linear-algebra

[outer goblet]

hm ok now i get waht to do

only missing to find how to do it

wait

so if i get roots to be

1, 1, and d

its algebraic multipicity is 3?

[lavish jewel]

an alg mult is something each eigenvalue has

in general here, you'd say the eigenvalue 1 has multiplicity 2, and the eigenvalue d has multiplicity 1

and if d = 1, the eigval 1 has mult 3

[outer goblet]

so basciailly i have to show that there are 2 possibilities where the matrix is diagonalisable

d=1, a=b=c=0
algebraic multiplicity = gemetric multiplicity

or

d != 1, a=0, b,c=anything
algebraic multiplicity = gemetric multiplicity

but how do i show this

[barren sentinel]

why is U made up of eigenvectors in SVD?

[wintry steppe]

how would i use eigenvectors to solve this question? the answer is in black, just got no clue how that came to be

[lavish jewel]

_b, note that eigenvectors and eigenvalues satisfy that Ax = lambda x

[wintry steppe]

yup got that part

[lavish jewel]

if you arrange all of the eigenvectors into columns of a matrix P, and the eigenvalues as the elements of a diagonal matrix D, then AP = PD

[zinc timber]

eigen vectors of AA' yes

[lavish jewel]

since multiplying a matrix by a diagonal matrix from the right scales the columns of the matrix

i.e. the product PD scales each column of P by the corresponding element on the diagonal of D

[wintry steppe]

okok, just need to process that for a sec lol

[barren sentinel]

why tho

[zinc timber]

did u just write the answer without knowing how to solve it?

[zinc timber]

let _b finish

[barren sentinel]

bruh

i have been asking this like 3 times but no one answered. but alright

[wintry steppe]

im trying to grasp these concepts as best as i can, like i want to understand eigenvectors and eigenvalues fully is there a ressource you guys learnt from or just practise and youtube? (and school ofc)

[wintry steppe]

oh im done go ahead

[lavish jewel]

then for beasts question, first simply assume the SVD of A exists

so that A = USV^T

then look at AA^T and A^T A

note that AA^T = USV^TVS^TU^T = U(SS^T)U^T, which is of the form QDQ^T and is then the eigenvalue decomposition of AA^T

i.e. U contains the eigenvectors of AA^T

now rinse and repeat for A^TA

(also recall that U and V are orthogonal [or unitary] matrices)

[torpid socket]

Hello

So for this question I just need o find this 1 0 0 1

right?

[zinc timber]

for given A, yes

[worldly bear]

yes so you row reduce the matrix to the identity matrix, and make note of the row operations you use

[torpid socket]

Yeah thanks

[tacit cypress]

since the process of taking a limit is a linear transformation, what would the representative matrix of that transformation look like for a given function and a point?

[teal grotto]

this needs to be made more precise

first of all, whats your vector space? the space of continuous functions from R to R?

[tacit cypress]

yeah, just like how youre able to construct a differentiation matrix

[teal grotto]

im only familiar with that being done for the space of real or complex valued polynomials with degree less than or equal to n

[tacit cypress]

it should be possible for an arbitrary continuous function though, right

[teal grotto]

unless you restrict yourself to a finite dimensional vector space of functions, you wont be able to look at what the matrix representation will look like

like, you need a basis to do this

whats a basis for the space of continuous functions from R to R?

(dont try to find this, lol)

[wintry steppe]

an element of {B : B is a basis of C(R)}

[outer goblet]

can someone tell what do i need to google to find out how to get rid of xy in conic section formulas?

like this

from what i understand i have to somehow get rid of xy to be able to solve

but idk how

[wintry steppe]

complete the square?

[outer goblet]

wym

[vale mauve]

How do you convert expression written in Sum of product(SOP) form to Product of Sum (POS)?

like this one:
AB+ CD

can u simplift it for pos(product of sum)

I need help

[dim lotus]

can someone explain to me how to solve this question?

[ionic laurel]

In any given linear transformation, if the dimension of the kernal was to be not zero, would 0 be an eigenvalue of T? If so, can someone explain why?

[slow scroll]

if the kernel is not zero, then there is v != 0 such that Tv = 0 = 0v, so 0 is an eigenvalue

[teal grotto]

i would start by trying to write any polynomial of degree at most 1 in terms of the given basis

quick question, what is the ground field here? the real numbers?

[dim lotus]

yes it is real numbers

[teal grotto]

okay. after you do this

try writing the derivative of a polynomial in terms of the specified basis

then ur pretty much there

[ionic laurel]

oh shit

thank you

that was a very simple explanation

[dim lotus]

ight thank you

[teal grotto]

the converse of this is also true

[ionic laurel]

if 0 is an eigenvalue then the kernal cannot be nonzero

and the kernal must be greater than dimension 0

is that what you mean? @teal grotto

[teal grotto]

0 is an eigenvalue if and only if the kernel is non-trivial

[ionic laurel]

right

[teal grotto]

since eigenvectors cant be zero

[ionic laurel]

yep

[desert lava]

anyone knows how to draw vectors in 3d online?

[empty ibex]

can someone help prove this for me

[teal grotto]

A is diagonalizable, so there is an eigen-basis {a1, a2, a3} of R^3 consisting of eigen-vectors of A
(that is, for each 1 <= i <= 3, Aai = di * ai for some eigenvalue di of the eigen-vector ai).

for an eigenvalue d of A, we have det(A - dI) = 0.

now evaluate the matrix polynomial at each of the eigen-basis elements

[empty ibex]

Thanks!

[desert lava]

?

[teal grotto]

i think mathematica let’s u do this

[wintry steppe]

just appeal to cayley-hamilton :^)

[teal grotto]

A is similar to a diagonal matrix D by A = P D P^{-1} with P being equal to [u1 u2 ... un] and the ith diagonal of D is lambda_i. @empty ibex

matrix multiply Ax = (P D P^{-1})x with x a vector represented in the standard basis

[wintry steppe]

can someone explain how this is the answer

i understand how to get the eigenvectors and eigenvalues, but what calculations do you do to get the matrix, like [-2+i][-1+i]

[zinc timber]

solving for eigen vectors

[worldly bear]

what eigenvalues did you come out with?

bc if you didn’t get -3+i and it’s conjugate, you should start there

[proven oracle]

Hey guys, i have question here. It'd be great if you can take a look at it asap, it's something easy but i couldn't figure how to do.

[zinc timber]

,w eigen vlaues [[-6, 5],[-2, 0]]

[stoic python]

Results provided by WolframAlpha

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

Requested by Ryuzaki#5323

[zinc timber]

'asap' is sus

[proven oracle]

A coordinate system K'(x',y') has an origin that is shifted by v(vector) = (2,4) compared to the coordinate system K(x,y) and is rotated by the angle - pi/6.

First i need to make a sketch with the two coordinate systems and then calculate how the coordinates (x,y) of the coordinate system K are calculated from the coordinates (x',y') of the coordinate system K' and vice versa.

[proven oracle]

it has nothing to do with a test or a quiz i promise

i read the script of the lesson many times and couldn't find sth similar to this question

i know how to rotate a system by an angle, obviously, but it doesnt make sense to shift the origin of a system by a vector. Like, is the new origin the at the point of 2,4 ?

is it that simple, just determine the new origin as the point 2,4 and then rotate that new system by the angle, then sketch them both ?

[vocal isle]

Hey folks, a positive definite matrix, A, must satisfy the following criteria:

1. it's eigenvalues must all be > 0
   I want to use this definition to prove that xT A x > 0 for all x (except x=0)

I thought at first that this would be a proof:
lambda v = A v (definition of eigenvalues)
vT lambda v = vT A v (pre multiply by vT)
lambda vT v = vT A v (lambda is a constant scalar so bring it out the front)
lambda | v |^2 = vT A v
Then since lambda > 0 and since |v|^2 > 0 then vT A v > 0

however this proof is only true for v = eigenvector. I need to prove it generally using ANY x (except x = 0). Any ideas on how I can do that from knowing 1) ?

[teal grotto]

you're done

A is symmetric, so it has a basis of eigen-vectors

[lavish jewel]

to be fair, A need not be symmetric

idk what definition of definiteness they are using though

[vocal isle]

I don't think a positive definite matrix needbe symmetric. For example [2,0; 2,2] is positive definite and not symmetric

But even if A was symmetric, I'm unsure how that proves it works for any x. I realize that if it's symmetric then its eigenvectors are all orthogonal, mening that you can write any x = c1 v1 + c2v2 + c3v3 ...

oh wait i think i got it

[teal grotto]

my b. they arent in general orthogonal

[vocal isle]

if a matrix A is symmetric, then it's eigenvectors are orthogonal. I just summarized Gilbert Strangs video on youtube here.

[lavish jewel]

yea if the mat is symmetric this is pretty straightforward

[teal grotto]

i dont think this is true

1 1
-1 1

this is just a modified version of the matrix you gave me

its "positive definite" according to your def

but this has no real eigen values

wait shit

thats backwards

ughh

[lavish jewel]

right, the eigenvalues being real nonnegative is for the symmetric case

i think in general that need not be true

[vocal isle]

notice all > 0

[teal grotto]

,w eigenvalue decomposition of {{7, -2, -4}, {-17, 40, -19}, {-21, -9, 31}}

[stoic python]

Results provided by WolframAlpha

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

Requested by coycoy#4831

[teal grotto]

positive eigen values

[vocal isle]

yes and not symmetric

[teal grotto]

,w multiply {-48, -10, -37}{{7, -2, -4}, {-17, 40, -19}, {-21, -9, 31}}{{-48}, {-10},{-37}}

[stoic python]

Results provided by WolframAlpha

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

Requested by coycoy#4831

[teal grotto]

boom

not true

[vocal isle]

woah

[teal grotto]

stolen from here lmao https://math.stackexchange.com/questions/83134/does-non-symmetric-positive-definite-matrix-have-positive-eigenvalues

[vocal isle]

now i'm confused

lol

[teal grotto]

i just wanted to verify

the converse of what you are trying to show is also not true

idk who gave you this problem, but they need to get their ish together lmao

[vocal isle]

haha thanks for the counter example!

https://www.youtube.com/watch?v=xsP-S7yKaRA&t=1731s&ab_channel=MITOpenCourseWare

[teal grotto]

yea. would just check with whoever assigned the problem. A being symmetric is kinda important

[vocal isle]

i'm surpised because at 1:04 in the video he says "each one gives a test for positive definite matrices"

[teal grotto]

theres a big "Symmetric"

in the top left corner of this thumbnail

[vocal isle]

oh fuk

[zinc timber]

[vocal isle]

yeh ur right

[teal grotto]

[vocal isle]

i feel dum

[winged prairie]

anybody know how some good resources for calculating the rank of the matrix

[lavish jewel]

pos def is almost always connected to symmetry

[vocal isle]

is that because if you have xT A x

[lavish jewel]

not necessarily, but it's the most useful

[teal grotto]

thats why i was surprised when u guys said it didnt have to be symmetric

[vocal isle]

and you can write A in terms of symmetric parts and non symmetric parts A = A_sym + A_notsym

[lavish jewel]

it doesnt have to be

[zinc timber]

btw you can turn a non-symmetric matrix to a symmetric matrix that will act the same as a quadratic form

[vocal isle]

then xT A_notsym x = 0 ?

that might not be true idk

[zinc timber]

it is true

if A is anti symmetric then A has 0's in the diagonals and a_ij=- a_ji

that cancels out all the xy terms so result is 0

[teal grotto]

nvm not helpful there

[vocal isle]

so would i be correct in saying this:

A matrix, A is positive definite (regardless of whether its symmetirc or not) if xT A x > 0
And you can prove that the symmetric component of A (A_sym) will have the eigenvalues > 0

[teal grotto]

wait, how do you know its anti symmetric? is that like a standard decomposition of square matrices? into its symmetric and anti symmetric parts?

dude what a beast

[teal grotto]

okay, i believe this now

[zinc timber]

row reduction/wolfram alpha

[winged prairie]

ty

[lavish jewel]

SVD lol

[zinc timber]

SVD by hand

why sully me @teal grotto

[teal grotto]

i feel like you have had to do this before on a test

[lavish jewel]

i have

[zinc timber]

I didn't tho

never computed a svd by hand in my entire life

[teal grotto]

you should receive cash compensation for your loss

[lavish jewel]

a 2 part problem where they gave some properties of the kronecker product and ask used 1. to find the general form of the svd of the kronecker product of 2 matrices, and 2. apply it on some 3x3 matrices

the matrices were kinda simple though

[teal grotto]

i see

[lavish jewel]

then again, you are bound to make mistakes writing out a product of 3 9x9 matrices

[zinc timber]

tho I had to calculate the zeros of a 5x5 non linear system by hand

and then the eigen values for them

[lavish jewel]

that's some

[zinc timber]

wanna see the solution?

had to put them here and calculate the EVs and their signs

for the stability

[teal grotto]

for the stability!

[zinc timber]

stability of the fixed point

[teal grotto]

where do you even start tho

[zinc timber]

set all them equal to 0

[zinc timber]

what is the dimension of the subspace $S = { AB-BA | A, B \in M_n(\mbb{C}) }$

[lavish jewel]

does AB - BA have any special structure?

[zinc timber]

like it has trace zero

[lavish jewel]

that already means that dim S <= N^2 - N

[zinc timber]

why not N^2-1?

[lavish jewel]

because there are N diagonal elements

wait

nvm

[zinc timber]

yeah so n-1 elements can be arbitrary but the last one has to be -sum

[lavish jewel]

i misread what you wrote

[zinc timber]

yeah Ik it's <= n^2-1, just don't know what

[lavish jewel]

mhm

[teal grotto]

is the converse true?

every matrix with trace zero can be written as AB - BA?

[zinc timber]

that's my question to begin with

[teal grotto]

um

[zinc timber]

probably no

[teal grotto]

whats the dimension of the space of matrices with trace zero

this is like the only property that we have

im gonna run with it lol

[zinc timber]

n-1

[teal grotto]

um

[zinc timber]

one thing I have is $\text{char}(AB)=\text{char}{BA}$

[zinc timber]

characteristic poly

[teal grotto]

i dont agree with this

what is ur reasoning

[zinc timber]

n^2-1

lol

[teal grotto]

yea

okay lol

that makes more sense

what... whats a basis for this space

i have a hunch that they are the same space

[zinc timber]

basis of matrix of tr 0 $e_{ij}$ with 1 at the $ij$ th place and (n,n) element is -1 if i=j

something like that

[teal grotto]

erm

this not nice

what about

um

isnt that just all matrices of the form E_{ij} with i != j and E_{ii} - E_{jj} with i != j?

[zinc timber]

[teal grotto]

wait

[zinc timber]

if you want the basis I can write them out for you

[teal grotto]

no wait

this should be a basis too

[teal grotto]

this one

[zinc timber]

same as this

I was fixing the last element and letting others be arbitrary

[teal grotto]

okay then i just wasnt understanding what you wrote here

um

[teal grotto]

we just gotta write each of these bad bois as something in the form AB - BA

[zinc timber]

lol that's one way to do it

[teal grotto]

alr

waht about E_{ij}

[zinc timber]

ahahha

F

[teal grotto]

LETS GOO

i knew it

[zinc timber]

it feels weird that it should be n^2-1

like every tr 0 matrix can be written as AB-BA

[teal grotto]

i fkn love these dimension counting problems

wait, why dont we need to show that E_{ij} has to be expressed like that

[zinc timber]

https://people.eecs.berkeley.edu/~wkahan/MathH110/trace0.pdf

[teal grotto]

what is a commutator. eff. why is there so much i dont know

[zinc timber]

cool indeed

[zinc timber]

[A, B] = AB-BA

[lavish jewel]

the matrix commutator is precisely this expression 😛

[zinc timber]

commutator has 0 trace is trivial

but the other way around is kinda surprising to me

[zinc timber]

we are considering E to be basis of Mn, not AB-BA there

[teal grotto]

wait, so just by knowing that we have a set of AB - BA with trace zero that spans S, we're done?

huh?

[zinc timber]

we know AB-BA has trace zero

[teal grotto]

yes

thats clear

[zinc timber]

since all trace zero matrix is a commutator, they have same dim, i.e. n^2-1

[teal grotto]

sry this is redundant

[zinc timber]

learning something new every day, cool

only if I could focus on abstract part/topology

[teal grotto]

the topology part >>>

[zinc timber]

[teal grotto]

why sad cat for topology

[wintry steppe]

sl(n) is the set of commutators

sl = [gl, gl]. i wonder if this has any significance in lie theory

[teal grotto]

how

[wintry steppe]

sl(n) = lie algebra of SL(n) = trace zero matrices

sl(n) is the tangent space to SL(n) at the identity matrix :^)

you're learning manifold theory so im allowed to use these words

[teal grotto]

oh i thought u were just being lazy and writing sl(n) instead of SL(n) lol

[wintry steppe]

if A is a matrix and e is a really really super small number, so small that I + eA is still in SL(n), then 1 = det (I + eA) = 1 + e(trace A) + o(e). simplifying and dropping higher order terms you get trace A = 0, so the linear approximation to SL(n) at the identity is just the traceless matrices

not at all a rigorous proof but it's a fun way to see why this should be true (and you can basically make it rigorous using the matrix exponential)

[wintry steppe]

writing uppercase for a lie group and lowercase (fraktur) for its lie algebra is standard notation :^)

[zinc timber]

[teal grotto]

e(trace A) and not just trace(A)? sry that was dumb

how do u get that trace(A) = 0 tho, if e —> 0

[wintry steppe]

divide by e and take e to zero. o(e) means something with o(e)/e -> 0 as e -> 0

or just pretend the o(e) is gone and divide

lol

[zinc timber]

ngl I hv no idea

[wintry steppe]

not a dumb question, it's actually an important identity. try to show that the derivative at t = 0 of det(I + tA) equals trace A

then use the definition of the derivative

[zinc timber]

ok so I get linear approximation of det SL(n) at I is tr(A)

what's next?

[wintry steppe]

what's next is me falling asleep it's almost 6 am and i have class at 2

[teal grotto]

i’m in the same situation as u but now i’m calculating derivatives

[wintry steppe]

i have the luxury of being comfortable with failing my courses

[zinc timber]

[wintry steppe]

i am thriving

here's a hint for the derivative i just gave

if f(t) = det(A - tI) is the characteristic polynomial of A, then write det(I + tA) as t^n det(A - (-1/t)I)

fuck

i had two things in mind and convinced them both incorrectly

combined. autocorrect moment

you get what i mean

[teal grotto]

okay, so hint is to look at the characteristic polynomial and rewrite it

[wintry steppe]

also: trace = sum of roots of characteristic polynomial

that's the shortest proof i know of the derivative identity

[zinc timber]

what after the identity

that one I have done before

[wintry steppe]

[winged prairie]

can somebody help me do this, i keep getting that the rank is 2 but im defo doing something wrong

[zinc timber]

[wintry steppe]

what comes after is the rest of lie theory. but ill stop clogging up the channel now

[teal grotto]

probably messing up SVD calculations by hand

[teal grotto]

nah but fr that’s the right answer

look at the rows instead

[winged prairie]

wait the rank is 2?

[teal grotto]

third row is a linear combination of the first and second row

[winged prairie]

wolfram alpha gives me 3

[teal grotto]

first and second row are lin independent

[winged prairie]

ah

[teal grotto]

can’t be

[winged prairie]

no way i copied it out wrong

been doing this problem for an hour

[teal grotto]

lol

[winged prairie]

didn't notice this tho

[teal grotto]

yea. row rank = column rank too

[winged prairie]

ye

ty

[teal grotto]

np

[zinc timber]

?

[teal grotto]

they add and subtracted u1

[vocal isle]

I can't figure out how to do this, can someone help me out here? I know I can write x = c1 v1 + c2 v2 where v1 and v2 are orthogonal eigenvectors. But I don't know how I can then prove xT A x > 0

[teal grotto]

looks like a typo tbh

should be u2 - u1 in the second red underline

it should be pretty obvious

W is closed under scalar multiplication

[zinc timber]

?

[vocal isle]

follow the reply I made on myself to see full context 🙂

[zinc timber]

why don't you just use EVD

as I have told you can take symm part of A, so no harm in considering A is symmetric in the first place

@vocal isle

even you can assume A to be diagonal since every symmetric matrix is diagonalizable and has orthonormal eigen vectors

then assume $A$ has a -ve eigen value, so take the corresponding eigen vector $v$... and your analysis follows

[zinc timber]

btw you saying 'all eigen values are > 0' must also mean you are taking for granted that 'eigen values exist and are real' right?

[vocal isle]

thanks Ryu. I don't know what EVD means 😦

and yes, in this case i'm assuming A is symmetric

[lavish jewel]

eigenvalue decomposition

[vocal isle]

oh i know this! That's just when you diagonalize a matrix by

ok so that means if A is symmetric then I can write

A = E D ET

where E = [v1;v2;v3] (consisting of eigenvectors)

[zinc timber]

yes

and v1, v2, v3 are mutually orthogonal

[vocal isle]

but I don't see how this proves that if eigenvalues are all positive and if A is symmetric then xT A x > 0

I can write A = E D ET

xT A x = xT E D ET x > 0 ???

i also know that

v1T A v1 > 0

v2T A v2 > 0

and i also know since eigenvectors are orthogonal that v1T v2 = 0

[zinc timber]

@vocal isle you following?

[zinc timber]

vector*

[vocal isle]

hmmm

so let q = v2T A v1

if i premulltiply by v1

then i can get v1q = v1 v2T A v1

hmm, that doesn't get me anywhere, i don't know how to simply that 😦

[zinc timber]

you don't need to do that

$Av_1 = \lambda_1 v_1$

[vocal isle]

this i agree with though

[vocal isle]

yep, thats the defintiion of eigenvector and eigenvalue

[zinc timber]

so $v_2^TAv_1 = v_2^T (\lambda_1 v_1) = \lambda_1 (v_2^Tv_1) = 0$

[zinc timber]

that's somewhere for you?

[vocal isle]

ah nice yep i follow that

[zinc timber]

so for any $v \in V$ you can write it as $v = c_1v_1+c_2v_2$ and so $v^TAv = (c_1v_1)^TA(c_1v_1) + (c_2v_2)^TA(c_2v_2) + (\text{rest are zero})$

[zinc timber]

which gives you $\lambda_1 c_1^2+\lambda_2 c_2^2$ which is positive as $c_i^2 \geq 0 $ and $\lambda_i > 0$ and $v\neq 0$

[zinc timber]

you following?

also do you know what is a inner product?

[vocal isle]

inner product = dot product? I know what that is

so your v = c1v1 + c2v2. How did you get the next line?

[zinc timber]

ppl sometimes distinguish between them, idk what definition you are using so I'm assuming you know what that is

[vocal isle]

how did you expand that out?

[zinc timber]

compute v^TAv by hand

they'll distribute componentwise

[vocal isle]

this makes sense

hmmm

[zinc timber]

clear?

[vocal isle]

let me try and digest this step

[gray dust]

the dot product is an example of an inner product

[zinc timber]

don't make me write the steps

[vocal isle]

in getting to your answer i can only kinda get there

here [E] = eigenmatrix

[zinc timber]

just use that fact that $[E]^TA[E] = \mqty[\dmat{\lambda_1}{\lambda_2}]$

[vocal isle]

OH!

[zinc timber]

$A[E] = [E] \mqty[\dmat{\lambda_1}{\lambda_2}]$

[zinc timber]

and so $[E]^TA[E] =[E]^T [E] \mqty[\dmat{\lambda_1}{\lambda_2}]=I,\mqty[\dmat{\lambda_1}{\lambda_2}]=\mqty[\dmat{\lambda_1}{\lambda_2}]$

[zinc timber]

hope all clear now?

[vocal isle]

woah, what a crazy proof. I can't beleive how easily that comes to you. But for me this is a mindblow moment haha

[zinc timber]

also I would like to add that any positive definite matrix (symmetric?) induces a inner product on the space by $$\ip{a}{b}_A = b^TAa$$
what you showed $v^TAv > 0$ is just the requirement that $\ip{a}{a} \geq 0$ and $0$ iff $a=0$

[vocal isle]

I guess I don't know what the inner product is. I've never seen < > brackets before

a would have assumed that meant a dot b

[zinc timber]

nvm then

[vocal isle]

but it's kinda like a dot b but with a matrix A inside 😮

[zinc timber]

you'll prove in again sometime

[zinc timber]

it is, A is +ve definite symmetric

[vocal isle]

hey folks, so I've finally finished my adventure into learning the basics of linear algebra. Something I should have done a long time ago. here are my quick summary notes:

if anyone could quickly skim these notes and let me know if there is a wild error, I would be super happy 🙂

[winged prairie]

yo guys im having trouble understanding multilinear forms

does anybody have a concrete example

of a mapping from like V^n -> F

[dusky epoch]

a typical example is the determinant

[dark brook]

This is probably a bit of troubleshooting, but I just can't seem to get it right.
So we've got an function $f(t) = \alpha_1 (t^2 - 1) + \alpha_2 t + \alpha_3$ and 4 sets of points (-1, 0), (0,4), (1,-2), (2,2). I want to find the determinant of $(A^T A)$, which I got to be 80 with my calculations. $ \begin{pmatrix} 4 & 2 & 2 \ 2 & 6 & 6 \ 2 & 6 & 10 \ \end{pmatrix}$. And If I take the inverse of that, I get $\begin{pmatrix} \frac{3}{10} & - \frac{1}{10} & 0 \ -\frac{1}{10} & \frac{9}{20} & -\frac{1}{4} \ 0 & -\frac{1}{4} & \frac{1}{4} \ \end{pmatrix}$. In another assignment question, they say that $(A^T A)^{-1}$ is $(A^T A)^{-1} = \frac{1}{\det(A^T A)} \begin{pmatrix} 20 & -20 & 0 \ -20 & 36 & -8 \ 0 & -8 & 24 \ \end{pmatrix}$, which is not what I get. I've tried to redo the calculations and I don't seem to find that answer sadly. Is there anything I'm missing?

[stoic python]

HrJonas

[zinc timber]

what the hell is A

[dark brook]

Ax = b, so A is all the entries from the points and taken the function in consideration.
I've said $A = \begin{pmatrix} 1 & -1 & 0 \ 1 & 0 & -1 \ 1 & 1 & 0 \ 1 & 2 & 3 \end{pmatrix}$

[stoic python]

HrJonas

[winged prairie]

does anbody know some good resources to understand this

cuz the book doesn't have an examples

its a chapter about multilinear forms

[zinc timber]

^

[dark brook]

Yea you get the same as me. The following question asks me to explain why the determinant can be written in the above mentioned format, which I can't get with my calculations

[lavish jewel]

no need to go as far as the determinant

the dot product of 2 vectors is multilinear

[lavish jewel]

(bilinear)

[zinc timber]

well if they are not the same, how are you supposed to explain why they are same

WTF is that username

[dark brook]

No idea, it makes me wonder if I've done a wrong calculation, but I can't seem to find it

[zinc timber]

Also I probably think what they are referring to is the matrix inversion using the classical adjoint

[lavish jewel]

another example is the sum of the entries of a vector

[worldly bear]

why do i not have an eigenspace?

isn’t that just the eigenvectors?

[dark brook]

Hmm. Can I somehow get the "classical adjoint"?

[zinc timber]

the matrix of the cofactors

probably look it up

[dark brook]

I'll look into it, thanks!

[winged prairie]

i haven't studied them yet

[zinc timber]

^

[dusky epoch]

that's a 1-linear form

[lavish jewel]

sure, even if the vectors are in R^1, T is linear in each argument if the others are kept fixed

[winged prairie]

i don't understand why this is true

maybe its cuz i have not studied multivariable functions yet

[lavish jewel]

they just applied the definition of linearity to each of the arguments

i.e. it is linear in the first argument, and also in the second

hence bilinear

it's the same as your previous image says

[winged prairie]

but why can u just do that

it doesn't make sense to me

[lavish jewel]

this is a definition, drunk

only some maps satisfy it

those that do are said to be bilinear

in general, you can't just do that

[winged prairie]

ok ty

and how does this relate to the dot product

[lavish jewel]

the dot product satisfies these properties, and is therefore bilinear

[zinc timber]

yes, over R

[lavish jewel]

you can test it yourself by looking at the sum form of the dot product

[winged prairie]

how do u define the dot product as a linear map

[lavish jewel]

over R, as ryu says.

[gray dust]

we talk of eigenspace of each eigenvalue

[lavish jewel]

something like

$D(u,v) = \sum_{n=1}^N u_n v_n$

[stoic python]

Kangaskhan Edd

[lavish jewel]

for u and v in R^N

now you can test that D satisfies the definition of linearity both for u and v

[winged prairie]

ok ty

[shut orbit]

So I got this question about finding the basis for span but I dont understand whether to make each of the set into a row or a column? is it depends on the question or is there any regulation in the usage of "()" and "[]"?

[lavish jewel]

no standard notation for it, as far as i know. it depends more on how you did it in your lecture

as for the row and column question, it actually doesn't matter, but the procedure is slightly different depending on how you do it

if you set the vectors as rows, you can find a basis directly by keeping the rows after row reducing

if you set them as columns, you row reduce and note the pivot positions, then use those to pick out columns from the original matrix pre-RREF

[shut orbit]

oh okay thankyou very much

[dark brook]

@zinc timber Sorry for ping. I tried quickly to find the adjoint matrix of A^T A, but it did not give the correct one - the numbers were there, but they were in a weird reversed form

This is what the computer gave me as the adjoint.

I found the error. Silly me. Everything was reversed from the beginning. I thought it was how it should be done, but apparently not

[winged prairie]

guys looking back at the dot product. Is it defined to be a multilinear form by this defintion because if i take any scalar in R and replace it with another scalar which is being used in the dot product (Imagining that we are going from R^2 to R), then it will still remain a linear functional?

also could you give me an example of something that is not a multilinear form? thanks a lot guys

[gray dust]

f(x)=1

[winged prairie]

why?

[lavish jewel]

it's not linear, for one

can't be linear in several arguments if it only has one argument and it's not linear

guess that falls in line with my usual "proof by lmao look at it"

[winged prairie]

I know you will think im stupid but i don't get any of this multilienar stuff and im going crazy

the way i interpret this

[winged prairie]

is like this

[lavish jewel]

before we look at this

bring up the definition of linear

just linear form

[winged prairie]

T(x+y) = T(x)+T(u)

y*

[lavish jewel]

that's not all

[winged prairie]

T(cx) = cT(x)

and it goes from V to F

[lavish jewel]

and in one line, T(cx + y) = cT(x) + T(y)

[winged prairie]

yes

[lavish jewel]

ok

now let's say there is another linear transformation T(u, v)

[winged prairie]

ok

[lavish jewel]

we say T is linear in the first argument if T(cx + y, v) = cT(x,v) + T(y,v)

[winged prairie]

yes

[lavish jewel]

T could also be linear in the second argument following a similar def

and if it is linear in both, then it is bilinear

[winged prairie]

ok makes sense up to here

[lavish jewel]

ok, now let's say we have a T(x,y,z,w)

and if it is linear in each of the 4 arguments, it's multilinear

[winged prairie]

ok

makes sense

[lavish jewel]

technically the previous 2 were also multilinear

just special cases

[winged prairie]

ye

the only thing i don't get is how this relates

to the definition above

are they the same thing?

[lavish jewel]

yes

it's exactly the same thing

your image says "if we pick any of the arguments, T is linear in it"

they call one such argument x_i

[winged prairie]

ok

thanks man

[bold sun]

hey i have a really random question would cycling 3 rows be classed as an elementary row operation? cuz ik swapping rows are but not sure about that one?

[lavish jewel]

sounds more like associating 2 elementary row operations

merely by convention

[upper badger]

Write with the base 5
5^3 (5^4 + 3)

Anyone mind giving me a helping hand

[lavish jewel]

that's not linalg

[bold sun]

hmm okk so it aint?

[upper badger]

Really, its in my linear algebra book, what part of math is it?

[lavish jewel]

i'm not sure where that is encountered first tbh

[lavish jewel]

and yeah, it isn't. it's a more general permutation

[upper badger]

alright thx m8

[lavish jewel]

maybe #prealg-and-algebra , dany

[bold sun]

okk well i got that wrong then lol...Thank you for the help though 🙂

[sinful valve]

any example of this definition

not really sure what it means to have a vector space of functions?

[dusky epoch]

it's a vector space in which the vectors are functions with a fixed domain and codomain

[sinful valve]

So rather than just lists its functions mapped to inteveral of whatever S is

[dusky epoch]

S is the domain

as a somewhat silly example, if you have f, g ∈ R^R given by f(x) = x^2 and g(x) = 3x

you have (f+g)(x) = x^2 + 3x

and (10f)(x) = 10x^2

[sinful valve]

yeah that makes sense i guess, so yeah the domain is S and it holds arbitray functions f(x) etc, that themselves map to elements in F

so any function that does not have the entire domain of S is not define in that vector space ig

[gray dust]

its also not in F^S if its codomain isnt F

[sinful valve]

yeah true

[random axle]

Finally finished the matrix algebra chapters. Thanks for all your help guys. Hopefully won't have to return back here for another year

[wintry steppe]

here's a solution (and definitely not the only way to do it): if the characteristic polynomial of $A$ factors over $\bC$ as $$\det(A - tI) = (-1)^n(t - \lambda_1)\cdots(t-\lambda_n),$$ then
\begin{align*}
\det(tA + I) &= t^n\det\left(A - \left(-\frac{1}{t}\right)I\right) \
&= (-t)^n \left(-\frac{1}{t} - \lambda_1\right)\cdots\left(-\frac{1}{t} - \lambda_n\right) \
&= (1 + t\lambda_1)\cdots(1 + t\lambda_n) \
&= 1 + t(\lambda_1+\cdots+\lambda_n) + o(t)
\end{align*}

[stoic python]

TTerra

[wintry steppe]

i wouldn't be surprised if you can also solve it using the density of diagonalizable matrices over C

that'd be a quick and cool proof

[tired fossil]

hey, I have a question, what would be the eigen vectors of I subscript 2, so [1,0//0,1]? I get eigen value, 1, multiplicity 2

but what would the eigen vectors be?

[wintry steppe]

you tell me

what vectors v satisfy Iv = v?

[tired fossil]

so i know det(xI-I) gives you [x-1, 0//0,x-1]

so your eigenvalue is 1

[wintry steppe]

you're right. so what vectors v satisfy Iv = 1v? what are the eigenvectors?

[tired fossil]

so becaus etheir are less eigen values than multiplicity, there is no eigen vector

?

[wintry steppe]

no

what vectors v satisfy Iv = 1v?

[tired fossil]

0?

[wintry steppe]

what is Iv?

I is the identity

I_2(v) = ?

[tired fossil]

[1,0//0,1]

?

[wintry steppe]

what is $$I_2\begin{pmatrix}v_1 \ v_2\end{pmatrix} = \begin{pmatrix}1 & 0 \ 0 & 1 \end{pmatrix}\begin{pmatrix}v_1 \ v_2\end{pmatrix}$$

[stoic python]

TTerra

[tired fossil]

1

[wintry steppe]

no

how do you multiply a vector by a matrix

do you know what an eigenvector is?

[tired fossil]

[v_1,0//0.,v_2]

[wintry steppe]

nope

[tired fossil]

is what the right side gives you

ophhh

wait

its any vector

[wintry steppe]

yes

[tired fossil]

i went barin daid

screw me

[wintry steppe]

because all vectors v satisfy I_2 v = 1 v

[tired fossil]

dead*

makes sense okay, thank you

totally forgot !x=Ax

Ix*

[sinful valve]

wait according to that definition i sent earlier they say that R^2 for example is considered as function from {1,2} > R, whats all that about

[wintry steppe]

a function {1, 2} -> R is basically the same thing as a pair of real numbers

1 maps to the first element of the pair, 2 to the second

[wintry steppe]

is this from linear algebra done right by sheldon axler?

[sinful valve]

yes

[wintry steppe]

my condolences

[sinful valve]

what?

[wintry steppe]

lol it's a good book

[sinful valve]

yeah i just wanted to fill in shit i didnt know about linear algebra so i decided to go over whole thing from start

[sinful valve]

so R is a set of functions defined so that the domain of them is {1,2}

i dont get that really though its kinda weird

R^2 ive only though of as pairs of numbers in R like coordinates

[wintry steppe]

and i'm telling you they're the same thing

it's just a funny way of looking at tuples

$$(a_1,\dots,a_n)\in\bR^n$$ is the same thing as the function $${1,\dots,n}\to\bR,\qquad i \mapsto a_i$$

[stoic python]

TTerra

[sinful valve]

so 1 and 2 point to values in R

like 1 and 2 are just placeholders or indexes?

[wintry steppe]

indices
yes

that's all there is to it

[sinful valve]

yeah ight

thats kinda odd tbf but yeah

[lavish jewel]

your notation there messed me up for a bit

i was like wtf is i, there is no i there

[wintry steppe]

[sinful valve]

yeah did u mean 1 to i > R?

[wintry steppe]

no

i meant exactly what i wrote

i goes from 1 to n

[sinful valve]

okay xd

yeah

of course

is the i maps to a_i linked to the bottom bit then

[teal grotto]

woul not have ever thought they were dense

[wintry steppe]

crazy fact huh

i'll try to come up with some intuition for it in a moment

i would guess something like: all matrices can be put in JNF over C, now perturb that a bit and you get something diagonal

[torn stag]

write down the equations and solve for the blank

[ionic laurel]

Hi I have a question.

If we have a real matrix that has complex eigenvalues with lambda = a +- bi, to create a matrix A = PCP^-1, P and C can be based on a+bi and a-bi or only a-bi?

The book theorem says a-bi but I am also able to setup PCP^-1 based on a+bi

Can anyone explain why this is the case?

Here's the theorem

[torn stag]

@charred root There could be more than one solution. But in your problem I think that solution is the only one.

[spring pasture]

Two eigenvectors can correspond to same eigenvalues , that may be the case here

[ionic laurel]

What do you mean

If a-bi can setup PCP^-1

why not a+bi?

Because the matrix multiplication works out still no?

[spring pasture]

I didnt say that , I'm saying check if both the eigenvector correspond give same eigenvalues

[ionic laurel]

Oh

No they do not

Usually the signs just flip

Let me give an example

[spring pasture]

Ok

You can show your working on it

[ionic laurel]

Idk how to use the bot to create a matrix

So i’ll just send a picture of the problem

[spring pasture]

$\begin{bmatrix} 1&0\\0&1\end{bmatrix}$

[ionic laurel]

$\begin{bmatrix} 1&-2\1&3\end{bmatrix}$

[stoic python]

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

[ionic laurel]

oh

i cant delete that

thats wrong btw

[spring pasture]

Just send a photo np

[ionic laurel]

$\begin{bmatrix} 1&-2\1&3\end{bmatrix}$

[stoic python]

eswag

[ionic laurel]

okay so this matrix we're told to create PCP^-1

the eigenvalues are 2 +/- i

[spring pasture]

Ok

[ionic laurel]

And C has a format of $\begin{bmatrix} a&-b\b&a\end{bmatrix}$ while P has a format of [Real Imaginary]

[stoic python]

eswag

[ionic laurel]

so if we take eigenvalue 2-i we get C as $\begin{bmatrix} 2&-1\1&2\end{bmatrix}$ and P as $\begin{bmatrix} -1&-1\1&0\end{bmatrix}$

[stoic python]

eswag

[ionic laurel]

if we take eigenvalue 2+i we get C as $\begin{bmatrix} 2&1\-1&2\end{bmatrix}$ and P as $\begin{bmatrix} -1&1\1&0\end{bmatrix}$

oh i messed up the matrix format

[spring pasture]

Edit message

[stoic python]

eswag

[ionic laurel]

yeah that looks about right

[ionic laurel]

The answer keys says that these are the answers

[ionic laurel]

But why cannot this set be an answer?

why does the a-bi eigenvalue form the right answer but not the a+bi

[spring pasture]

Should work , what problem are you getting

[ionic laurel]

why does the theorem say it as a-bi is my question lol

i guess its wrong?

[spring pasture]

They must've done only one case

[ionic laurel]

gotcha

right

but both cases work

[spring pasture]

Yeah

[ionic laurel]

considering a+bi is just a-(-b)i

okay i just wanted to make sure that I was right

Thank you sam

[stoic python]

it's Sam

[ionic laurel]

right

[still lodge]

has anyone seen a proof of cayley hamilton that uses cyclic generating functions

i think that's what they were called

but my prof did that proof today and i highkey did not follow

[zinc timber]

send the proof maybe?

[wintry steppe]

Does this look like im on the right track to proving this?

[zinc timber]

follow the convo, lol

[teal grotto]

lol

shorter proof:

x^TAx = x^TM^TMx. qed

[zinc timber]

proof by "scroll up"

[wintry steppe]

Is this for me or...

[teal grotto]

both are for you

[wintry steppe]

Oh thank you

[teal grotto]

this is the short proof i was talking about

the convo ryu linked is closely related

[wintry steppe]

Yeah but I think we are supposed to be proving that the "sum of squares" is > 0

[teal grotto]

this accomplishes your goal

[wintry steppe]

I know but sadly my teacher is a goof

Actually I see now that I could pretty much just copy what he has for a nxn case and just add alot of dots and n's

[teal grotto]

no

dont do that

lmao

[wintry steppe]

Why its the same thing isnt it

[zinc timber]

it is

[teal grotto]

your goal is to prove this statement right

[wintry steppe]

Yeah but he wants it proved in this specific way for a nxn case

[teal grotto]

x^TM^TMx = (Mx)^TMx = <Mx,Mx> > 0

for all non-zero x

[wintry steppe]

So he literally wants me to write out 2 nxn matrices M and M ^T and then prove it this way

[teal grotto]

why on earth

would anybody

[wintry steppe]

murder thine professor

i want to type more lie group stuff here later

Because hes crazy

Ive had multiple people in here say my teacher is doing problems in the dumbest way possible

Another example of his style of writing out formulas

[zinc timber]

pls do

[teal grotto]

i mean, i guess this one is alr

[wintry steppe]

Yeah well , some coding people would say this is a very very very stupid way to write out Big O

[teal grotto]

oh the big O part yea

[zinc timber]

not really

to me that's perfectly valid thing to do, for cs ppl, maybe not that much

[wintry steppe]

I think the guy I showed was just mad cause he prefers the way cs majors do it over math

[wintry steppe]

yeah this lmao

a quickie: let $A \in M_{n\times n}(\bR)$ and let $\varepsilon > 0$ be such that $I + \varepsilon A \in O(n)$. then $$I = (I + \varepsilon A)(I + \varepsilon A^T) = I + \varepsilon(A + A^T) + o(\varepsilon)$$ so $A + A^T = 0$. that is, the linear approximation to $O(n)$ at the identity matrix is the set of skew-symmetric matrices

[stoic python]

TTerra

[zinc timber]

[teal grotto]

[zinc timber]

I+ϵ A ∈O(n) feels weird

[wintry steppe]

Honestly it kinda pissed me off too cause for our first problem I was like , oh easy 2 loops its 2n. Then we proceeded to spend 30 min doing it the "math flop count " way for a result of 2n

[zinc timber]

epsilon better

[teal grotto]

\varepsilon > \epsilon

[zinc timber]

ε<0

[wintry steppe]

Thanks tho guys, I think imma just add lots of dots and n's and hope he accepts it

[wintry steppe]

good, it should

[teal grotto]

so, to get that A + A^T = 0, you have to subtract I from both sides, divde by epsilon, and then you're like, 0 = A + A^T + o(1)

but the o(1) part has to be zero?

[zinc timber]

u can do it in a savage way, expand everything use every inequality

[wintry steppe]

as epsilon -> 0 yeah

[wintry steppe]

ill expand , add uncessecary white space, and infnite dots

but this is all non-rigorous so i wouldn't think too hard about the technical aspects lol

infinite*

[zinc timber]

in what field do I learn these things?

manifolds?

[wintry steppe]

but it's a nice handwavy way to see why some things should be true!

manifolds, lie theory, something like that

c squared is learning manifold theory right now so i'm tempted to dump some rigor on them

[zinc timber]

[teal grotto]

im just a child, please lmao

[wintry steppe]

i show no mercy to even children

[zinc timber]

I only know the definition

we do have manifolds next sem tho

[teal grotto]

literally struggling to work out these little o approximations lol

[zinc timber]

[wintry steppe]

don't think about them too hard

the "proofs" im giving aren't even rigorous

[zinc timber]

dymn this actually makes sense why dimension of commutators should be n²-1

and for dim of skew symmetric matrices, do I just subtract 1 from dim of O(n)?

[wintry steppe]

the dimension of the set of skew symmetric matrices will equal the dimension of O(n)

[zinc timber]

not the tangent space?

[wintry steppe]

tangent space has the same dimension as the thing it's tangent to

[teal grotto]

why is this not o(e^2) instead, sorry for getting caught up on this, i just want it to make sense

[zinc timber]

oh lol,

[wintry steppe]

$$(I + \varepsilon A)(I + \varepsilon A^T) = I + \varepsilon (A + A^T) + \underbrace{\varepsilon^2 AA^T}_{=o(\varepsilon)}$$

[zinc timber]

so Sl(n) has dim n²-1

[stoic python]

TTerra

[wintry steppe]

Dont worry totally makes sense

[zinc timber]

where are the dots?

[wintry steppe]

Oh yeah crap

lemme add those

then its valid

Pretty sure I did my cross multiplication wrong anyway

[zinc timber]

nah

[wintry steppe]

Or do I fail Linear algebra again

[zinc timber]

that'll just be $\sum_{i=}^n \sum_{j=1}^n \sum_{k=1}^n
x_i m_{ik}m_{kj} x_j$

[lavish jewel]

what are you trying to do

[zinc timber]

prove M'M is positive def in a hard way

[wintry steppe]

Yes this

[lavish jewel]

can't use similarity to a diag mat on an orthonormal basis?

[zinc timber]

no

[lavish jewel]

well

i would just factor then

[zinc timber]

that's what his prof demands

[lavish jewel]

how about x^T M^T M x

then substitute w = M x

so that the original expression is w^T w = 2 norm(w) squared

[teal grotto]

this is the way i suggested, but their prof apparently hates his students

[lavish jewel]

and the 2 norm is already pos def

ah

so what are they supposed to do

the prof specified a bad method to use?

[teal grotto]

thye specified to write out the matrix mult

[lavish jewel]

the matrices ur multiplying are wrong btw, you didn't transpose the left one

[lavish jewel]

this one here

that's M^2, not M^T M

[wintry steppe]

Perfect

[zinc timber]

not aligned

[wintry steppe]

yes thats a good point

Lmfao

[wintry steppe]

and he better like it

[lavish jewel]

i would honestly show it for 1 element and say the rest follow immediately

[wintry steppe]

Sadly I cant

[lavish jewel]

i mean, you can't show it for all infinitely many

[wintry steppe]

I also am new to LaTex so this is super fun

[lavish jewel]

so you're doing it at some point

[zinc timber]

think this should be true no?

[lavish jewel]

i propose to do it after the first one

you also squared the matrix ryu

[wintry steppe]

Yeah after this step imma just say .. and so on

[lavish jewel]

swap the indices in one of the two ms

[wintry steppe]

Then write > 0 at the end pretending I know what Im talking about

[lavish jewel]

you'd want x_i m_ki m_kj xj, i'm pretty sure

@zinc timber

compare to

[wintry steppe]

All jokes aside is this how you take the dot product of a 2x2 matrix again I actually forgot lol

[lavish jewel]

the arrows you put there compute the elements 1,1 and 2,1 of the result of M*M

1,1 in red, 2,1 in green

it's also not a dot product, python/numpy just has shitty nomenclature

[wintry steppe]

oh yeah then I need 2,1 2, 2

No I just men I always found the dot and cross product super confusing by hand

[lavish jewel]

sure but it's not the dot product of a 2x2 matrix

[wintry steppe]

Yeah its for nxn, but im being stoopid

[lavish jewel]

no

that's not what you're computing

there is no such thing

[wintry steppe]

wut

[lavish jewel]

there is an inner product for two square matrices

but that is not this

you are simply multiplying 2 matrices

[wintry steppe]

yes

[lavish jewel]

that is not a dot product

[wintry steppe]

oh

[lavish jewel]

it's just multiplication

[wintry steppe]

I have failed again

[lavish jewel]

inner products yield an element of the base field

the inner product would rather be trace(M^TM)

or, well, AN inner product

[teal grotto]

whats an example of a vector space over a field with an inner product, where the field isnt Q, R, or C

[lavish jewel]

i have no idea :x

[teal grotto]

hmph

i think like field extensions of Q work

but idk. i dont think you can have one over finite field

[wintry steppe]

Can I die yet

I'm literally doing this to save 5% of my grade

[dusty pond]

Hi! Could someone help explain this to me please?

Thank you!

[dusky epoch]

what part do you want explained?

[wintry steppe]

Well it seem like you have a popup window ontop of your work, usually this happens when you hover over certian text

[dusty pond]

the solution is one of those

but i dont understand how to approach it

[dusty pond]

honestly most of the problem

it's fine if you cant! imma go to help session tomorrow

[dusky epoch]

do you know what an invertible operator is?

[dusty pond]

a little bit

[dusky epoch]

this is a yes/no question

you either know what an invertible operator is or you don't, there's no inbetween

[dusty pond]

well yes but im not the best at applying it

[dusky epoch]

okay, so can you tell me in your own words what we mean when we say "T is invertible"?

@dusty pond ?

[dusty pond]

nvm...

[wintry steppe]

[dim lotus]

Lol

[lavish jewel]

why do you use hyenas for your pfp tteppa

[dusky epoch]

@dusty pond so that's a no?

[wintry steppe]

ill post it in ivory edd

[dusky epoch]

is that a "no i can't do that" or a "i don't need your help anymore"

[dusty pond]

sorry, i can't do it. I dont think. I have T = Id_V or T = 0_V

[dusky epoch]

okay, so let's back up a bit

do you know about invertible functions?

outside the context of linear algebra

(the answer options are "yes", "no" or "i don't need your help anymore stop badgering me")

[static sparrow]

hello everyone, hm...

I have a math problem that I can't solve

Hope everybody help please

rhombus in the exercise, its meaning is the number when divided (the total number of students) by the number 5.
Eg:
My student ID: 2017478
(2 + 0 + 1 + 7 +4 +7+8 = 29 remainder 4)
so rhombus is number 4.

[wintry steppe]

mom

[dusky epoch]

??

[static sparrow]

English translation: There exists a matrix X satisfying X^2 = (k - 5)I3
Please explain?
k : is a rhombus.

I know how to do it when k > 5 but don't know how to do it when k < 5

sorry my english seems poor. Because I'm Vietnamese

[dusky epoch]

the "please explain?" part is unclear

are we supposed to prove that it's true for all k?

[static sparrow]

As far as I know, the remainder when divided by 5 is 0,1,2,3,4

so the left side always has a negative number

[dusky epoch]

???

what do remainders have to do with this?

oh, nevermind, it's the student ID thing

[static sparrow]

must prove that that X exists a matrix or not? When k < 5

[dusky epoch]

so you're asked to determine whether there exists X such that X^2 = -I

you said k=4 in your case

[static sparrow]

that right!

Too bad I don't know how to get there.

Thank you so much for helping me

[dusky epoch]

i didn't even begin to help in any way

it's only now that i even know what the problem is

anyway: consider what eigenvalues X might have based on X^2 = -I

[static sparrow]

I'm so sorry that my English has to use Google Translate

If you refuse to do this, that's okay too. I also really appreciate that enthusiasm.

[spring pasture]

The hint is already given

[static sparrow]

😘

[static sparrow]

yes yes

[zinc timber]

yes, get's me everytime

oh i see it was M'M

[lavish jewel]

you, mrsystem and tteppa tripping me up with indices

[zinc timber]

[zinc timber]

since when did a rhombus mean 'k'?

[wintry steppe]

is it time for me to post my riemannian geometry work again?

[zinc timber]

yes

[wintry steppe]

indices...

[static sparrow]

sorry some k < 5 for readability

[static sparrow]

or you can call it a rhombus.

[zinc timber]

lol no, next sem maybe

[wintry steppe]

what i posted is just calc 3

nothing fancy just the chain rule a million times

[zinc timber]

I had a introductory course on tensor calc, have dealt with similar things before

[lavish jewel]

yeah i gotta go back and change this to black once i start writing the new part

dat \blue macro

[static sparrow]

excuse me

[zinc timber]

you said k <5? X²=(-ve)I right?

[static sparrow]

yes yes

[static sparrow]

thank you very much

[zinc timber]

for what?

[static sparrow]

determine if there exists any matrix X satisfying the above condition or not? If you please decake because why does the matrix X

[zinc timber]

can X have complex entries? or only real entries?

[zinc timber]

real then

[wintry steppe]

interesting notation

[zinc timber]

ok, so tell me does there exist a real number x with x²<0?

[zinc timber]

that rhombus is k btw

[static sparrow]

yes

[static sparrow]

Hic hic at Vietnam University, that symbol is quite familiar

[dusky epoch]

@zinc timber ||it is important that X is a 3 by 3 matrix. its size affects the answer||

[zinc timber]

|| don't think so, but u r welcome to correct me if I'm wrong ||

[dusky epoch]

||there exists a 2 by 2 real matrix whose square is -I||

[zinc timber]

||hmm true||