic horizon]

ohh thank you

[haughty berry]

the product of the eigenvalues gives +- the determinant no?

[lavish jewel]

no +-

[haughty berry]

But the last coefficient in the characteristic polynomial of A (of size nxn) is det(-A)=(-1)^n det(A), and that’s equal to the product of the eigenvalues, no?

[lavish jewel]

what?

[haughty berry]

Lol sorry.
So if A is an nxn matrix,
The last coefficient in the characteristic polynomial of a matrix is going to be (-1)^n * det(A).
The last coefficient of a polynomial is also the product of all the roots of the polynomial. So (-1)^n * det(A) is the product of all the roots of the polynomial.

[lavish jewel]

there's a sign difference there depending on how you define the char poly

are you looking at lambda I - A or A - lambda I

[haughty berry]

But does it matter? Because you’re still only dealing with the product of eigenvalues

[lavish jewel]

this is the one i'm referring to

the result differs by precisely (-1)^n depending on what you call the char poly

since det(cA) = c^n det(A)

[haughty berry]

I’m confused. How did they get rid of the (-1)^n

[lavish jewel]

they multiplied it into each (lambda - lambda_i)

[haughty berry]

O yeah whoops

[lavish jewel]

i'm guessing you were using lambda I - A in your char poly

[haughty berry]

Right, but why would that change anything?

[lavish jewel]

because of this

it flips the sign of the eigenvalues

[haughty berry]

Oh yeah doi

My bad

[haughty berry]

I got that I just got mixed up, I was multiplying by the negative eigenvalues ((-lambda1)…(-lambda_n)), not the positive ones

My bad

[lavish jewel]

no prob. i think books that go very in depth into the polys use the other def anyway

(i wouldn't know)

[haughty berry]

Eh it seems that both ways have their benefits

[zinc timber]

my writing tab is finally here

[haughty berry]

[lavish jewel]

oh sweet

[dapper gorge]

Is there a standard notation for writing the span of a set but without including the zero vector? I was thinking maybe $\textup{Span}(S)^*$ works

[stoic python]

Croqueta

[dusky epoch]

$\mathrm{Span}(S) \setminus {0}$

[stoic python]

Ann

[dusky epoch]

wait so hold on

i'm not sure i understand what you're going for

[stoic python]

Croqueta

[dapper gorge]

Just realized what I said doesn't make any sense, ignore me

[zinc timber]

literally the complement subspace of span(S)?

nah nvm

[plush solstice]

I'm after some help involving y=mx+b
I have a webcam detecting my face and it's tilting degrees, and I am trying to determine which half of the face certain landmarks fall on, but if the face is tilted, that just confuses me further. I know it's cartesian, but this subject eludes me. #constraints
It's like rotating the entire graph
(Then i have to convert it to python lol)

[zinc timber]

maybe share an example of what you are talking about, it's too abstract

[plush solstice]

k sec

just wanna mark each point as on the center-line, to the right of it, or to the left. But when the face is tilted, my brain just melts trying to mathematically compensate for it, drawing a blank.

[hearty rapids]

https://youtu.be/Q4pgK_n_5Ik

um i am so confused rn

oh nvm he never said explicitly that B is the inverse of A

[zinc timber]

so you want to know which side of line a point of interest lies?

[hearty rapids]

oh wow my brain is shot

2(2) + 3(-1) is not a -1

it is a 1

[zinc timber]

do you have the eq of the line? maybe you can approximate?

[plush solstice]

for all points in a loop, yes. there are points that are 0 on the X axis, though, so they wont return left OR right.

[zinc timber]

you can calculate (y-mx-c) for each of your point then check if it's +ve or -ve

[plush solstice]

well here's the thing

[hearty rapids]

if i can say B = A^-1 can i also say that A^-1 = B?

[plush solstice]

the x and y axises ( i cant plural rn) rotate with the face, but the paired coords for each point are gridded along the canvas pixels.

so rotation is a huge obstacle

make sense?

[zinc timber]

honestly, no

[plush solstice]

your face is the grid, but the points' (x,y) are translated along the canvas grid instead of your now-rotated graph (face)

[zinc timber]

even if the canvas rotates you still have the line right?

at least 2 point on the line is enough

[plush solstice]

they're moving with my face

and the centerline is moving

from which to determine

i mean

[zinc timber]

nope doesn't make sense

[plush solstice]

I can easily get the angle of a vertical line drawn on my face

right down the center

I was thinking that line was the Y axis for the dots

ugh my brain hurts

hey das

[hearty rapids]

like could i put a matrix and then on the exponent put another matrix

it's a dumb question

matrix raised to a matrix

[plush solstice]

they were talkin about matrixes and tensors and shiz either in genchat or maybe it was the python disco i saw it in

[hearty rapids]

hey jok3r

[plush solstice]

so to simply my question, how do you determine what side of a line a point is, even if that line is on any angle?

[hearty rapids]

bro what is this

[plush solstice]

its face detection plus lines plus dots plus RDJ

[sick sandal]

yes you can

check matrix exponential*

[plush solstice]

THIS
https://www.youtube.com/watch?v=0TD9EQcheZM

I want to list if each point is in, on, or outside the "region" (constraint)

but when i rotate the face, it's like rotating the graph the points are drawn on.
Also the screen grid starts at 0,0 being top left, not centered, which is (I think) the big problem I'm trying to solve, along with remembering constraints lol

[hearty rapids]

woah determinants are so cool

sort of like a modification of a graph algorithm

you recommend anything for linear programming?

any particular youtuber?

[plush solstice]

throw the term into youtube just like you do google

"YouTube it"

[hearty rapids]

gotcha i'll find something

man orgo chem tutor is a god

[plush solstice]

the vid i shared is conveniently named Linear Programming xD

[hearty rapids]

i had a feeling he had a vid for it

i'm actually using his vid for determinants too

[plush solstice]

nice

[hearty rapids]

bc strang is uh

um

confusing

at least to me he is

[exotic horizon]

if matrix A*B is invertible, does that mean that matrix A and matrix B are also invertible ?

[wintry steppe]

A and B don't even have to be square (in which case they of course aren't invertible), but if you further assume that they are, then it is true

[hearty rapids]

are they asking abt this concept

https://youtu.be/OVJpEYMm7o0

[wintry steppe]

no

[hearty rapids]

oh

whoops

[wintry steppe]

they're asking about the converse question: "if AB is invertible, then are A and B invertible?"

[hearty rapids]

i see

[wintry steppe]

if A and B are square, det(A)det(B) = det(AB) is non-zero since AB is invertible, so det A and det B are non-zero, qed

[hearty rapids]

i was going to attempt the (AB)^-1 proof but a source said that the determinant had to be larger than zero

and i didn't know what a determinant was

so i'm practicing that

[wintry steppe]

if A and B are not square, it's false, and a counterexample is easy to find

[hearty rapids]

is it just me or does linear algebra feel tedious

and it's very easy to make like simple small math mistakes

even with just row operations

[exotic horizon]

oh i see thank you

[exotic horizon]

i always make mistakes like this...

[hearty rapids]

or i just write one element wrong in the matrix and i catch it like 5 steps later and i'm like whoops

[sick sandal]

it doesn't have to be
you can find more consistent methods

[fringe fjord]

If you're doing endless exercises of "solve this linear system", then it can very well feel tedious.

One thing to keep in mind is that you could also solve the same system by writing out the equations in full for each row, with variables and coefficients everywhere like you'd do in high-school algebra, and that would be no less tedious and error-prone.
The systematization of linear algebra really makes such things easier to solve, problem for problem, but that also means that it's now feasible to attack larger problems, so you might not feel the progress when you're continually exercising right at the edge of your capability.

[hearty rapids]

yeah that's what i quickly realized

once you know how to turn a system of linear equations into an augmented matrix and get it into reduced row echelon form or row echelon form enough times

doing it excessively just leads to burnout

good that i don't do it excessively

i use pomodoro lucky for me it works

[quartz karma]

I'm working on a computational linear algebra project and stumbled upon this quote:

"we can consider an equivalent condition for the uniqueness of the solution as m>=n and the square
block matrix formed by the first n rows and the first n columns of the reduced echelon form of
[A b] has a special form (condition (4)) – a form has to be identified based on the fact that A
has a pivot position in every column."

What would this form look like? I've been trying to figure it out and nothing is coming to mind

[proud gate]

Can someone point me toward a proof of this statement?

[round granite]

every vector in V can be written as a linear combination of the basis vectors, so write it as v = a_1e_1 + a_2e_2 + .. + a_ne_n

then use linearity and simplify f(v)

alternatively you can set 2 functions st they have the same image on a basis of V and then show that their difference is identically 0

[wintry steppe]

Hey! Is there a trick to solve this quickly using Gauss Jordan Elimination method? It's an example in the textbook and the first thing that they did is to divide row 1 by 1/3 to get 1 in the far left. Is it that we should aim to get the 1 along the diagonal first and then try to get the 0's after?

[mortal isle]

Is the way I’m presenting this correct?

[wintry steppe]

Thank you so much

[halcyon spindle]

Basically row reduction gets you row echelon form of the matrix, so you can easily do a back substitution. If desired you can go further to get it to reduced row echelon form, so you can read the solution from the augmented matrix.

Reference linear algebra done wrong, it free online.

[wintry steppe]

Okay :) thank you very much.
Much appreciated 🙏

[wintry steppe]

Gonna check it out

[twilit flower]

Hey guys, does anyone have the pdf Linear Algebra for Everyone by Gilbert Strang?

[proud gate]

I suppose the same proof work on any dimension right? Even infinite, assuming it is Hamel basis?

[slow scroll]

Yes. It’s essentially the same. As soon as you know how to define a function on a basis, you can extend to the span of that basis by taking linear combinations. It’s well defined and unique since a basis is linearly independent

@proud gate

[proud gate]

Just curious, is this also a result following the Hahn Banach?

I feel like it is not if we can't guarantee g is linear

[proper cradle]

True or false?

[wintry steppe]

if you had a non-zero proper subspace like that then you'd be able to find a non trivial factor of the characteristic polynomial

ponder that

[proper cradle]

So eigen space is proper subspace for this transformation?

[vestal spire]

I think proper here just means not the entire space, otherwise its trivially true

[wintry steppe]

that's what proper means, yes

[proper cradle]

This is definition of invariant subspace right?

[wintry steppe]

yes

[zinc timber]

yes

[proper cradle]

Is eigen space invariant ?

[wintry steppe]

eigenspaces are invariant, yes.

[zinc timber]

what do you think?

[proper cradle]

So in this case eigen space will be improper or just zero space?

[proper cradle]

How to prove or disprove?

[slow scroll]

are you asking if this result requires hahn banach to prove?

[proud gate]

Yes

I mean, is there a proof where we can simply apply HB

The constructive proof doesn't seem to require HB

[slow scroll]

Of course if ur vector space is not a banach space, there is no proof using Hahn Banach. If you have a Banach space then possibly in some very roundabout way. Its not super clear because in the assumption of hahn banach, you start with a linear functional defined on a subspace, but in this you start with a set map on a basis

its an interesting question though

because I don't think the proof of hahn banach uses that every v.s. has a basis (which is equivalent to the question you asked), and both results rely on the axiom of choice in some way

[proud gate]

That is fair, though I don't recall HB require Banach space to be true. As long as we have a linear space and a control Minkowski functional, we are good. But there's a couple of ways of stating HB..

[proud gate]

Ye..

[slow scroll]

actually im wrong, the statement that every v.s. has a basis is not equivalent to the question you asked. But if you modified it slightly to say "there exist a subset X of a v.s. V such that every set map g : X --> W induces a unique linear map f : V --> W which restricts to g" then it would be.

wait is the extension we get even unique in hahn banach?

[proud gate]

Hmm not necessarily I think.

Probably with some extra regularities on the v.s.

convexity maybe

Yeah if we are really going to us HB, the proof is going to be unnecessarily indirect

[zinc timber]

all eigen values of f(0) are 0 and for f(1) is 1, so for them to morph continuously their eigen values must also morph continuously, but A^2=A forces eigen values to be 0 or 1

[proper cradle]

but co-domain is not compact right

[zinc timber]

what's co-domain?

[proper cradle]

set of matrices of order 2 in reals which is idempotent

[zinc timber]

ok but that doesn't have to be compact (idk if it is or not)

the thing you are hinting is probably that continuous image of compact set is compact, but that has nothing to do with co-domain, rather the image

no it won't help

[proper cradle]

yep got it, then how to disprove it?

[zinc timber]

isn't this enough?

[proper cradle]

i didn't understand this one

g(x)= {0 when f(0), 1 when f(1). This is not continues that's why?

[zinc timber]

eigen values of f(0) and f(1)

[proper cradle]

ok

[orchid delta]

How to find the missing value in a matrix so that the operator on R^2 is self adjoint?

is there any generalised way?

[torpid harbor]

https://media.discordapp.net/attachments/268882317391429632/935079364293447690/20220124_132103.jpg

[zinc timber]

it has to be symmetric so yes

given you are given enough values

[zinc timber]

what is your question?

[wintry steppe]

How does one represent a 2x1 vector that contains complex numbers i.e $u = [\frac{a+bi}{c+di}]$

[stoic python]

SuckyDucky

[lavish jewel]

wdym by represent? plot them?

[wintry steppe]

Yes

[lavish jewel]

you don't

[wintry steppe]

Why?

[lavish jewel]

it requires 4 dimensions

[wintry steppe]

Is it because complex numbers themselves are vectors?

[lavish jewel]

they behave like vectors in some sense, yes

[wintry steppe]

that can be represented by 2d vectors?

[lavish jewel]

mhm

[wintry steppe]

Then what is hte difference between complex numbers and 2d vectors?

[lavish jewel]

complex numbers have additional structure

[wintry steppe]

Are they just a "subset"?

They do?

[lavish jewel]

on their own, 2d vectors have no multiplication operation

complex numbers do

[wintry steppe]

but nor does a complex?

[lavish jewel]

you can multiply 2 complex numbers

[wintry steppe]

oohhh

you multiply by foil

ohh

[lavish jewel]

there is no standard f: R^2 x R^2 -> R^2 that we call "multiplying two vectors in R^2"

and the definition of a vector space doesn't require it

so complex numbers have additional structure

[wintry steppe]

ah

Thank you so much

[eager geyser]

Okay if we have an Ax=0 and we reduce it to this problem above, I've noticed that it implies the 3rd column of the original A is 2 x 1st + 3 x 2nd. Why is this true? I'm also trying to convince myself that the lack of pivot position means it's not an independent column.

[hearty rapids]

wow transpose is so cool

i can go from a 2 X 3 matrix and flip the dimensions making it a 3 X 2 matrix

[wintry steppe]

ikr

I just started learning and it solved one of my errors in my neural network

np.array(matrix).T in python

[haughty berry]

The transpose is… cool? That’s a first

[zinc timber]

welp

probably this should be R but not sure

nvm P is false

probably Q false as well

nvm answered

[hearty rapids]

excellent

[wintry steppe]

👍

[dim magnet]

Let C be a linear code given by a generator matrix G ∈ Mat(k ×n, F_2) and let A∈GL(k, F_2) be an invertible k×k matrix, show that AG is also a generator matrix of C

can somebody help me with this

[zinc timber]

coding theory?

[wintry steppe]

hey all how would i be able to do these?

[outer goblet]

can someone check if my anwser to 21 is sufficient

click open original to zoom in on text

i dont know how to explain it better than just saying that

a set of all real numbers, has to contain whatever number the set converges to

[dim magnet]

uhm yeah ...... im having a difficult time finding anybody in that cs-math branch i hope i find somebody here

[zinc timber]

can if you can briefly explain what "generates" here is then maybe someone will be able to help

[clear iron]

Hey, can anyone help me understand ?
consider the following statement,
For all Vector spaces V,W,
for any Linearly Independent set {v1,...,vk} contained in V, and set{w1,.....,wk} contained in W , there exists such linear transformation T: V-->W such that
T(v_i)=w_i
I am convinced this is wrong , and yet the answer is true

[zinc timber]

not readable to me

[outer goblet]

you can click on origianl and zoom in

wait let me scan

[zinc timber]

it is true, T(v_i)=w_i will give you the desired LT

[zinc timber]

no it's not enough

[outer goblet]

bruh

i hate proofs

[zinc timber]

you are supposed to show that ${x_n} + c{y_n} \in V$ that means you must show addition and multiplication of elements of V are also in V

[zinc timber]

which is a real analysis exercise

[dim magnet]

so the rows of a generator matrix form a basis for linear code and the linear code is the row space of the generator matrix

[outer goblet]

yeah but im not supposed to do question 20 lol, just 21, so i think im allowed to assume that stuff in qst 20 is proven

[zinc timber]

you have to show that {x_n+cy_n} converges

which you didn't show

[outer goblet]

hm

[clear iron]

but a linear transformation cannot be defined in such manner for EVERY spaces
consider V,W such that dimW<dimV
the set {w1,....,wk} will be linearly dependent by definition,
so there are at most dimW vectors in the set that are linearly independent, lets call them{w1,.....,w_i}
I can map T(v1)=T(w1) and so on till T(vi)=T(wi)
however if I map T(vi+1)=T(wi+1) it will mean that vi+1 is linearly dependant on the first i vectors

[outer goblet]

but how would i do it

why wouldnt it converge lol

[clear iron]

since the set {w1,....,wk} is not necessarily linearly independet.

[zinc timber]

you already wrote that the set has same length

and they are LI

so..

[clear iron]

No it wasn't stated

wait

I mea j

the sets are the same length, W's set isn't linearly independent

[zinc timber]

{u1, u2..} is LI

[clear iron]

it doesn't necessarily mean that W

[zinc timber]

so they form a basis? and their image determines where the other vectors should be mapped?

[clear iron]

say W is R3 and V is R4

then i can take V's basis, and any 4 vectors from W

[zinc timber]

is k>n?

like is this matrix full rank?

[zinc timber]

THE SETS YOU HAVE WRITTEN HAVE SAME LENGTH, IF THEY DON'T THEN THE STATEMENT IN GENERAL IS FALSE

[outer goblet]

can you help a bro out pls :(

[zinc timber]

Linearly Independent set {v1,...,vk} contained in V, and set{w1,.....,wk} contained in W
both end at k, which means both have length k

[zinc timber]

not saying it will not converge, it will and it's a real analysis exercise

[dim magnet]

yeah but idk if the matrix is full rank

[zinc timber]

if you have already proved that $\lim x_n = x$ and $\lim y_n = y$ then $\lim (x_n + y_n) = x+y$ and $\lim cx_n = cx$ then it shouldn't be any problem for you

[outer goblet]

so i have to say that the sum of two converging numbers will converge, and that the scalar multiplication of a converging number will converge

[zinc timber]

wrong message tag

[dim magnet]

ah ok

[outer goblet]

real analysis???? bruh the class is called linear algebra 2

[zinc timber]

ok so I don't get this, say $G = \m{ 1 & 0 \ 1 & 0 }$ and $A = \m{0 & 1 \ 1 & 0}$ then $AG = \m{ 1 & 0 \ 1 & 0}$ so clearly the row spaces of $G$ and $A$ are not equal @dim magnet

[zinc timber]

wrong mat mul wait

[outer goblet]

😔

[zinc timber]

yeah that's real analysis, not LA2

[outer goblet]

bruh, not only are the lectures not even realted to the homework, and now youre telling me its not even the correct subjects wth???

smh my uni is a joke

[dim magnet]

$AG = \m{ 0 & 1 \ 0 & 1}$

[stoic python]

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

[dim magnet]

wtf i suck at latex

[clear iron]

wait I took a look at your questions

honestly we had the same type here

and that's in Linear Algebra 1 lol

[zinc timber]

@dim magnet ok so this ans may not satisfy you but

the code $C = {G^Tx| x\in F^k}$, now $A \in GL_k(F)$ and so code generated by $AG$ is $(AG)^Tx = G^TA^Tx = G^T (A^Tx) = G^Ty$ so the code are same

[zinc timber]

since A is invetible, x for all and Ay for all y is F_x^k so they are same

[clear iron]

I understand that
but it still baffles me... just because they have the same length doesn't mean both these sets have LI, because I have no information about the dimension of the subspaces. can I write up an example of what I mean?

[zinc timber]

I'm not saying w set is LI

only requirement is v set is LI

(thought it's not necessary)

[zinc timber]

showing the sum and prod of convergent series is again convergent is a thing you learn in real analysis. You can still use the result to conclude your V is a subspace but, also why are they teaching you LA2 before teaching ANALYSIS1?

[outer goblet]

Im a physics student lol, i think they just let me assume that part or smth

[zinc timber]

ok then yeah

proof by assumption

[clear iron]

smh

you just map it one by one

[outer goblet]

sin x = x

[clear iron]

[zinc timber]

did you just realize where you went wrong?

[clear iron]

yes

[zinc timber]

happens

[clear iron]

I assumed you can't have T(vi)=T(vi+1) or some dumb shit like that

[hearty rapids]

so for linear programming and markov matrices is in #10 do you need to know everything before it

[zinc timber]

maybe not all of it but yeah pretty much

[hearty rapids]

oh dear

the kids in my mth 040 course

i pray for them

[zinc timber]

LPP doesn't require eigen values shits, only row elimination is enough

[hearty rapids]

i see

LPP?

[zinc timber]

but MC requires all

Linear Programming

[hearty rapids]

oh ok

yeah i'm not worried i've been looking at this stuff over break

i'm interested in trying to apply this stuff to ml

so i wanted to learn it anyways

[zinc timber]

[hearty rapids]

i know ordinary least squares is how linear regression works

so the linear algebra behind it should be interesting

this server is so cool

woah

"Description:
Fall, Spring
Matrix Algebra, systems of linear equations, linear programming, Markov processes, and game theory. Applications to business and the biological and social sciences are included."

so the prof will be applying linear algebra to game theory

interesting

i think i am in a much better spot than other students so that's good

[wintry steppe]

Guys... i passed the exam... thank you all

[dapper gorge]

An improper rotation in R^2 is just a reflection, right?

[dapper gorge]

Congratulations!

[wintry steppe]

Well a reflection can be a rotation. If you do it in a square for example and you rotate it on one of his side, you can define a rotation in R^3 in R^2 yes. I think that is just a reflection

[wintry steppe]

Thank you

[hazy haven]

is there a necessary condition so that G is a subspace of H if F+G is a subspace of F+H with F, G and H being subspaces of E a vector space

[fringe fjord]

Um, where would the necessary condition fit into that statement? And how are F, G, H quantified?

[lucid glacier]

I seem to be having a hard time coming up with exercises. Does someone have some maybe less standard, constructive exercises regarding eigenvalues and eigenvectors?

(Non-computational)

[nocturne jewel]

Prove Cayley-Hamilton

[lucid glacier]

I could do cayley hamilton for diagonalisable operators tbh

that might be good

wait

idt they got to char poly

[nocturne jewel]

if $A^3=A$, what are the possible eigenvalues?

[stoic python]

Mosh

[nocturne jewel]

Just go through your textbook for problems tbh

[lucid glacier]

the class isn't using a textbook

i've already scoured lang, artin and axler for problems

[wintry steppe]

Just did lots of course questions on dot notation with some proof questions

I now know what I hate the most in maths

PROOFS

[dapper gorge]

Then you hate maths lol

@lucid glacier not too difficult, but fun

Maybe you can check Putnam problems, there are many for linear algebra

[lucid glacier]

that might be a good idea. Don't want to kill them tho

thanks for the suggestion!

[dapper gorge]

also, maybe you can check Evan Chen's napkin, it is not meant to be a rigorous textbook obviously, but the exercises are somewhat original (the ones form linear algebra at least aren't anything like the textbooks I have read, and the text is great imo, although not as a primary source, obviously)

[wintry steppe]

@lucid glacier you might be able to find some problems at the past course website when i took LA2, http://www.math.toronto.edu/payman/mat247/main.html

assignments with full solutions (at least for the main problems) and tons of problems

friedberg's book may also have some stuff

[dapper gorge]

@lucid glacier This is something it came to my mind: (prove or disprove) let $A\in M(n,\mathbf{R})$ be such that it's rows are in arithmetic progression. Then the rows of $A^k$ are also in arithmetic progression for all $k\in\mathbf{N}$.

[stoic python]

Croqueta

[dapper gorge]

So, I dont know if the answer is trivial or not (haven't thought about it), but fun nontheless I'd say

I hope it's true

Friedberg book is mostly computational or inmediate stuff from my experience. I'd just say that when you encounter a theorem or a proposition, don't read the proof, first try to prove it on your own. And even if you have read the proof, sometimes it is fruitful to try other approaches

[wintry steppe]

computations are good

[dapper gorge]

but he asked specifically for non computational problems

[wintry steppe]

ok

i didnt scroll up

[lucid glacier]

thanks!

that sounds interesting

btw this isn't for me lol, i'm giving a supplementary course in linear algebra

i'm just looking for novel exercises that my students have probably not seen before

though I definitely agree with your advice

[dapper gorge]

and if it is generally false, say for which sizes of A it is false (because it is trivially true for n=2 lmao)

oh ok

[lucid glacier]

you could try and start by characterising the JCF of such a matrix perhaps, but I don't think you'd have a very strong characterisation

might be wrng

wait nvm this is over R so JCF doesn't always exist

[dapper gorge]

ye

Would you say is true btw?

[wintry steppe]

the characteristic polynomial can be factored into linear terms and irreducible quadratics, so maybe you can put it in a nice rcf and work with that?

[wintry steppe]

troll them by putting a bunch of computations on the exam

if you're giving a mainly theoretical course

[dapper gorge]

no pls

[lucid glacier]

thankfully i'm not writing the exam

but they were warned there might be computations

[wintry steppe]

alternatively, maybe you can argue there's no loss in working over C instead, where you have jcf. don't want to think about it though

R

[dapper gorge]

what problem are you referring to?

[wintry steppe]

the one you posted, probably

idk i didnt read too carefully

[dapper gorge]

oh ok

[wintry steppe]

mostly responding to shin's remarks on it

[dapper gorge]

I'll think about it one day tbh

would you say it's true though?

[wintry steppe]

i dont know

haven't given it any serious thought so i'm hesitant to say yes or no

[nocturne jewel]

UofT moment

[lucid glacier]

what

[wintry steppe]

it's an okay place

the weather kind of sucks in the winter

[nocturne jewel]

ik, just didn't know you went to UofT lol

[dapper gorge]

I did a little experimentation on wolframalpha once on this (just some minutes, and wolframalpha not good for this), and for 3x3 matrix arithmetic progresisons were preserved for the powers I tried, but not only that, the ratios of the different rows (I'm not sure how you call them, the d's in the progression a,a+d,a+2d,a+3d,...) exhibited nice relations

[wintry steppe]

unfortunately, i do, mosh

[zinc timber]

@lucid glacier http://univ.tifr.res.in/gs2022/Files/GS2020_QP_MTH.pdf
you can find some brilliant LA questions here

[dapper gorge]

it was trivial guys lmao

I was about to go to sleep but I decided to see if it was trivial

In fact, a much stronger statement is true

You don't need to write that many details, you can do it in one sentence. Apply product definition and that's it literally

[dapper gorge]

In fact, you don't even need A to have rows in arithmetic progression lmfao

[ripe shale]

Can someone explain to me the second equality. I don't get how M(Tv_k) equals M(T)., k

This is 3.64

I think the second step would imply M(Tv) = M(v)

[wintry steppe]

this is errata

https://linear.axler.net/LADRErrataThird.html

3.64 should say M(Tv_k) on the rhs

@ripe shale

[minor echo]

So linear algebra is basically

Going back and forth on a number line

[zinc timber]

[minor echo]

This server is so nice

I’m happy I found it

I can finnaly not fail my exams

Gotta study hard

[royal escarp]

unfortunately, i am not versed in linear algebra. ill let the experts take your request

[ripe shale]

Thanks!!!!!!

[ripe shale]

Gonna have to go back and take a look at all the previous errors

[ripe shale]

Side effect is that now whenever I don’t understand I just think it’s an error 😂

[round thicket]

is anyone available to help me with some intro to linear algebra hw

[wintry steppe]

if you have a question, just ask

someone might help

[round thicket]

i have solved the matrix into rref

and am stuck on what to do next

[lavish jewel]

where did the a go

[wintry steppe]

When can we use gradient descent method to solve a problem ?Like problem to minimize a equation?

[zinc timber]

when function is smooth enough (C^2) and you are willing to do it numerically

[wintry steppe]

can you elaborate a little more

This is the question

[zinc timber]

they are asking you to use conjugate grad method

[wintry steppe]

oh sorry

can someone help me with this problem

[lavish jewel]

that's not enough, ryu

if by "solve" they mean find a global minimum, the function has to be convex

strictly convex if you wan it to be unique

that makes the grad = 0 condition necessary and sufficient

[wintry steppe]

so if grad=0 problem can be solved by conjugate gradient method ?

[lavish jewel]

that is exactly the opposite of what i wrote 😛

[wintry steppe]

how do i find gradient ? and if its not 0 that means its not convex and therefore it can be solved by conjugate gradient method ?

[gray dust]

$\grad f=\br{\pdv{f}{x},\pdv{f}{y}}$

[stoic python]

RokabeJintaro

[wintry steppe]

this should be (0,0) ?

[lavish jewel]

convexity is related to the second derivatives

so you will need to check those

[wintry steppe]

oh

[strange eagle]

helpB(

[round thicket]

its in the first equations left of the equal sign

[dusky epoch]

do you know how to verify whether or not a particular number satisfies an inequality?

actually do you still need help with this?

oh and this probably doesn't belong in this channel but rather in #prealg-and-algebra

[lavish jewel]

yes, i'm asking where it went after RREF

[zinc timber]

if they mean global minima then yes,

[round thicket]

is someone able to help me find the linear combination of basic solutions?

I am able to get this far but get confused when doing this part

[low heart]

Can somone help

I don't understand this

or atleast gimme a link of some sort to help me out

im taking a pretest

[lavish jewel]

i think this is better suited for #prealg-and-algebra , but i think you are looking here for something like the "vertical line test". you can look that up and read about it

also khan academy probably has some good content on this

@low heart

[low heart]

@lavish jewel Alr thanks alot

[lavish jewel]

you can also benefit from checking out several definitions of what a "function" is

[round thicket]

@lavish jewel do you know how to do this?

[acoustic crescent]

Heyo

Im doing some high performance programming and i need to optimize a function for multiplying matrices together, in my book it says that multiplying a matrix A by the transpose of another matrix B is faster than just multiplying matrix A and B normally, why is this?

[lavish jewel]

can you give more context? maybe show a snippet from the book

[acoustic crescent]

@lavish jewel

[lavish jewel]

the only thing i can immediately think of is how memory is allocated

the rows of 2D arrays are often memory adjacent

it's not inherently an algorithmic thing, since usually matrix multiplication is O(mnp)

for matrices of size m x n and n x p

[acoustic crescent]

Yes it must be a memory thing in C , BUT at least on my computer its actually slightly slower than the naive matrix multiplication algorithm

[lavish jewel]

do you mean the built in matrix mult or another you coded yourself?

[acoustic crescent]

one i coded myself but using the well known naive one found on the wiki, its an excersize
cause obviously if i were to use a library it would be very pre-optimized

[lavish jewel]

mhm. so what they say is that instead of, say, 3 for loops where the last loop iterates over rows of A and columns of B, you instead create the arrays A and B^T, and the innermost loop iterates over the rows of both A and B^T. is that how you did it?

[acoustic crescent]

if you know C i can send a pic of the code, that may be easier than me trying to explain

but yes, instead of A and B its A and B^T

[lavish jewel]

last time i did any C was like 8 years ago, but you can try lol

[acoustic crescent]

Where B is the transposed matrix

[lavish jewel]

it might have to do with how you're indexing

since you use a single index instead of [][]

[acoustic crescent]

im storing it in 1D arrays, but yeah maybe when i grab the B[p * n +j] index i should use m instead of n

[lavish jewel]

aha then that's the problem, i think

there is basically no difference in that case

or it's slower because now you jump around in B

whereas actual 2D arrays have memory adjacent rows

here you're jumping around inside of B thanks to p*n

at least that would be my thinking

the goal in transposing B is that the index p is used directly in both A and B^T, instead of some multiple of it

so that you can go through all m memory adjacent positions before jumping to some other place in memory

that's my naive take at a glance

[acoustic crescent]

Does this algorithm work for non square matrices?

[lavish jewel]

yes

as long as they share that inner dimension

here A is n x m, and B is m x k

[acoustic crescent]

n = 500;
m = 200;
k= 500;

this wouldnt work for example

[lavish jewel]

that would work

if they are all different it would also work

that's just how matrix mult works

from wikipedia

[zinc timber]

the way I think of it is $m\times \overset{collapse}{\fbox{ n \quad n}} \times k = m\times k$

[acoustic crescent]

my mult function was wrong

[hearty rapids]

i see

so

lu decomposition is basically an extended form of gauss-jordan elimination

[hearty rapids]

organic chem tutor has an excellent vid on this

https://youtu.be/2spTnAiQg4M

[zinc timber]

yes kinda

you just keep track of the row operations using the L matrix

[hearty rapids]

sorry not gauss-jordan

just gauss

yo trev tutor is nice

i'm still confused by this tho

[lavish jewel]

what troubles you about it

[hearty rapids]

how do i disprove whether or not something is a vector space?

[lavish jewel]

you test its definition

[hearty rapids]

i have tons of axioms that they gave me

so in this example

i can set x and y to anything i want?

[lavish jewel]

you can, but you have to show the axiom holds for ALL x and y

[hearty rapids]

oh i see now

i must have that x*y = 0

[lavish jewel]

all x and y that satisfy the conditions of the set

[hearty rapids]

i see

[lavish jewel]

and the accompanying operations

[hearty rapids]

it's a mathematical proof

[proper cradle]

take u= (1,2) and another one is v= (-2,-1), u+v=(-1,1) and this is not in W but u and v in W.

[hearty rapids]

i understand now

thanks guys

i really feel like my general intuition for this stuff is improving

like i used to struggle w dot product and stuff

i find this stuff really cool

i feel like every day i am closer to applying this to ml

are you kidding me psuedoinverse is that simple?

[twilit flower]

Hey guys, does anyone have the pdf Linear Algebra for Everyone by Gilbert Strang?

[nocturne jewel]

Tried libgen?

[hearty rapids]

https://tenor.com/view/wait-thats-illegal-halo-meme-gif-14048618

jk

[dapper gorge]

If A is a linear operator on a vector space over C such that there exists an orthonormal basis consisting of eigenvectors of A, then any matrix representation of A with respect to any basis is unitarilly equivalent to a diagonal matrix?

[zinc timber]

what do you mean by unitarilly eqv?

like in similar or itself is diagonal?

if you mean itself diagonal then it's false

[dapper gorge]

I mean that there exists a unitary matrix $Q$ ($Q^*Q=I$) such that $P=Q^*AQ$

[stoic python]

Croqueta

[dapper gorge]

where P is diagonal

oh like in similar

for any matrix representation of A

[zinc timber]

oh oke

do you want answer or hint?

[dapper gorge]

I'm reviewing the change of coordinate stuff now

I believe it is not true, but that's why I'm confused

hint

I'm having trouble understanding this proposition specially from the point of view of A as a linear map, if what I said is false it is not making sense to me, as you have to make an arbitrary choice for a basis first. Maybe its true, anyway I'm reviewing the stuff

[zinc timber]

that's the spectral theorem

[dapper gorge]

it's part of it I suppose

[zinc timber]

ye

[zinc timber]

that's a good enough hint I would say

[dapper gorge]

the spectral theorem? I haven't read it yet

[zinc timber]

||try showing your operator is normal ||

[zinc timber]

oh so you can't use it?

[dapper gorge]

no

so

[zinc timber]

ye think of change of basis then

[dapper gorge]

I can prove if A is a normal operator, then there is an orthonormal basis consisting of eigenvectors of A, I have no problem with that

But I have a little bit of trouble understanding the counterpart for the matrix

yes

[zinc timber]

since your operator already have an orthonormal basis, change the basis to it and you'll have a diagonal

[dapper gorge]

yeah

[zinc timber]

if your operator is normal then it's matrix wrt any basis is also normal

normality is a property of the operator so remains invariant as you change the basis

my argument is too long

basically idea is you compose the unitary matrix in a way that conjugation with it gives you diagonal matrix

[twilit flower]

Yes, couldn't find

[dapper gorge]

So, take any basis $D$ and take $B$, the orthonormal basis consisting of eigenvectors of $A$. Denote by $A_D$ the matrix representation of $A$ with respect to $D$ and by $C$ the change of coordinate matrix. Then clearly $P=CA_DC^{-1}$ where $P$ is diagonal. So $C$ must be unitary? The text makes this claim, but $C$ is dependent on $D$ (which is just an arbitrary basis) so it is not that obvious to me. But this is saying that unitary equivalence is invariant under similarity? I'm reviewing some of the stuff of the text, I'll probably find an answer there.

[zinc timber]

actually wait

[stoic python]

Croqueta

[zinc timber]

wrt any basis is strong statement

because normality is not obvious

[dapper gorge]

maybe I didn't understand something and it is not wrt to an arbitrary basis ? idk

[zinc timber]

so the thing probably bothering you is that the adjoint P* is not the conjugate transpose of P wrt arbitrary basis

it's conjugate transpose iff underlying basis is orthonormal

try showing no matter what basis you choose $P^{-1}AP$ is again normal

[zinc timber]

so from spectral thm, it's unitarily equivalent to diagonal matrix

that is, say $X=P^{-1}AP$ then show $X^X = XX^$ where $A$ is diagonal

[dapper gorge]

yeah right just realized

So I think I solved my confusion. The correct statement is: suppose $A$ is a NORMAL linear operator on a (finite) complex inner product space $V$. Let $\beta$ be any orthonormal basis for $V$ and let $A_\beta$ be the matrix representation of $A$ with respect to $\beta$. Then, there exists a unitary matrix $Q$ and a diagonal matrix $P$ such that $Q^*A_\beta Q=P$

The confusion I had was with the identification of the operator A with a matrix, which must be done with an orthonormal basis I suppose.

btw

im an idiot, id idn't put the whole hypothesis

[stoic python]

Croqueta

[dapper gorge]

Q is a change of basis matrix, and it changes coordinates from an otrhonormal basis to an orthonormal basis. So I should just show that any type of matrix basis change of that type is unitary

[brave cliff]

I need to row reduce the matrix above to row echelon form, I just wanna know if my second step looks good

[lavish jewel]

looks like you successfully swapped the rows

[brave cliff]

Yea

Idk, this problem seems tricky and I've made progress, but what I really want to know is, if there's a way to look at it and tell what rows need to be swapped. Like is there a swapping algorithm I could follow to set it up before solving?

[lavish jewel]

when doing it on paper there is no difference

just depends on which operations you are more comfortable with

[brave cliff]

I see, thank you

[serene tide]

is this all thats needed?

im not sure how else to interpret the question but this seems a little simplistic

[wintry steppe]

looks about right to me

[serene tide]

guess im just overthinking it

[dapper gorge]

Self-adjoint operators in complex spaces aren't that important, because it would require that the eigenvalues are real. Is that right?

[zinc timber]

Not important??

[dapper gorge]

haha

well idk

Since the eigenvalues would be real, it would be like a real space

that's what I was trying to say