linear-algebra

[barren sentinel]

i dont even know if its correct 😂

[lavish jewel]

i know you did, i had you blocked for like 4 days cuz you were asking annoying stuff without reading

[barren sentinel]

now that i look at it

probably isnt

[lavish jewel]

and i see i should do it again

[zinc timber]

[dusky epoch]

do it all over again

[lavish jewel]

like i'm surprised you did anything at all there if you don't know how to write the equation for a plane

[zinc timber]

[wintry steppe]

does anybody know for a website that will solve this problem? i am really stuck. wolfram gets everything expressed in terms of a and d but doesnt show steps (although i paid for a mobile version).

[barren sentinel]

what?

[barren sentinel]

i do

[lavish jewel]

that is indeed the best you can do. it does not have a unique solution

[barren sentinel]

just slight mistakes i think, ill go over them now and fix i guess 😂

[lavish jewel]

you could technically pick any two variables as free parameters tho

[barren sentinel]

i was reading, i just couldnt find what i was looking for, still didnt

regarding the transpose

[wintry steppe]

yep but i dont know how to do it, cant express everything in terms of a and d

[zinc timber]

fr watch the MIT lectures on LA

[zinc timber]

@barren sentinel

[teal grotto]

did somebody say transpose

[barren sentinel]

i asked u for one small thing and u gave me a whole book of equations and proof and went, just read the 1000 pages and find out what u need

like.. bruh

[zinc timber]

i saw what happened yesterday

[lavish jewel]

oh my god beast, the thing has something called "index"

at worst you'd have to skim over like 20 pages of a chapter

[barren sentinel]

yeah as if it was at all helpful

i went to the derivatives part and couldnt find what i needed

[lavish jewel]

ok i'm just blocking you again, i give up on you entirely

[barren sentinel]

ok bro do it

[teal grotto]

bruh💀

[barren sentinel]

but u cant call that shit "helping"

other than that u did help me

[lavish jewel]

😌

[zinc timber]

well you can only fill a cup upto the level it's supposed be filled

[lavish jewel]

now replace the cup with a strainer

[barren sentinel]

omg u are such a dick

[teal grotto]

do i dare ask what you wanted to know about the transpose?

[barren sentinel]

best not to probably

but basically when i was differentiating for ANN backprop

i needed to transpose sometimes

i still dont get why

i get that the dimensions dont line up, but i didnt find any proper proof for it

so i did basically what @lavish jewel did and cited matrix cookbook and told the people to just figure it out by themselves 😂

[teal grotto]

example?

[barren sentinel]

tho i had done that even before edd told me about the cookbook

[teal grotto]

i’m out

nope

[barren sentinel]

this is simpler

[barren sentinel]

each one of these has outer products

(essentially transpose)

[teal grotto]

i would literally rather do ryu’s test than try to read that stuff lmao

[lavish jewel]

the answer to all of these questions is "write it as a sum and use the chain rule componentwise"

[barren sentinel]

fair

[teal grotto]

just not fun at all

[zinc timber]

there's a part B left if you want to help

[barren sentinel]

[teal grotto]

dude i have no idea how to do optimization problems

[barren sentinel]

bruh

u just

optimise

simple as that

[zinc timber]

i'm just gonna copy, thus is not even my field of interest

[teal grotto]

would just be more fun to try that than whatever beast is doing

[barren sentinel]

what u read here is optimisation

[barren sentinel]

[barren sentinel]

done*

i dont ever fucking want to do that shit again

[teal grotto]

i want no parts

[barren sentinel]

u are using up arrow here as if u get it 😂 @teal grotto

[barren sentinel]

integration by parts tho 💀

[barren sentinel]

later on u gotta integrate this shit for baysian neural nets

[teal grotto]

rearrange some stuff and spam chain rule component wise seems like a reasonable answer lol

[barren sentinel]

yeah pretty much

thats what matrix diff is in nutshell

especially for neural nets

[wintry steppe]

Edd I managed to solve the problem. Thanks

[barren sentinel]

the thing crashes when i try to convert it to PDF 💀

[lavish jewel]

cool

[barren sentinel]

anyways imma go eat and then continue with 13

[dusky epoch]

why are you doing it in word in the first place

[teal grotto]

exactly what i was thinking 💀

[barren sentinel]

ala project

[lavish jewel]

since we're looking at nasty derivatives of (multi)linear transformations

[analog nacelle]

Why do it in latex when you can do it in word 😎

[barren sentinel]

btw mines not a linear transformation

[barren sentinel]

i did do it in word

[barren sentinel]

oh u are asking why in word

its better imo, more readable when requiring copy pastes

harder to copy paste code

but editing it is kind of glitchy

would be nice if u could just copy paste latex from its output

[teal grotto]

i imagine if someone were to read this aloud it would sound like something like “gugog fifu gougugu fufogugi…”

[barren sentinel]

or if word wasnt as glitchy

[lavish jewel]

sounds like a use case for macros

[barren sentinel]

bruh

[lavish jewel]

if you need to repeat stuff a lot, you usually just make a macro

like \myfunction{x}{y}{z}

[barren sentinel]

what kind of model is this?

[barren sentinel]

hmm might try that

stil easier to copy paste and work with word imo

[lavish jewel]

it's a linearized version of the helmholtz equation in 2D

[teal grotto]

just as easy to cntrl v cntrl p in latex but macros are faster and there for that specific reason lol

[wintry steppe]

i dont know anything about matrices so could someone tell me

if i have matrix

1 2
3 4

is that same as vector (1, 2, 3, 4) in R^4?

[barren sentinel]

not as easy tbh

[lavish jewel]

not the same, it under certain conditions, equivalent

[barren sentinel]

ah

thermodynamics

yeah i dont engage in that shit

[lavish jewel]

not thermodynamics

[barren sentinel]

comp sci major

[barren sentinel]

then?

[dusky epoch]

the space of 2 by 2 matrices is isomorphic to R^4 but there is no one canonical isomorphism

[lavish jewel]

imagine googling wrong

ultrasound propagation

[barren sentinel]

bruh im gonna block u

[lavish jewel]

please do

PLEASE

[barren sentinel]

never seen a bigger dick in my life

[lavish jewel]

that's what s..

[wintry steppe]

if i have to find intersection and sum of matrix subspaces can i translate matrices to "R^n vectors" to write down equations to find intersection?

[teal grotto]

i…

[lavish jewel]

yes, you can

[wintry steppe]

thanks!

[lavish jewel]

it can make it easier to handle matrices when treating them as vectors in some vector space

[teal grotto]

edd did u write all of this? or are these just images from somewhere?

[lavish jewel]

i wrote it

[teal grotto]

sheeesh

[lavish jewel]

what's more

[teal grotto]

this can’t just be like, for hw right?

[lavish jewel]

as part of taking derivatives later on, i literally do by hand the thing beast keeps complaining about

[zinc timber]

i do lpp by hand

[lavish jewel]

the proof is in the annex

[teal grotto]

dude that’s hilarious

[lavish jewel]

that's why i find them so annoying. i explained it like 3 times and they keep answering "i know, but..." and immediately showing that they in fact don't know

then providing references, but index too stronk

[zinc timber]

tho this looks like more differential geometry than optimization

[lavish jewel]

i don't know any differential geometry

or not that i'm aware

[zinc timber]

me neither, that was the point

[teal grotto]

wb like inverse/implicit function theorems or basic manifold stuff?

[lavish jewel]

not by that name

[teal grotto]

i see

[barren sentinel]

whats ipp?

[lavish jewel]

i can wave my hands and say "neumann series!" though

[teal grotto]

i’m gonna laugh and pretend ik what neumann series is

edit: i do know what this is after googling it lol

[barren sentinel]

gonna take diff geo next sem i think

[zinc timber]

btw why us it $\pdv{x^T}$ and not $\pdv{x}$?

[barren sentinel]

or adv stats

not sure which

[lavish jewel]

i use denominator layout

since these are vectors, the derivatives will add in extra columns

so as you follow the rows in the matrix, you get derivatives w.r.t. different variables

no reason other than i like it in that order, and for whatever reason i find the notation clear

[zinc timber]

yeah so it's for convenience? tbh now i also feel like using it

yeah it's clearer this way, nice!

[lavish jewel]

this is also what keeps tripping beast over, but now they will never know 😌

[zinc timber]

[teal grotto]

does anybody know of any links to a constructive proof of spectral theorem? it feels way to important to application for people to just not have a constructive proof and only rely on EVT

[zinc timber]

for reals you can look into Jacobi's iteration, it's not a proof but numerical algorithm to compute EVD

there's also housholder reflection method

[teal grotto]

this sounds like a proof to me lol

it gives u a complete EVD of a real symmetric matrix?

[zinc timber]

yeah

basically it show that for any 2x2 block you can multiply with some orthogonal matrix and make the non-diagonals zero

if you apply it iteratively on each non diagonal element you get a diagonal matrix and the orthonormal basis

[teal grotto]

why would this not be the main proof of “every real symmetric matrix is diagonalizable”

[zinc timber]

because other one is easier

[teal grotto]

oh no

[zinc timber]

you looked it up?

[teal grotto]

yes

i um

[zinc timber]

[barren sentinel]

btw edd, the normal vectors of the plane was not orthognal to column 1 of matrix in the first place

[teal grotto]

thank you for letting me know the names of the methods that exist for engineers to use so our buildings don’t topple over. i will not however be trying to become one of the 7 people in the world who understand it

[lavish jewel]

givens rotations 😌

[zinc timber]

if you think about it, it just diagonalizes a 2x2 block at a time

[zinc timber]

isn't that what jacobi uses?

[wintry steppe]

Guys how to find basis for vector space that contains even polynomials of degree 5 or less such that p(-1)=0?

[zinc timber]

example, take x, then x(-1)=-1, so you add +1 to it and you get one element x+1

do it with x² and xⁿ

[barren sentinel]

basically 0, (x-a)(x+1), (x-a)(x-b)(x+1),...

-1 is always a root

[barren sentinel]

eh i guess u could do that too

[wintry steppe]

how do we know they are even? p(x)=p(-x)

[barren sentinel]

its not asking for even functions i think

just even powers

[teal grotto]

no, i think it’s asking for even functions

[barren sentinel]

but then i wonder if certain properties of vec space will be satisfied

[wintry steppe]

sorry my bad, it says p(x)=p(-x) and i translated that as even

[barren sentinel]

i guess its the same thing, my bad

[barren sentinel]

yeah even powers and p(x) = p(-x) should be the same

so ur basis would be 0, x^2 +1, x^4 +1

(at least i think so)

[teal grotto]

0??

what basis do u know that has zero

a set of vectors containing the zero vector isn’t linearly independent

[teal grotto]

and this really depends on the field

[barren sentinel]

the fuck do i write in C?

[barren sentinel]

yeah even polynomial

[teal grotto]

like in the field with two elements, every polynomial is even

[teal grotto]

0 cannot be an element of a basis

[barren sentinel]

why?

[teal grotto]

in any vector space

[teal grotto]

this

[barren sentinel]

nah it can be cuz

[teal grotto]

bro

[barren sentinel]

0 + x^2 is not 0

[limber sierra]

mate

[teal grotto]

fam

[limber sierra]

0 + 0x²

[barren sentinel]

yeah but thats the condition of linearly independent, u cant just multiply all by 0

[limber sierra]

??

[teal grotto]

crucially: 1•0 + 0x^2

[barren sentinel]

u gotta find non trivial solutions

[limber sierra]

the coefficient of the 0 can be anything

which gives you nontrivial solutions

[limber sierra]

150354090354095 * 0 + 0x² is a nontrivial solution

[barren sentinel]

hmm then how do u include even powers?

1 i guess

[teal grotto]

u don’t because the field is important here

[barren sentinel]

but then p(-1) is not = 0

[teal grotto]

my b namington

[barren sentinel]

anyways can someone explain what i should do here?

[barren sentinel]

yeah that one makes sense

in part C?

[teal grotto]

just do it lol

[barren sentinel]

do what tho?

like it is the range space, its right there

do i have to write some fundamental proof or something?

[teal grotto]

show that every vector in the range space can be written as a linear combination of those two vectors

[limber sierra]

how do you know that?

[barren sentinel]

cuz those are the columns

column space = range space

[teal grotto]

not true

[barren sentinel]

column space is span of those two vectors

sorry range space*

[limber sierra]

it probably wants you to just verify the claim then

so show that every element of A's range is a linear combination of those 2 vectors

[teal grotto]

it’s literally a two liner

[wintry steppe]

so in such questions i just find polynomials degree 1 to 5 that satisfy conditions?

{x^2 + 1, x^4 +1}

is that my basis of polynomials degree 5 or less, p(x)=p(-x), p(-1)=0? 2D?

[limber sierra]

(and also that theyre linearly independent but thats obvious)

[barren sentinel]

the fuck?

[limber sierra]

is what i said unclear

[barren sentinel]

i have already done that and thats so obvious

it cant be just restating it 😂

[teal grotto]

part c is literally saying here’s an “alternative way to say this if u didn’t notice already and here’s some easy points to grab”

[limber sierra]

it probably just wants to emphasize why column space = range space is true

[barren sentinel]

bruh 😂

[limber sierra]

by making you work through a specific example

[barren sentinel]

yeah probably

so like an elementary proof

why is it true btw?

[teal grotto]

because matrix multiplication nice

[barren sentinel]

bruh

[teal grotto]

if i have Mx, then i can write is as
m1x1 + m2x2 + … + mnxn which literally follows from matrix multiplication, where n is the number of columns of M, mk is the kth column vector of M, and xi is the ith component of x

[barren sentinel]

yeah got it thnx tho

[teal grotto]

sure

[barren sentinel]

idk how i should write it exactly tho

[teal grotto]

just like that

wdym lol

[barren sentinel]

but is the vector R2 column vec or row vec?

[teal grotto]

wot bruther

[barren sentinel]

bruh like..

idk how to show

lol

[teal grotto]

your matrix

is M

[lavish jewel]

"yeah lol it's obvious"

"but why?"

bruh

[barren sentinel]

yeah

[lavish jewel]

here, from 2 hours ago

[teal grotto]

any element y in the range space, by definition, comes from an element [x1 x2]^T so that M [x1 x2]^T = m1x1 + m2x2 = y

there’s not much else to it

m1 and m2 are the columns of M

[barren sentinel]

yeah i get that, i just dk how to write it exactly

[teal grotto]

and don’t say “u got it” until u actually read it and tried to understand it

[barren sentinel]

in the book

[teal grotto]

I JUST MFKN WROTE IT FOR U

[barren sentinel]

but i cant write that on a peice of paper

can i?

[teal grotto]

i wrote it in two lines

[lavish jewel]

this is definitely a troll

[teal grotto]

just now

[lavish jewel]

should i ban them

[barren sentinel]

like ill need to write the matrix A

and then apply it on that

anyways ill figure it out thnx

[teal grotto]

u just wrote the matrix A right here

[barren sentinel]

or at least try to i guess

1 1
1 0
1 1

is the matrix

[teal grotto]

no, just say a1 is the first column, a2 is the second column

[barren sentinel]

alright fair 🤷‍♂️

[wintry steppe]

could someone tell if that is correct?

[barren sentinel]

seems like it

[teal grotto]

fair? have u never heard of the concept of abbreviation?

[barren sentinel]

bruh chill

ill write it as that thnx

[teal grotto]

no it’s not correct. p(-1) != 0

for all of those basis elements

unless ur working over the field with two elements

[barren sentinel]

so u do something like i said i guess?

[wintry steppe]

{x^2-1, x^4 -1} is that basis?

[barren sentinel]

(x-a)(x-1) and (x-a)(x-b)(x-c)(x-1)

but then idk how to write it as a basis 😂

[teal grotto]

i have to think about it for a minute before i give u a half ass answer lol

[barren sentinel]

can someone double check?

a transpose dot a =
2 1
1 2

?

[teal grotto]

use an online calculator

[barren sentinel]

😂

[teal grotto]

we’re trying to help mate bruh

[barren sentinel]

i was mostly asking for the dimensions but ok

yeah go ahead @teal grotto

[teal grotto]

(n x m) times (m x p) has dimension n x p

[barren sentinel]

yeah k cool thnx

hopefully its correct

[teal grotto]

p(x) = p(-x) for all x in F. p(-1) = 0 = p(1), so -1 and 1 need to be roots of every polynomial in this space.

since p(x) - p(-x) for any polynomial in this space is identically the zero polynomial, u can’t have any odd degree terms popping up in p.

this leaves u with p having only x^0, x^2 and x^4 terms.

so p(x) = a_0 + a_2 x^2 + a_4 x^4.

p(-1) = a_0 + a_2 + a_4 = 0

so a_0 = -a_2 - a_4

now p(x) = a_2 x^2 - a_2 + a_4 x ^4 - a_4

[barren sentinel]

what do i do in h) i)?

[teal grotto]

so p(x) = a_2(x^2 - 1) + a_4 (x^4 - 1)

[teal grotto]

all in all, yes this is a correct basis for the reasons above

[barren sentinel]

the columns of V are of size 2

so im confused

[teal grotto]

Sigma is just the diagonal matrix of eigen values

[barren sentinel]

yeah i got that

i think i understood

its asking Avi

so multiply A with each column in V

[wintry steppe]

thank you!

the next question asks what is sufficient and necessary condition for natural number n so that Pn (polynomials degree n or less) and our subspace with basis (x^2 - 1, x^4 - 1) share same complement in P5 (polynomials degree 5 or less)

do you know how to approach that?

[teal grotto]

basis x^2 - 1 and x^4 - 1 right?

[wintry steppe]

yes, sorry

[teal grotto]

do u have an inner product on Pn?

[wintry steppe]

sorry dont know what is an inner product

[teal grotto]

like a dot product

[wintry steppe]

unfortunately we never mentioned it

probably will later

im a beginner

[barren sentinel]

the fuck

if its obvious how do i verify?

[wintry steppe]

one complement is span of {1, x, x^3, x^5}? right?

[teal grotto]

what does it mean when u say “Pn and our subspace with basis … share the same complement”

[teal grotto]

i get that this is a complent to our subspace with basis …

[wintry steppe]

so vector space is P5

and we have subspace W (for which we found a basis) and Pn

we must find subspace X such that

W + X must me P5
and their intersection only zero vector

and

Pn + X must be P5
and their intersection 0

sorry no

i edited

[teal grotto]

okay thank you for clarifying

[wintry steppe]

we try to find sufficient and necessary condition on number n so that such X exists

[teal grotto]

i think you want it to be n = 4 2 1 right

[wintry steppe]

do you agree with this

[wintry steppe]

with this

[teal grotto]

yes

if n = 1, then Pn is isomorphic to W

[wintry steppe]

complement of P2 is {x^3, x^4, x^5}?

[teal grotto]

yes

[wintry steppe]

sorry can you elaboeate why n = 2 i am confused

[teal grotto]

god damn zero indexing

Pn has dimension n + 1. u need n = 1. my bad

for the isomorphism to work out

[wintry steppe]

complement of P1 is {x^2, x^3, x^4, x^5}?

sorry could you elaborate why n = 1

and we never mentioned isomorphism unfortunately

[teal grotto]

it’s the only choice that works

we have that
dim(W + X) = dim W + dim X = 6

dim(Pn + X) = dim Pn + dim X = 6

so 2 = dim W = dim Pn = n + 1

(+ here is of course a direct sum when “adding” vector spaces)

we also get that dim X = 4

[wintry steppe]

thank you i get it

and what would be example of X?

[teal grotto]

i honestly don’t know. it’s making me think that there is either a problem with my understanding/explanation or there is a problem with the question

[barren sentinel]

how do i show/prove this?

that singular values of A = sigular values of A transposed?

[teal grotto]

singular values of A are the square roots of the non-negative eigen values of A^TA

singular values of A^T are the square roots of the non-negative eigen values of
(A^T)^T A^T = AA^T = (A^TA)^T

u just showed that a matrix and it’s transpose have the same eigen values in ii.

apply that to A^TA

[barren sentinel]

i copied proof here

but im not sure why determinant is invariant to matrix transpose

[teal grotto]

det(A) = det(A^T)

[barren sentinel]

yeah

but do i gotta prove that too?

[teal grotto]

this is because of… some stuff

[barren sentinel]

yeah i know, but thats the hard part 😂

[barren sentinel]

lmfao

[teal grotto]

um. ugh, they didn’t just let u take this as a given?

[barren sentinel]

idk

its vague aF 😂

[teal grotto]

just take it as a given

ur working with SVD

[barren sentinel]

lemme post my proof

[teal grotto]

it should be a known that det A = det A^T

[barren sentinel]

https://documentcloud.adobe.com/link/track?uri=urn:aaid:scds:US:f49086a6-f1f7-4e7b-8c88-8d68d71b4f13

this my proof

btw do i have to show its nonzero? in ii)?

[teal grotto]

no, it just asks for non-negativity or not non-negativity

[barren sentinel]

(A^TA)^T = AA^T?

[barren sentinel]

wdym?

[teal grotto]

no, this is wrong. should be (A^TA)^T = A^TA

[wintry steppe]

thanks @teal grotto

[teal grotto]

my b was looking at 3

[barren sentinel]

yeah

this shit

just?

[lavish jewel]

one way would be to recognize that A = R_n R_n-1 R_n-2 ... R_1 I, where R_i are elementary row operations. then using that det(AB) = det(A)det(B), and that (AB)^T = B^T A^T, it follows directly. the elementary row operation matrices all have nice properties, such as being diagonal/upper triangular in many cases, or equal to their own transpose, which simplifies all the intermediate dets to det(R_i) = det(R_i^T)

[barren sentinel]

use what i found in i)?

[wintry steppe]

A and B are subspaces of V, dim V = n, A is not a subset of B, B is not a subset of A. dim A = n - 1 = dim B. what is dimension of intersection of A and B?

[barren sentinel]

but in i)

i didnt write any reasoning for it

i just did it for one and found it to be true

[teal grotto]

B = A^T = (USV^T)^T

simplify and tell me what u get

[barren sentinel]

U^T S^T V?

[teal grotto]

yea. no. but S is rectangular diagonal

so S^T = S

[lavish jewel]

V S^T U^T

[teal grotto]

wait

it should be VS^TU^T

[barren sentinel]

what?

[teal grotto]

i just assumed whatever he wrote was going to be correct bruh. it shouldn’t be this difficult

[zinc timber]

[lavish jewel]

the S needs the transpose if it's rectangular. not if ur doing economy size svd

[teal grotto]

sry for the mistakes. but whatever, this proves a, and hence proves iv

[outer goblet]

Let A be a 3x3 matrix, with det(A)!=0. Find all vectors v R3 that are orthogonal with all three column vectors of A.

[barren sentinel]

yeah ok thnx

[outer goblet]

so if you do the dot product equals 0, i end up with

A^t *v=0 as the solution space

[barren sentinel]

so this is same as 1 right?>

[teal grotto]

yea

[barren sentinel]

k cool

[outer goblet]

but the solution to the question says that since A is invertible, so is A^t, and thus only the nullvector that fullfils it

but i dont see the connection, any help?

[barren sentinel]

what do i argue here? 😂

[lavish jewel]

akintos, you could alternatively ignore that and simply find the solution space through elimination

set up your augmented matrix [A^T | 0]

then when you do gaussian elimination, you will end up with [I | 0], meaning the only solution is the 0 vector

[outer goblet]

yeah but im not given the matrix

its ambigious 3x3

[lavish jewel]

ah, then yes, it is simply a direct consequence of being invertible

that A is invertible means the columns are linearly independent

[barren sentinel]

man this part is hard..

[teal grotto]

it doesn’t matter because u can row reduce any matrix, since it’s invertible, you will end up with the identity as rref just as edd pointed out the first time

[outer goblet]

aaa

[lavish jewel]

because they are linearly independent, b a_1 + c a_2 + d a_3 = 0 by definition only has coefficient 0s as solution

and for the transpose as well

[outer goblet]

yeah i forgot about this theorem

ait

[wintry steppe]

could someone take a look at this

[barren sentinel]

what does this notation mean?

[stable kindle]

i would guess it's asking for the coefficients of v1, v2 such that they sum to x

like

[barren sentinel]

is that it?

[stable kindle]

ok so x = a1v1 + a2v2

[barren sentinel]

a =2 and b = 3

[stable kindle]

ok so then it would be [2, 3]

[barren sentinel]

do i write something more?

[wintry steppe]

@teal grotto thanks!

i tried like this

dim(A+B)=n-1 + n-1 - dim intersection

how did you find dim of intersectionm

[stable kindle]

but vertical

[barren sentinel]

oh ok

thnx

[teal grotto]

it’s not zero, that’s what i want it to be but it’s not in all cases lol

[barren sentinel]

u sure thats correct?

like what does the notation even mean?

for future reference?

[stable kindle]

it's like

the vector in the subspace with those bases

[wintry steppe]

what?

[stable kindle]

the vector expressed in those bases

[wintry steppe]

im confused

[teal grotto]

wishful thinking

[stable kindle]

dim A+B = dim A + dim B - dim A&B

if neither is a subspace of the other, what can you say about dim A+B

[wintry steppe]

id say its direct but dont understand why

[stable kindle]

?

no

[wintry steppe]

oh sorry

[stable kindle]

ok if neither is a subset of the other, what can you say dim A+B isn't

[barren sentinel]

Th

this good?

[wintry steppe]

sum A plus B isnt direct?

[stable kindle]

?

no

[wintry steppe]

sorry i am confused

[barren sentinel]

then?>

[stable kindle]

ok so if A was a subspace of B

what would the dimension of A+B be

[barren sentinel]

B?

[stable kindle]

which is?

[wintry steppe]

dim B

[stable kindle]

which is?

[wintry steppe]

n - 1

[stable kindle]

ok

and similarly if B was a subspace of A

then A+B would also have dimension n-1

so if neither A or B are a subspace of the other

then A+B does not have dimension n-1

what dimension must it have

[wintry steppe]

n?

[stable kindle]

yes

because it has to be bigger than n-1, but the max is n since it's all in V

so dim A+B = dim A + dim B - dim A&B

so now you know enough

[wintry steppe]

sorry i am not 100 percent sure why it has to be bigger than n-1

[stable kindle]

A is in A+B, so A+B has at least dimension n-1

but dim A+B can't be n-1, we just showed that

so it has at least dimension n

[wintry steppe]

so dimension of sum is never smaller than dim A + dim B if A not subset of B or B of A?

wait

wrong

[stable kindle]

dimension of A+B is never smaller than dim A

[wintry steppe]

thank you very much!!!!

[zinc timber]

?

B =-A then?

[stable kindle]

we're talking sum of subspaces

[zinc timber]

aha,

[wintry steppe]

i got dimension of intersection is n - 2

[stable kindle]

ye

[wintry steppe]

great, thanks

Have to check when (0, 1, 1), (1, 0, 1), (x,x,1), (1,1,y) form a generating system for R^3.

I got that x!=0.5 or y!= 2

(then we have three linearly independent vectors and dim of R^3 is 3 so thats it)

for that values of x and y 3rd or 4th vector cant be expressed as a linear combination of 1st and 2nd vector

is there anything else i must check?

[barren sentinel]

ho would i write integral on polynomial as a linear transformation?

[teal grotto]

if you know that any collection of 3 linearly independent vectors in R^3 is spanning then ur done. otherwise, i would show that this families of vectors is spanning

[barren sentinel]

this seems incorrect

[teal grotto]

like, indefinite integral of p?

[barren sentinel]

yeah

should it have 0 in first row?

[teal grotto]

are u taking the +C from integration to be zero?

[zinc timber]

indefinite integral is not one-one because of the +c

[barren sentinel]

assume so

this is the question

[zinc timber]

what you can do is define the transformation like this instead

$$
(Tp)(x) = \int_0^x p(t) \dd t
$$

[zinc timber]

[barren sentinel]

idk man this is the question

[teal grotto]

it’s not well formed lol

[barren sentinel]

yep

and then they fucking us over with it

anyways

assume it to be 0 for now

orr...

[zinc timber]

is that from a book or an assignment sheet

[barren sentinel]

what if i just include c

on the top row?

[barren sentinel]

assginment sheet

[zinc timber]

yeah thought so, tell yr professor to make the modification I had mentioned

[barren sentinel]

This fine?

[zinc timber]

No

[wintry steppe]

X = {(a+1, a+b+2, 2a-b+c+3, a+b+c+4)}

is X subspace of R4? i am completely confused. a b c are reals

[barren sentinel]

why>?

[zinc timber]

C is arbitrary constant, not all C's are the same

[barren sentinel]

ik

but just for representations sake

is it fine?

[zinc timber]

you can't treat is like a constant you add with every term

[barren sentinel]

ok ill assume c to be 0?

[zinc timber]

already told you, it's mathematical gibberish

[barren sentinel]

ok

ill assume c to be 0

[zinc timber]

you can't assume that given the question specifically talks about indefinite integral

[zinc timber]

^^

[barren sentinel]

i added a small note

c must be assume to be 0 or another constant, here c is assumed to be 0

[teal grotto]

forget about the constant thing i just realized it wouldn’t make T linear

[barren sentinel]

what?

[teal grotto]

u have to do ryus way

[barren sentinel]

yeah thats what i was confused by

[barren sentinel]

whats that?

bro i dont have time

[teal grotto]

the way that he just explained

[barren sentinel]

and i cant scream at prof rn

[barren sentinel]

i cant just change the fkin question

so just tell me what i should write for now, ill discuss with sir later

[zinc timber]

at the corrent form, that's not even a question

[barren sentinel]

bruh doesnt matter

[zinc timber]

it's some random sequence of words

[teal grotto]

just don’t mention anything and make the constant zero if ur really that pressed

it’s the only way the question makes sense

[zinc timber]

or you can try one thing, show that T is not linear by taking
$p(x) =0$ and show that

$$T(p)=\int p(x)\dd x=\int 0 \dd x =C \overset{?}{\neq} 0$$

[teal grotto]

T isn’t even a function at this point

[lavish jewel]

can you show the original problem, noname?

[zinc timber]

generalized function

[shy meteor]

9x1 + 6x2 − 12x3 + 6x4 = −3
6x1 + 4x2 − 8x3 + 2x4 = 6
2x2 + 4x3 + 6x4 = 0
−3x1 + 4x2 + 7x3 + 18x4 = −10.

That is the stuff given

[lavish jewel]

all right. well, i think homogeneous is the wrong word, since homogeneous means that Ax = 0

and you don't have 0 on the right hand side

maybe you meant consistent?

[teal grotto]

i think they’re trying to determine whether or not it is a homogeneous system

[lavish jewel]

i'm not sure what you mean, maybe you can help them out, then

[shy meteor]

Yes exactly or if it is inhomogenous and prove it then ...

[teal grotto]

edd u already helped them haha

[lavish jewel]

i guess what was meant then was for them to write this as a matrix-vector equation and show that the rhs is nonzero?

[shy meteor]

Yes I also guess so, I simply thought that is way to easy 😂

[teal grotto]

homogeneous is just all the constants are zero

[shy meteor]

And that is why I tried to do it more complicated :D. Anyway thank you !!!

[teal grotto]

not all the constant terms are zero

so it’s not homogenous

[shy meteor]

Got it, as I said I simply wanted to make it more complicated, because I thought, that can't be it 😂 , THANK YOU!!!

[zinc timber]

ig you can use a complicated argument by showing that (0,0,0,0) is not a solution of the system, so it can't be homogeneous

[shy meteor]

Yes, or you could try to show that it is not homogeneous, by setting up a homogeneous one

[lavish jewel]

proof by "lmao look at it"

[teal grotto]

if only this was a valid proof method

[zinc timber]

I would have written a 'obviously' argument

[lavish jewel]

not very far off from "to begin, assume that 6 = 0. but this is a contradiction, as 6 \neq 0, and so the system is not homgeneous"

[teal grotto]

i’ve been doing this on my calc hw lately and just saying stuff like, “it’s way to squiggly over here to be continuous, look at this desmos plot”

[lavish jewel]

lol

[zinc timber]

Thome's function is cont at irrationals

[wraith monolith]

Hi, guys, why if i take the projection map to the orthogonal complement of the subspace spanned by a,b,c, then i got F(a)=F(b)=F(c)=0?

[zinc timber]

what's 'k'?

a field?

[wraith monolith]

yes, a field

[dusky epoch]

i think they meant to say there exists a nonzero linear function F: k^3 -> k...

anyway

a, b and c are by construction orthogonal to the subspace being projected onto.

[wraith monolith]

yes, i think that too. But does not orthogonal mean <a,F(a)>=0, how does this implies F(a)=0

[dusky epoch]

<a, F(a)> does not even make sense

you cannot take the inner product of a vector and a number

[wraith monolith]

oh yeah, so the function F is taking the bilinear form of a with vector in the orthogonal complementary subspace?

[dusky epoch]

...you could say that

[wraith monolith]

thanks!

[winged prairie]

yo guys i have a question about the annihilator. Im having trouble wrapping my head around the fact that the annihilator of a subspace can have more elements that just the 0 linear functional

why is this the case?

[barren sentinel]

whats this matrix?

containing?

[winged prairie]

nvm its wrong

idk

lemme think about it

[barren sentinel]

how do u solve for a transformation matrix?

[winged prairie]

The matrix would be 2x2, where the first column contains (0,1) and the second column contains (1,0) i think

because every vector in R^2 is written as a1(1,0) + a2(0,1)

and if we do this transformation

the basis vectors will just get flipped so we will have a1(0,1) + a2(1,0)

[winged prairie]

so u end up with this

[gray dust]

annihilator of {0}

[winged prairie]

wdym

[dusky epoch]

0 is annihilated by everything.

[winged prairie]

but its also a vector space right

but i thought all vector spaces only have 1 linear function which maps it to 0

[gray dust]

recall the def of annihilator

[winged prairie]

set of linear functions belonging to a vector space V, that map every element in the subspace to 0

functionals*

[gray dust]

functionals are in V*

[winged prairie]

ye sorry

[gray dust]

in other words such functionals are 0 when restricted to that subspace

any functional is 0 when restricted to {0}

[winged prairie]

oh that makes sense

[gray dust]

so V* is the annihilator of {0}

[winged prairie]

ok ty

[gray dust]

np

[barren sentinel]

is that right?

cuz thats what i did too

[winged prairie]

i think

[teal grotto]

the cool way is commutative diagrams

which is secretly what u guys are doing

[barren sentinel]

how do i do this?

what does it mean basis set using eigenvectors ?

[teal grotto]

exactly what it says. a basis of eigen vectors. an eigen basis.

[barren sentinel]

so is it just the set consisting normalised eigenvectors?

[teal grotto]

that form a basis yea.

[barren sentinel]

and how do i do h?

[teal grotto]

V ——> W
^ ^
| |
R^n ——> R^m

take the standard basis in R^n

map each e_i to ur new basis vector v_i in V by the map g. this goes up the left side. look at where T sends v_i and write in in terms of w_j, the basis in W. this goes across the top by T. the go down the right side by h, mapping each w_j to e_j. the matrix will be the composition h o T o g

[zinc timber]

f_3 = sin(x)?

missing a bracket f((...) ?

[stoic python]

Yeetus

[barren sentinel]

can someone help with b)?

[wintry steppe]

If A is a subspace of B, is dimension of A intersection B = dim B?

[teal grotto]

no

A intersect B = A

[zinc timber]

You can write $A$ in the form $$A = \mqty[-v_1^T- \ -v_2^T- \ \vdots \ - v_n^T - ]$$. So $x\in\mathfrak{N}(A)$ will look like $$
Ax = \mqty[-v_1^T- \ -v_2^T- \ \vdots \ - v_n^T - ] x = \mqty[v_1^Tx \ v_2^Tx \ \vdots \ v_n^T x ] = 0$$

[zinc timber]

but if want a more LA approach, there's one thing you can try

[wintry steppe]

if dim A > dim B

is dim(A+B) >= dim A?

[zinc timber]

$\innerproduct{x}{ A^Ty} = \innerproduct{Ax}{y} = 0 \qquad\forall y\in \mathfrak{R}(A)$ and $x\in\mathfrak{N}(A)$, so $x\perp y \forall y\in \mathfrak{R}(A)$

[zinc timber]

obviously

[hard drum]

Don't need the hypothesis lol

[wintry steppe]

how would you prove that 3 dimensional vector space has infinitely many 2 dimensional subspaces?

[teal grotto]

it doesn’t

necessarily

only if the field is infinite

[wintry steppe]

sorry, we assume that

[zinc timber]

then yes

[hexed burrow]

I'm working with the momentum and position operators $p_x$, $p_y$, $p_z$ and x,y,z. I'm wondering why
$$yp_{z}zp_{x}-zp_{x}yp_{z}=yp_{x}[p_{z},z]$$
is true, because if I expand out the RHS I get:
$$yp_{x}[p_{z},z]=yp_{x}p_{z}z-yp_{x}zp_{z}$$
And I didn't think I could just rearrange the order

[stoic python]

june

[wintry steppe]

do you know how to prove it?

[zinc timber]

think about the spaces $\cos(\theta)y+\sin(\theta)z=0$, for $-\pi \leq \theta \leq \pi$

[zinc timber]

probably a physics server question

[hexed burrow]

Ah ok I'll move it there thanks!

[teal grotto]

frick

i should be able to do this lol

[zinc timber]

[zinc timber]

do what?

[teal grotto]

this

mate was typing again

[zinc timber]

u can always add ϵ

[teal grotto]

bruh

[zinc timber]

also fr any vector in 3D , u can look at the orthogonal complement

[teal grotto]

what if it’s Q^3

[zinc timber]

that's what I did here

[ionic meteor]

hi could maybe someone help me with a math problem?
if u guys have the time would apreciate that so much
am at HELP-25 btw sorry for interrupting

[zinc timber]

epsilon trick

[wintry steppe]

In R4

if subspace A is generated by (1,0,0,0), (0,1,0,0) and (0,0,1,0) and B generated by (1,0,0,0) and (0,1,0,0) is dimension of ther intersection = 2, and of sum equal 3?

And if A is generated by (1,0,0,0), (0,1,0,0) and (0,0,1,0) and B generated by (0,0,1,0) and (0,0,0,1), is dimension of intersection = 1 and dim of sum = 4?

[zinc timber]

span {(1,0,0) (0,1,\epsilon)}

[zinc timber]

ye

[teal grotto]

okay well those are all the fields. R C and Q lmao

[wintry steppe]

sorry but im literally dying

i know i have too many questions but you are my last hope. sorry if i am boring and my questions stupid

[teal grotto]

nah never a bore lol

[zinc timber]

ngl you are literally asking the same question in different format

[teal grotto]

which one lol

[zinc timber]

dim A+B

[teal grotto]

LOL

[wintry steppe]

thanks Ryu

[barren sentinel]

why does the transpose notation come in here?

but i think i wrote that

hopefully its correct

i just wrote it as let rows of A be ai, ... an

then ai ... an * x = 0

but legit this guy made us prove that same thing 3 times

like b c and d is the same thing

https://cdn.discordapp.com/attachments/540211747613704221/914559685544255508/unknown.png

what would i possibly write differently in b c and d

[barren sentinel]

i think i did c wrong

it doesnt imply its in null space of A i guess

cuz if x is over a higher dimensional field as compared to rows in A,

then it can be orthoganal while not being in null space i guess

[zinc timber]

actually they are same

[zinc timber]

fyi

[wintry steppe]

subspaces A and B

if vectors from basis of A cant be written as a lin combination of vectors from basis of B and vectors from basis of B cant be written as a lin comb of vectors from basis of A, what can we say about subspaces A and B?

[fringe wadi]

they are incompatible, which means they has not intersection but vector 0

i think

[wintry steppe]

guys how can i tell if {(0,1,1), (1,1,2)} and {(1,0,1), (0.5,0.5,1)} span the same subspace?

[sick sandal]

you can write each vector in the 1st as a linear combination of the vectors of the 2nd and vice versa

[wintry steppe]

hi, so to calculate the eigenvectors, am i following the right steps? at the bottom here i did reduce row echelon, now all i have to do is find the four x values?

matrix A is in green with the corresponding eigenvalue which equals 1

or is there a faster way of calculating eigenvectors idk

[wintry steppe]

thanks. so they do span the same subspace.

[fringe wadi]

can someone help me with his proof?$$λ^mdet(λI_n + BA) = λ^ndet(λI_m + AB)$$ with dimension of A nxm and B mxn

[stoic python]

centuryegg

[fringe wadi]

with non zero λ

[wintry steppe]

<@&286206848099549185> sorry to ping but do you think you can help with my eigenvector question?

[agile bronze]

not sure if you did the right row echelon as in your third matrix your row 1 and row 2 are multiples of each other?

but the procedure is fine, in the end you'd find vectors that are perpendicular to all rows

[wintry steppe]

ahh ok thanks for the point about perpendicularity i didn't know that

[wintry steppe]

this exact question has been posted at least 3 times in this server now

word for word

#groups-rings-fields message
#linear-algebra message
#groups-rings-fields message

[stoic python]

Yeetus

[teal grotto]

remind me what annihilator is again?

okay

and f^T is just the map that sends phi to phi circ T?

[frank blade]

phi circ f, yea.

At least, I think it is.

Should be the only thing that makes sense.

[teal grotto]

yea

assuming it is, just evaluate f^T(phi) - phi at w for any arbitrary w in W

it should follow rather quickly from defs

nope that’s correct

you have just shown that for any w in W,
f^T(phi)(w) - phi(w) = 0

so f^T(phi) - phi is in the annihilator of W

yea haha

[tardy canopy]

Hey so my shsat test is in 4 days and I am not too good in geometry

Can someone help me out

Thanks

[teal grotto]

so, just assuming finite dimensional stuff here, let w_1,…,w_m be a basis for W. extend this to a basis w_1,…,w_m,w_m+1,…,w_n of V.

let p_i be the i-th dual basis vector, which maps w_i to 1 and w_j to 0.

then look at f^T(p_i) - p_i. it says that p_i(f(w)) = p_i(w) for all i and for any vector w in W. the idea is that p_i picks out the ith coefficient of any vector when expressed in the basis we chosen

so yes. i think with this info, we should get that f(w) = w for all w in W

since x = y if and only if phi(x) = phi(y) for all phi in the dual space (and we just showed that phi(x) = phi(y) on a basis of V*, so the result should follow)

[winter harbor]

Kind of busy now, but look up Sylvester's determinant theorem.

[teal grotto]

its not quite an injectivity thing

its more like an identification thing

[proud compass]

If I have a cholesky decomposition A=(T')T
Would [(T')^-1]A(T^-1) = I?
As in, could the components of a cholesky decomposition's inverse be applied to the original matrix this way?
I have a similar question but with the cholesky components replaced with the inverse of the symmetric square root. That one I'm fairly sure is also true

[zinc timber]

it's unclear what u r trying to write, use latex

[zinc timber]

still need help?

[proud compass]

It's not super time sensitive, I'll reword it later probably

[zinc timber]

it looks like you asking whether $(L^T)^{-1}AL^{-1}=I$ or not. answer is yes, and it's easy to verify

[proud compass]

yep!

It's not something I need to verify strongly, but knowing it's true helps, thanks

[zinc timber]

Here's the proof if you are interested-
\begin{gather*}
A = L^TL\
\implies (L^T)^{-1}AL^{-1} = (L^T)^{-1} L^TLL^{-1} = [(L^T)^{-1} L^T][LL^{-1}]=I\cdot I = I\end{gather*}

[zinc timber]

a sufficient condition is I think U is a subspace of V, or V is a subspace of U

[zinc timber]

https://www2.math.ou.edu/~dmccullough/teaching/slides/maa2010.pdf

[dry pulsar]

Hi guys

I have $F$ a Field, and i have to show that $M_{m \times n}(F) \simeq F^{m} \otimes F^{n}$
I try for $u=\left(u_{1}, \ldots, u_{m}\right) \in F^{m}$ and $w=\left(w_{1}, \ldots, w_{n}\right) \in F^{n}$ I defined $\psi: F^{m} \times F^{n} \rightarrow M_{m \times n}(F)$
such that $\psi(u, w)=\left(u_{i} w_{j}\right){m \times n}$
this is $\psi(u, w)=\left(\begin{array}{ccc}u{1} w_{1} & \ldots & u_{1} w_{n} \ \cdot & & \ \cdot & & \ u_{m} w_{1} & \ldots & u_{m} w_{n}\end{array}\right)$
SO $\psi$ is bilineal and then there is an unique lineal application $L: F^{m} \otimes F^{n} \rightarrow M_{m \times n}$ such that $L\left(t_{i j}\right)=\psi(u_i, w_i)$ where $u_i,w_j$ are elements of basis of $F^m$ and $F^n$ my question is how can i show that L maps a basis from $F^m \otimes F^n$ to a basis from $M_{m \times n} (F)$

[stoic python]

alef0

[worldly bear]

looking at this, what is p2 ect? char(A) = λn − tr(A)λn−1 + p2(A)λn−2

• · · · + (−1)n−1pn−1(A)λ + (−1)n det(A)

[zinc timber]

sum or the eigen values taking 2 at a time

[zinc timber]

or just look up sylvester determinant identity

[worldly bear]

thanks

[teal grotto]

why are u using tensor product instead of direct sum.

[gleaming knot]

$e_i\otimes e_j$ should map to a matrix with only one nonzero entry

[stoic python]

Icy001

[gleaming knot]

Work it out...

@dry pulsar

[teal grotto]

there’s quite a few ways to do this one

[dry pulsar]

its an excersive from a book 😄

and im trying to do it

[dry pulsar]

@gleaming knot this entrie will be $(a_{ij})$ right?

[stoic python]

alef0

[gleaming knot]

ya

[winged prairie]

does anybody understand the passage from the first equation to the second one?

[lavish jewel]

as i'm sure c squared can tell you, the transpose of a matrix can be defined elementwise as A^T_i,j = A_j,i

i.e., by swapping the indices

so what they did was swap indices AND transpose, which is the same as transposing twice

i.e. doing nothing

so the two expression are equivalent

the reason they swap the order when multiplying is twofold

first, it's valid because multiplication of scalars does not depend on the order of the multiplicands

second, by putting the C term first and the A term second, you'll recognize that the sum is of the form of the definition of matrix multiplication

[winged prairie]

oh ye cuz if you don't swap them you can't do the matrix multiplciation

thanks

[outer goblet]

i had to find orthonormal eigenvectors for a given matrix

[zinc timber]

2D?

[outer goblet]

ye

and i got u and v

but the solution says that its v and w

so i wonder if eigenvectors can be like "rotated" to the other side?

or is the solution just wrong?

like im supposed to find eigenvectos for this matrix

and i got

{(1,2),(-2,1)}

but the solution says its

{(1,2),(2,-1)}

is eigenvector the same if i just scale the eigenvector with a negative scalar?

or cant i

[zinc timber]

If $v$ is an eigen vector, then so is $-v$, in general the vectors of the subspace spanned by $S(v) = {\alpha v| \alpha\in \mbb{K} }$ are also eigen vectors.

[dusky epoch]

except zero

[zinc timber]

yeah

[barren sentinel]

can someone help me with this pls?

like i know that U has eigenvectors as columns but idk how to argue that or anything

[craggy finch]

l

[outer goblet]

can i scale orthagonal diagonoalized matrix too?

[barren sentinel]

this is correct right?

can someone pls explain this in less technical terms?

[echo nova]

someone knows spanish?

i dont understant two problems of my online book

[wintry steppe]

Can anyone explain the last sentence of first para ??

In proof

[outer goblet]

im supposed to find basis for R5, that contains 3 given vectors

[wintry steppe]

Channel busy

[outer goblet]

????

[wintry steppe]

Ask somewhere else

[outer goblet]

you cant have the whole channel for ursellf get off ur high horse

im supposed to find basis for R5, that contains 3 given vectors
so basicially how do i find 2 vectors that are linearly independant to the given 3 vectors?

[stable kindle]

lol

[stable kindle]

they just rearrange for v, move everything over to the right that isn't v and divide through by the coefficient

[outer goblet]

nvm i found the anwser myself

[stable kindle]

good for you

[wintry steppe]

No i got that .but they said s is linearly independent and one of u is equal to v .i didnt get this

[stable kindle]

that's not the last sentence lol

[wintry steppe]

Yeah sry 😅

[stable kindle]

i don't understand what you're confused about

so it's like

S is linearly independent

so none of u1, u2, etc. are the same as each other

it's v and something else

v and one of the u's

[wintry steppe]

Since S U {v} is linearly dependent so there is at least one a_i so that a_i != 0. So we dont know which a_i exactly . So how can we multiply by its multiplicative inverse(of v)

[stable kindle]

WLOG we can let it be a_1

here they're indistinguishable, more or less, so we might as well name them conveniently so that the i where a_i !=0 is 1

[wintry steppe]

My que is how do u know that coefficient of v is non zero ?

[stable kindle]

?

because it's linearly dependent

one of them is v

u_1 is v

ok well look

[wintry steppe]

Oh wait they mentioned that all a_1,a_2,.....a_n are non zero

[stable kindle]

nah, some of them can be 0

it's just that not all of them can be 0

anyway yeah

so

basically i misread it

they're taking S union {v}

ok wait

ok so basically yeah so if coefficient of v is 0 then you have all the other summing to 0 by themselves

so they would be linearly dependent

but S by itself is linearly independent

so v must be involved, it must have a non-zero coefficient

that's it

[wintry steppe]

Oh okay got it

Thanks for ur precious time!!

[stable kindle]

i coulda done this 10x faster if my brain was working ngl

[wintry steppe]

😂

[outer goblet]

can you diagonalize non symetric matricies?

[stable kindle]

yes, sometimes

[teal grotto]

to get an example, start with, a 2 by 2 or a 3 by 3 diagonal matrix, then do change of basis with some random basis. it’ll probably end up being a non-symmetric matrix afterwards that you then know is similar to a diagonal matrix

[outer goblet]

find all a b c d such that A is diagonalizable

this is the one im struggling with

i tried showing that

|lambda I - A| = 0

but im kinda stuck

[zinc timber]

nice problem

[lavish jewel]

well the char poly is straightforward, since it's a triangular matrix

the eigenvalues should be 1, 1, and d

so it's more about finding what the eigenvectors look like, and find the cases where they become linearly (in)dependent

[zinc timber]

that gave me an idea

looking at the null space of A-I

it has to be 2 for A to be diagonalizable

[outer goblet]

no

[lavish jewel]

or dim 3 if d = 1

[zinc timber]

yeah

that will just give the identity mat

[lavish jewel]

btw what language is that? i can almost recognize some germanic stuff

[outer goblet]

this one has nullity 1

and is diagonalizable

[outer goblet]

norwegian

[lavish jewel]

ah i see

ryu means the nullity of A - I, not of the original A

[outer goblet]

ah

it has to be 2? not 0?

[lavish jewel]

i.e. the dimension of the eigenspaces

[zinc timber]

yeah and that's still 2

[lavish jewel]

if d =/= 1, yes

[zinc timber]

d = 0 here

[outer goblet]

wym 2?

like nullility 2?

[lavish jewel]

oh specifically for that case

[zinc timber]

the nullity of A-I=2

[outer goblet]

what does it have to be

[zinc timber]

everything

[outer goblet]

so it has nothing to do with it

[zinc timber]

you must have geometric multiplicity of the eigen values = algebraic multiplicity

for A to be diagonalizable

[outer goblet]

wait what was that

[lavish jewel]

what was what

[outer goblet]

algebraic multiplicity

[lavish jewel]

here, for instance, you have two eigenvalues that are 1

so their algebraic multiplicity is 2

[outer goblet]

never heard of that word

is that some other word for it?

[lavish jewel]

this means that, for the matrix to be diagonalizable, there need to be 2 linearly independent eigenvectors corresponding to this eigenvalue

idk, repeated eigenvalue?

algebraic multiplicity is kinda standard

[outer goblet]

i prolly just didnt know the english word lol

[lavish jewel]

in german it's geometrische Vielfachheit des Eigenwertes

idk if that helps

[outer goblet]

:8

:(

[lavish jewel]

no?

[outer goblet]

nopp

[lavish jewel]

it should be similar in norwegian

[outer goblet]

🧐

hm

[lavish jewel]

well, i wrote geometric there, shoulda been algebraic, but still

"numerosity"

algebraischen Vielfachheit

[outer goblet]

wait

so a matrix is diagnolizable if its vector space has nullility 2?

3x3 this case

[lavish jewel]

no

[outer goblet]

i mean

eigenspace

[lavish jewel]

it is diagonalizable if the geometric multiplicity matches the algebraic one for all eigenspaces

in this case, when d is not 1, you have an eigenvalue with algebraic mult 2, and so the corresponding eigenvectors must have a corresponding geometric mult

[zinc timber]

,w necessary condition for diagonalization

wasn't helpful

[lavish jewel]

in simpler words, for a matrix of size nxn, you need to have n linearly independent eigenvectors

ryu is suggesting one method. you could also just straight up find the eigenvectors of the original matrix instead

[outer goblet]

well but then the eigenvectors end up being

1 or d

[lavish jewel]

no

[outer goblet]

no

eigenvalues

i mean

[lavish jewel]

they end up in terms of all the varaibles

yes, the eigenvalues are 1 or d, i wrote that in my first message

[outer goblet]

yea

so if there is a non 0 number above 2nd leading 1, the eigenspace doesnt have 3 linearly independant vectors

[zinc timber]

that is true