linear-algebra

[noble swan]

Pfff, bad book

[zinc timber]

we actually extend the concept of basis to the 0dim space

so in some way it is correct,

[still lodge]

is this an appropriate restatement of cayley hamilton: the characteristic polynomial of a matrix A annihilates said matrix

[noble swan]

Gotcha, thanks for the help!

[zinc timber]

ig

[still lodge]

i mean wikipedia has every square matrix over a commutative ring (such as the real or complex field) satisfies its own characteristic equation.

now idk what a commutative ring is (yet) but it sounds at least close to interchangeable

im just tryna find my own wording so it's easier to grasp

[sinful valve]

right so yeah svd is calm tbf

where has this $M=AA^T$ come from

[stoic python]

shriller44

[still lodge]

https://math.stackexchange.com/questions/158219/is-a-matrix-multiplied-with-its-transpose-something-special

[zinc timber]

https://youtu.be/mBcLRGuAFUk

[lavish jewel]

AA^T and A^TA are symmetric, and therefore diagonalizable

[sinful valve]

so yeah the idea is just getting the same orthogonal matrices

as in same sense as eigenvalue decomp before this was mentioned

just that that eigenvalues and vectors are taken as squared to get singular values

wait nah S^2 and U/U^T is the eigenvalues n vectors of AA^T

but we are interested in singular values of course

presumably theres more than this just method to do SVD but yeah its calm

[still lodge]

B is a 5 × 5 matrix with minimal polynomial (x - 1)(x − 2)^2

is it possible to determine the characteristic polynomial of B from this info

it's over complexes btw

[zinc timber]

No

[still lodge]

there are a limited number of possibilities though right

[zinc timber]

it could be anything of the following
\begin{align*}
(x-1)(x-2)^4 \
(x-1)^2(x-2)^3 \
(x-1)^3(x-2)^2
\end{align*}

[stoic python]

Ryuzaki

[still lodge]

alright cool that's what i thought, just double checking :)

now if a 5x5 matrix has minimal polynomial x^3, then that must mean that the characteristic is x^5 right...?

bc the only eigen value would be 0

[zinc timber]

yes

can you also calculate rank the of matrix with this info?

[still lodge]

educated guess but wouldnt it be 3...?

actually wait

no yeah three

[zinc timber]

2

[still lodge]

why

i was looking at JCF

[zinc timber]

you need JCF for this

there are 2 missing EVs

so 2 1's

rest zero

[still lodge]

well if it's min polynomial is x^3 can't it have two possible JCFs

either two blocks of single 0's or a block w two 0's and a 1

[zinc timber]

char = x⁵

only one possibility

[still lodge]

ok then there's something wrong in my understanding can you elaborate pls

[zinc timber]

an eigen value of algebraic multiplicity m and geometric multiplicity k means there are k-1 blocks of 1x1 and one large jordan block of size m-k+1

let's take a simple example

say min = x² and chr = x³ then

$A=\mqty[ 0&0&0\0&0&1\0&0&0]$

[stoic python]

Ryuzaki

[zinc timber]

do you agree?

[still lodge]

yes

[zinc timber]

recognize the 1x1 and 2x2 blocks

yeah there is a 3x3 jordan block as you might have deduced

but remember that in the 3x3 block there are only 2 ones

number of ones = size of block - 1

so you get 3-1=2 ones, giving you rank 2

[lavish jewel]

i'm gonna have to learn this stuff at some point

but what are you calling rank here

[zinc timber]

rank of the entire matrix given min poly = x³ and char poly = x⁵

context was different here

[lavish jewel]

aight

[still lodge]

hmmm

[still lodge]

i did not deduce this, why is there a 3x3

does the min polynomial kinda tell us what the largest block can be

^^that could be very wrong

[zinc timber]

what do you think the blocks should be

[still lodge]

2x2 and 1x1

[zinc timber]

no together with the chr poly it tells the exact size

ig you can find the max block size but not the exact size

@still lodge no no I was talking about the x⁵ and x³ case

[still lodge]

ohhhhhh ok yeah i was gonna say

[zinc timber]

u did deduce right?

[still lodge]

yeah

[zinc timber]

ok then number of 1's = ?

[still lodge]

but if you have the 3x3 block in the 5x5 matrix, couldnt the remaining slot(s) be a 2x2 or 2 1x1s?

and if you have a 2x2 block you have another 1

[zinc timber]

yeah but min poly discards the 2x2 block

[still lodge]

why

[zinc timber]

no wair

it doesn't

[still lodge]

aha

x^2 * x^3 = x^5

[zinc timber]

x³ not x², my bad

yeah

[still lodge]

ok this was still helpful ty <3

[zinc timber]

there will be 2x2 block giving rank 3

you were right

whole time I was thinking it's x²

[wise oriole]

Can someone help me w this one

Question: Determine the pairs (a, b) ∈ R × R such that the system

respectively
(a) has no solution
(b) has a unique solution,
(c) has several solutions.
Show how you arrive at these conditions.

I found this but I don't know what to do now

[zinc timber]

can you tell me what is the condition for unique solution

[wise oriole]

(a,b) element R x R that?

[zinc timber]

no not that

the one involving the ranks of the augmented matrix

ok if you don't remember it then say our unknowns were x,y,z then the last row gives us

[wise oriole]

the question above that's the only thing we got

[zinc timber]

(1-b²)z=-(a+2)

when can a solution of this exist?

what values must b take?

[wise oriole]

1 and -1?

[zinc timber]

say b = 1 or -1 and a=0

what does that give you?

[wise oriole]

0?

[zinc timber]

0=-2

[wise oriole]

oh yh

ah

[zinc timber]

so you see how you are supposed to derive the conditions?

[wise oriole]

but where did you get the (1-b²)z=-(a+2) from

[zinc timber]

the last column of your matrix

[wise oriole]

(b) has a unique solution then a can't be -2 and 2 / b can't be -1 and 1 right

for (c) has several solutions: b=1/-1 and a=-2

(a) has no solution if a is not -2

is this correct

[zinc timber]

for unique solution, it doesn't matter what a is, all you need to care about is 1-b² is not 0

[wise oriole]

Then a can be everything

[zinc timber]

yeah

[wise oriole]

How can I say that symbolic is that a element R

[zinc timber]

you only need b ≠+1 or -1

a ∈R

[wise oriole]

Oké thanks a lot

[zinc timber]

your other conclusions are also wrong tho

[wise oriole]

Yeah?

[zinc timber]

when will u get infinite solutions?

ok tell the conditions again

[wise oriole]

(1-b^2)a=-(a+b) or b is 1 and -1

[zinc timber]

see

infinite solution means that the last equation is degenerate. like 0=0. which values of a and b give you 0=0?

that should be the condition for infinitely many solutions

it just means you get no additional information from the 3rd equation

[wise oriole]

b = 1 or -1 you get 0 and a = -2 or am I dumb

[zinc timber]

yeah

[wise oriole]

Yeah bcs I’m dumb or yeah that is correct

[zinc timber]

correct, lol

[wise oriole]

Then why is it wrong you said

[zinc timber]

bcz that's not what u hv written here

[wise oriole]

Ah oké Ty and is this correct

[zinc timber]

no solution if u get some kind of contradiction like 0= 1

b=±1 but a≠2 gives 0 = some non zero number

which is impossible

[wise oriole]

Then a can be everything but can’t be -2

[zinc timber]

ya

[wise oriole]

a ∈R / {-2} ?

[zinc timber]

hm

although it's\ and not /

[wise oriole]

yh but i don't have that symbol 🙂

and it was quick but it's right

or not

[zinc timber]

doesn't matter

ya it's right

[wise oriole]

maybe i have a quetion later but i'm ok for now thank you for your time

[ionic laurel]

hi i have a few questions

The matrix A and (B^-1)(A)(B) then they have the same set of eigenvalues for every invertible matrix B <- This is true correct?

If 2 is an eigenvalue of A, then A-2I is not invertible. (I believe this is true as well)

If two matrices have the same set of eigenvalues, then they are similar (True? Not sure)

[pallid rampart]

I found out that this is true if you let the vector space to be $\brc{f\in C^\infty([-\pi,\pi]):f(-\pi)=f(\pi)}$

[stoic python]

Whoever

[still lodge]

what's the difference between elementary divisors and invariant factors

[worldly bear]

ive found the the characteristic polynomial of a 2x2 matrix A is x^2 - tr(A)x+det(A), and that for a 3x3 matric A, we have x^3 -tr(A)x^2 -det(A)

Is there a similar way to find the coefficient of the linear term of the characteristic equation of the cubic or is it not possible

[winter harbor]

Yup, you can!

You write it in terms of the exterior powers of your matrix.

More specifically.

Let $A$ be an $n \times n$ matrix over a field $\mathbb{K}$ and denote $p_{A}(x) \in \mathbb{K}[x]$ its characteristic polynomial, then:
$$
p_{A}(x) = \sum\limits_{k=0}^{n} x^{n-k} (-1)^{k} \text{tr} \left(\bigwedge^{k} A \right)
$$

[stoic python]

MISTERSYSTEM

[winter harbor]

This is a consequence of something called the newton identities, which relates elementary symmetric polynomials and power sum symmetric polynomials.

This can also be derived by embeding your matrix into the splitting field of the characteristic polynomial and using something called vieta's formula.

You can do this because the minimal and characteristic polynomials are invariant wrt field extensions.

https://math.stackexchange.com/questions/131883/proving-that-the-coefficients-of-the-characteristic-polynomial-are-the-traces-of

https://math.stackexchange.com/questions/3983329/expansion-of-the-characteristic-polynomial-via-exterior-product

I found these two stack exchange threads which might give a more in depth explanation.

[worldly bear]

so I just need to learn how to compute the exterior powers of a 3x3 matrix and then I will be able to write the characteristic polynomial just by looking at it nearly?

[winter harbor]

Yeah

You actually just need the trace of the exterior powers tho

If the characteristic of your field is 0

there's an explicit formula

which relies on computing a certain determinant

but idk if it is worth it

[worldly bear]

I think it would be nice to just go straight to the characteristic polynomial without the typical steps of det(A - lambda*I) personally

[winter harbor]

There's an ''intrinsic way'' to define the characteristic polynomial of a given linear operator on a finite dimensional vector space without going into determinant stuff.

Axler presents this in a short ''paper'' entitled ''down with determinants''

which motivated him to write his book ''linear algebra done right''

but like

Computing characteristic polynomials like that is just...

why would you do that

The determinant is a tool made for computation

so like, if you really need to calculate something explicitly, go for it.

https://www.maa.org/sites/default/files/pdf/awards/Axler-Ford-1996.pdf

[worldly bear]

i found it

[winter harbor]

this is the paper I was referring to

ah, nice.

[pliant blaze]

hi does det(-A) == det(A)

[worldly bear]

i like his opening two paragraphs already

[winter harbor]

No

for instance

take the 2x2 identity matrix

the determinant of - I_2 is 1

[pliant blaze]

ahh ic

[winter harbor]

what is true tho

is that det(A) = c^n det(A) where n is the size of the matrix, i.e A is an n x n matrix.

so that det(-A) = (-1)^n det(A)

[pliant blaze]

yeah just remember that\

thanks

[still lodge]

If 𝑞(𝐴)=0, then 𝑞 is divisible by the minimal polynomial of 𝐴

why isnt this named lmao

this feels really important

@/MISTERSYSTEM ive done some problems since we last discussed this stuff and i think ive gotten better :)

[wintry steppe]

it's not named because it's basically the definition of the minimal polynomial

[winter harbor]

Yeah, TTerra is right lol

Remember how I said F[x] is a PID for F a field and we define the minimal polynomial of a matrix A as the monic polynomial that generates the ideal
Ann(A) = {f \in F[x] : f(A) = 0 } ?

That's the idea nitez

[winter harbor]

Nice

[wintry steppe]

monic!!!!!!!!!!!!!!!!!!!!!!!!!!

[winter harbor]

yeaaaaaaaaaaaaah

thanks

Was dumb for a sec there

[still lodge]

yeahhh i havent taken abstract alg yet so i only have a loose idea of what a field even is

but our original discussion prompted me to at least make note of it to look into over winter break

[worldly bear]

i feel better about not knowing what a field is bc 8 year old terance tau didn’t either 🥲

[still lodge]

terrence 2pi

[still lodge]

sorry to nag but

[still lodge]

nada?

[steel moon]

for #1, im having trouble to find two matrices that dont satisfy closed under addition / multiplication

[lavish jewel]

how about this

[steel moon]

what is that tool you are using?

[lavish jewel]

the scalar mult part is also easy. take an identity matrix of size 2n+1 x 2n+1 and multiply it by -1

it's octave 😛

[steel moon]

interesting, is it some sort of computational tool for math?

[lavish jewel]

indeed

a slow but free version of matlab, sort of

[steel moon]

oh cool

for the purpose of this problem, one counter example for a property would suffice right?

[lavish jewel]

for the purpose of any problem in which they ask if something is true, one counter example is enough to say it isn't

[steel moon]

gotcha

and for the 2nd problem, we should show each set is a subset of the other right?

So something like let s in Span(S). then s is a linear combination of each of the elements in S

[lavish jewel]

should be enough to show that the elements in S are linearly independent

then mention the dimension of P3(R)

[steel moon]

each of the elements in S have x^3 in it

hm i guess i should expand it out firsrt

to show the elements in S are linearly independent, i would need to show theres a non-trivial linear combination that equals the 0 vector

oh whoops i got the definitions mixed up

thats for dependant

[wintry steppe]

Is anyone using the latest calculator from Texas Instruments which is the Texas Instruments C version? The one with the python programming that was released in the fall? If so, does it support Linear Algebra?

[forest quiver]

My teacher meant "noncolinear" instead of "nonparallel" right?

Oh wait no he didn't if they were parallel it would become some sort of line equation

Wait,,, don't they mean the same thing with 2 vectors?

Ugh

[marble lance]

What does noncolinear mean for 2 vectors? If you seem them as points, then any 2 points lies on the same line. If you see them as directions and you just mean colinear as they point in the same direction, then that is the same as parallel. So I think the right word here is nonparallel.

[zinc timber]

yeah you can say that they are non- parallel

$\vec{v}_1 = c \vec{v}_2$ is also another way of saying they are co linear, so, if there is no contant c satisfying this, they are non co-linear

[stoic python]

Ryuzaki

[golden kindle]

Anyone know good websites to practise linear alg problems

[merry imp]

can i get a hint for this? i have an expression for the inverse of A-I by factoring A^k - I, but how can i use that to find the inverse of A+I?

[teal grotto]

(-A)^0 + (-A)^1 + ... = (I - (-A))^{-1}

[quaint steppe]

Hi everyone currently studying linear algebra
can someone further explain to me these example on why they are not invertible?

what does it mean 1 is not in the range?

[slow scroll]

It means that there is no x such that f(x)=1

In general a function is invertible if and only if it is bijective.

[quaint steppe]

but then x^2 is not surjective so it's not invertible?

[limber sierra]

right

if $f$ is a function $A \to B$, then $f^{-1}$ must be a function $B \to A$

[stoic python]

Namington

[limber sierra]

but as the image notes, $1$ is not in the range of the multiply-by-$x^2$ map

[stoic python]

Namington

[limber sierra]

so $f^{-1}(1)$ makes no sense

[stoic python]

Namington

[limber sierra]

hence an inverse cant exist

[quaint steppe]

okay i kinda understood it already for example 2, what does backward shift means?

[lavish jewel]

take all of the elements in the list and shift them one of the left

[quaint steppe]

does it mean like $R^3$ going to $R^2$ for example?

[stoic python]

jinichi

[lavish jewel]

if you start with (0,1,1,...) the transformation yields (1,1,0,...)

so they note that if you have a nonzero element in the first position, the transformation gives you 0, so it has a nontrivial kernel

[quaint steppe]

oh so basically the linear map backward shift rotates the input to the left and make the first input goes last?

[lavish jewel]

i don't believe they mean circular shift

just shift

[quaint steppe]

oh okay

Thank you very much

[quaint steppe]

I'll probably gonna ask more questions later.

[wintry steppe]

Guys, I'm dealing with a rotation matrix rotating vectors by an angle 30 degrees about the z axis. I've found one of the eigenvectors which is (0, 0, 1).

but is it the same as (0, 0, 1)^T ?

Transpose of that vector?

I'd just say that eigenvector and the transpose of it are the same thing.

(0, 0, 1)^T = (0, 0, 1)

[lavish jewel]

they're not the same thing, but they are both eigenvectors of the matrix

call your matrix R and this vector v, with eigenvalue lambda

then v^T R = v^T lambda, and R v = lambda v

[wintry steppe]

I mean yeah one of them is column vector and the second is a row vector.

[lavish jewel]

this is in general not true btw

but a rotation matrix is block diagonal, with a block that has just a 1, so it holds in this case, for this eigenvector

[wintry steppe]

How can I show that they are both eigenvectors of the same eigenvalue?

[lavish jewel]

apply the definition of eigenvector twice

once from each side 😛

you can also stop for a while to consider what it means for one matrix to be the inverse of another, and then assume your rotation matrix is diagonalizable

if AB = BA = I, and you consider the matrix product and a bunch of dot products, it might become clearer

[wintry steppe]

Should I infer any meaning from that?

[zinc timber]

orthogonal matrix doesn't necessarily mean rotation

[sinful valve]

what does he mean exactyl

the inverse is the same as the complex conjugate transpose

[zinc timber]

is that $AVV^{-1}=\hat{U}\hat{\Sigma}V^{-1}$?

[stoic python]

Ryuzaki

[sinful valve]

yes

its just another video i was watching on svd its really good just missing this one fact

[zinc timber]

looks like some kind of svd decomposition

[sinful valve]

yeah it is

https://youtu.be/EokL7E6o1AE?t=1679

[zinc timber]

you need to tell the context before I can answer

[sinful valve]

this is the timestamp exactly

but hes just constructing the SVD

its just the * thing im not sure about

its often transpose

[lavish jewel]

probably that U and V are unitary matrices

so their inverse is their complex conjugate transpose

[zinc timber]

Oh he already derived $AV = \hat{U}\hat{\Sigma}$ and then multiplying both sides by $V^{-1}$

[stoic python]

Ryuzaki

[lavish jewel]

right, and V is unitary, so V^-1 = V^H

[sinful valve]

yeah basically

[lavish jewel]

in some places this is written as V^*

[sinful valve]

unitary matrices i have never heard of ill have to check that

[zinc timber]

basically orthogonal but in the context of complex sapce

[lavish jewel]

it's the complex flavor of orthogonal matrices

[sinful valve]

why do other svd not talk about this version though

like explanations of it

[lavish jewel]

wdym

[sinful valve]

complex number version you mentioned

[lavish jewel]

they all mention that U and V are unitary. in the real case, they mention they are orthogonal

[sinful valve]

uh some didnt i dont remember but

yeah ill just see

is there any use case for the complex numbers in svd though?

[lavish jewel]

more than use case, you can't avoid it

when all the entries in the matrix are real though, the SVD will be as well

or at least it CAN be, since the svd is not unique

[sinful valve]

not unique as is in its general to any matrix you mean

[zinc timber]

means given a matrix, the u, v, s matrix will always be the same(upto permutation)

[sinful valve]

so its literally just

the same idea but a different way of inverting it

[zinc timber]

yes

sometimes written as U^* instead of U^H

well most of the time

[sinful valve]

yeah wikipedia said there is like 4 notations to use

that dagger looks cool tbf might use that

[zinc timber]

the dagger and + I use for pseudo inverse, never knew physicians use that for conj-transpose

lol

[sinful valve]

so in my guys video then

he has used a unitary matrix to include all real/complex basically

to make it general ig

cause my book only cares about real so it just say orthogonal

[zinc timber]

yeah

[sinful valve]

ight

[zinc timber]

you can get away with saying orthogonal if the space is real

[sinful valve]

when would you use an svd in complex space though

[lavish jewel]

whenever it's handy to use complex numbers

so usually engineering and physics applications

[sinful valve]

yeah thats fair in computer science i dont see many uses tbf

[lavish jewel]

also, the svd unique up to rotations

[sinful valve]

as far as i know at least

[lavish jewel]

unless you have repeated singular values, permutations won't show up

you might use it in CS as well, since stuff like rotations are easily done with complex numbers

at least in the more applied stuff like computer graphics

[sinful valve]

yeah this is for CG this content

the only complex numbers i cant think of are in the fast fourier transform

[lavish jewel]

then you'll need complex numbers

rotations, ffts, waves, etc

[sinful valve]

are you referring to quaternions

in cg

thats true tbf i havent done much on them but they seem like complex numbers

more of an extension of them to more dimensions

[lavish jewel]

i was talking more like rotations on a 2d plane, but sure

[sinful valve]

we dont need complex numebrs for that i guess

but it works either way of course

[lavish jewel]

you don't need quaternions either, strictly speaking

[sinful valve]

nah you dont

[lavish jewel]

but using them makes the computations easier

[sinful valve]

yeah its also gimbal lock or sumn

but i dont knwo much bout that

i just remember doing a non quat one you cant liek go all around without ebing cut off

like you get locked you cant freely rotate

[zinc timber]

yeah depending on the rotational system you use, it can happen

[winged prairie]

hello, I was lloking at the solution for this problem but i do not understand how they can map some of the basis vectors to 0 because wouldn't that make the set of vectors that get mapped to w linearly dependent?

[stable kindle]

wouldn't that make the set of vectors that get mapped to w linearly dependent?
why would it do that

wait ok no i think i see

[zinc timber]

you are looking at the finite version of the Hahn-Banach theorem

[winged prairie]

but is my intuiton corect?

[stable kindle]

i don't think it's quite correct

[winged prairie]

why

[stable kindle]

so why should it

make them linearly dependent

you can't map all the vectors that aren't in U to 0

but you can map the basis vectors that aren't in U's basis to 0

and then the remaining basis vectors, the basis vectors of U, determine the actual position of things

[winged prairie]

i don't get this

my idea is that by doing the mapping shown in the picture

u map over all the basis vectors of U + the 0 vector

[zinc timber]

yr intuition is correct

[winged prairie]

so why does that not make it linearly dependent

[stable kindle]

why would it make it linearly dependent

also why is the 0 vector important

you can get the 0 vector out of the basis vectors

[winged prairie]

because any vector can be written as a linear combination of another

[winged prairie]

why

[stable kindle]

it's just the linear combination of all of them

with coefficients 0

[winged prairie]

oh is it by the theorem which says that if u have a set of vectors that span a vector space

u can reduce it to a basis ?

[stable kindle]

i mean

what

[winged prairie]

if it has 0 it cannot be a basis

[stable kindle]

ok so

let v1, ..., vn be a basis

then consider the list 0, v1, ..., vn

[winged prairie]

ye

so..?

[stable kindle]

sorry back

[winged prairie]

dw

[stable kindle]

this list is linearly dependent because you can get a linear combination of those elements where not all the coefficients are 0, but they sum to 0

[winged prairie]

yes i agree

[stable kindle]

ok

so...

you can just remove 0

and then it's independent

it's like

why bother talking about 0

[winged prairie]

and you can remove the 0

[stable kindle]

0 always goes to 0

[winged prairie]

because of this

right?

[stable kindle]

and you can get it from the other elements

sure

so what

[winged prairie]

i was just trying to understand

that's all

ty

[stable kindle]

??

[winged prairie]

like u confirmed

[stable kindle]

ok if you got something from that that's good

[winged prairie]

ye

[stable kindle]

i don't know what's happening here lol

[winged prairie]

i was confused but now i get it

dw haha

i wanted to understand if removing the 0 was possible

[hard drum]

[stable kindle]

i don't understand

[winged prairie]

dw man

u helped me that is all

i will be back a lot these days

[zinc timber]

the heck q vegitable doing in liner algebra

@winged prairie also search Hahn-Banach theorem

for more info

[winged prairie]

ayt

ty

[winter harbor]

[winged prairie]

any hint on how to do this, i don't wannt look at the solution right away

[hard drum]

You can explicitly construct linearly independent vectors in L(v, w)

[winged prairie]

so i take a basis in V

then what

and map it to W

?

[hard drum]

I mean that's just a way to construct maps V to W so yes i guess

[still lodge]

whatd i do wrong here :(

probably something silly but i dont see it

[dusky epoch]

,w det {{1,-2,3,1},{0,2,4,-1},{2,0,1,2},{1,0,5,0}}

[stoic python]

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[dusky epoch]

,w det {{-2,3,1},{2,4,-1},{0,1,2}}

[stoic python]

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Requested by Ann#0413

[dusky epoch]

,w det {{1,-2,1},{0,2,-1},{2,0,2}}

[stoic python]

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Requested by Ann#0413

[still lodge]

🤦‍♂️

[dusky epoch]

oh i see where you went wrong

you forgot a minus sign

[still lodge]

i just copied a sign wrong

what if you wanted a 100 but your brain said (+)

[zinc timber]

u got a generous teacher ngl

[still lodge]

[fresh canopy]

I tried to use the Gramm Schmidt process to solve this problem, was that the correct approach?

[zinc timber]

can't read. what's the question?

[keen flame]

what would be the vector of function f(x)=cos(2x) when [-pi,pi]?

[fresh canopy]

Gimme a sec I'll just post a screen shot

[zinc timber]

????????????

[zinc timber]

and u hv to use GS orthogonalization fr tht?

hint: ||extend the basis to a basis of R⁴ then apply GSO||

[fresh canopy]

Thats the section it's in, so that's my assumption

ok I know I'm showing my L.A. iq here, but what do you mean extend the basis to a basis of R^4?

[zinc timber]

extend (v1,v2) witch is a basis of W to a basis of R⁴

[lavish jewel]

take u and v and come up with 2 more vectors such that all 4 are lin indep, and therefore span R^4

[zinc timber]

there are easier methods however

like $ \mqty[u^T \ u^T]\mqty[a \b \c \d] = 0$

...

tf?

[lavish jewel]

you left a space after the first $

[fresh canopy]

so I know if took u cross v, that would give me one more of the vectors I need to get to R^4, how would I get the 4th? could it just be arbitrary?

[zinc timber]

apparently I wrote 2 u's

[lavish jewel]

[zinc timber]

[lavish jewel]

sadly the x prod is only defined in R^3

[zinc timber]

when I copied it only one \ was copied

[fresh canopy]

ahh, I didn't even think of that.

[zinc timber]

like wtf

[quasi vale]

Is it true for operators that (ST)^n = S^n T^n?

[lavish jewel]

an alternative is to build a matrix with u and v as rows, and find its null space

ah

that's what ryu was getting at

[lavish jewel]

in general, no

only if they commute

[quasi vale]

what i currently have is $(T - \lambda I)^{n}$ and $(\lambda T)^{-n}$. Is it true that $(\lambda T)^{-n} (T - \lambda I)^{n} = ((\lambda T)^{-1} (T - \lambda I))^{n}$

[stoic python]

Sup?

[quasi vale]

You guys may answer Heathus first, I'm sorry to interrupt.

[zinc timber]

in this specific example it's true

[quasi vale]

Thanks!

[fresh canopy]

I'm going to find the null space given V and U are transposed, please feel free to help others while I work on this.

Then we can see what happens next!

[zinc timber]

that will be much more easier than extending the basis, then orthogonalizing

[fresh canopy]

ok good for my skill level that's the best choice 😉 Thank you

[true girder]

Hello I'm struggling a bit with homework problems involving eigenvalues and eigenvectors. I understand how to do the full problem conceptually like ik the steps but the part where u do det(A) = 0 and have to solve for 0 im really not getting anywhere.

the help channels appear to be only for pre-uni hw so this was my safest bet

[zinc timber]

post the question statement and your work

[true girder]

i skipped the first step of writing the question but the matrix is all 1's accept the diagonal is all 2's

[zinc timber]

so u r struggling finding the characteristic equation?

[true girder]

ye

my prof said to not multiply anything out but for everything after the 1st det its all single values

[zinc timber]

it's just routine calculation

[true girder]

ik but for some reason im struggling with this. like which route is best for solving it

is the very bottom correct?

[zinc timber]

,w char poly {{2,1,1},{1,2,1},{1,1,2}}

[stoic python]

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Requested by Ryuzaki#5323

[zinc timber]

looks like you've messed something

[true girder]

ok the first part of line 2 is correct somewhere below it i missed something

[fresh canopy]

Looks like finding the null space of U^T and V^T was indeed the answer! Thank y'all so much.

My question is, how did we know that finding the null space of their transposes would be the orthogonal compliment of the those two vectors? Is that just always the case?

[lavish jewel]

the null space is by definition orthogonal to the row space of a matrix

this is because you can write matrix-vector multiplication as the result of taking dot products between a matrix's rows and another vector

say you have a matrix [u v]^T, where u and v are column vectors. this gives you a matrix that is size 2 x n

now multiply that by some vector x of size n x 1

the resulting vector is of size 2 x 1, and the elements of the vector are u^T x and v^T x

u^T x = u dot x

so if all the entries of the resulting vector are zero, the vectors that satisfy this are orthogonal to both u and v

furthermore, if these vectors are linearly independent from u and v (and they are, because they're orthogonal to them) you can use them to extend u and v into a basis of R^4

and now consider the subspace spanned by u and v, and the one spanned by the vectors you just found

any pair of vectors where one vector is taken from one subspace and the other, from the other subspace, are mutually orthogonal

[fresh canopy]

Thank you so much for breaking that down. That makes so much sense!

[true girder]

ok i figured out where i went wrong

but now i dont remember how to solve for x^3

ik the quadratic formula for x^2 and x^4

but this is where im gonna struggle

[stable kindle]

so for cubics you just want to stare at it until you find a solution

they won't give you a cubic with awful roots

in a question

so for example -1 should work here

[true girder]

i need a formula / system or ill be staring til im handed an F

[stable kindle]

-1

basically

you try 0, 1, -1, 2, -2, 3, -3

and if none of those work then you cry

[lavish jewel]

isn't it 1?

[stable kindle]

fuck

ok it's 1

[lavish jewel]

well idk if you meant the factor or the root tbf

but lambda = 1 should make the poly 0

you can go from there. maybe divide the poly and then use the quadratic formula if needed

[true girder]

https://www.youtube.com/watch?v=Zca-gVo-cPU

this video looks good

ye my math brain died in high school so now im just praying that i pass any required math for my major

but thx for the help!

ok so i played with it and the video did not help at all.

but ye 1 works

but how do i get the rest

there arent any like terms and all i can really do is factor out a lambda

ok so the only thing i noticed from just finding the answer online is that since the answer contains a (x-4).... one of the factors is actually 4 so maybe thats a hint

[quaint steppe]

Practicing some exercise in the book. Is my proof correct? Is there a better proof?

[sick sandal]

can you elaborate further on your 1st proof im not sure i follow what you did,
as to how i would do it ||probably prove injectivity by assuming ST(u1)=ST(u2) and reach u1=u2 since injectivity is equivalent to invertiblility in finite dimensional spaces for some transformation(if dimemsions are equal for U and W )||

[quaint steppe]

this is new to me.That injectivity is equivalent to invertibility. For my first proof i just assume that since S is invertible and S(Tu) so the linear map here must also be invertible.

Idk if it's the right track.

[reef latch]

doesnt invertibility require bijection

[wintry steppe]

injectivity is equivalent to invertiblility in finite dimensional spaces for some transformation(if dimemsions are equal for U and W )
the spaces could be infinite-dimensional in this exercise, so this might not apply. you'd need to go the extra step and show surjectivity as well

[wintry steppe]

how does this prove that ST is also invertible? you haven't explained why ST has an inverse, you just wrote down the definition of ST and said "and it follows that this is invertible"

you need to elaborate

this proves both parts anyways, btw

[steel moon]

how would i go about creating a subspace in a?

[wintry steppe]

visualize it :^)

[steel moon]

we want to show every vector in R^4 can be written as a sum of an element in W and some subspace

how can you visualize R^4

[wintry steppe]

it might help to consider a hyperplane in R^3 first, where you can argue geometrically. then, see if you can transfer your reasoning over to a hyperplane in R^4

[wintry steppe]

weed

[reef latch]

3blue1brown

[steel moon]

😹

[reef latch]

https://www.youtube.com/watch?v=zwAD6dRSVyI

[sinful valve]

you know if i had a shape like on a graph right

with the original F already there

so i know the coordinates of start and end

how would i calculate transformation matrix to get there

[reef latch]

get the basis vectors?

[sinful valve]

i only have the coordinates of each vertex by the looks

vertices

[reef latch]

which vertices are that

basically you can construct basis vectors by drawing them on top of the F

[sinful valve]

[9.812, 10.1963302752294, 14.1855010660981, 13.4, 10.8298217179903, 10.6398305084746, 12.0611940298507, 11.7371879106439, 10.4931163954944, 10.2813455657492, 9.812],
[5.649, 8.39266055045871, 7.79957356076759, 6.91607142857143, 7.39059967585089, 6.74435028248588, 6.44328358208955, 5.95532194480946, 6.24530663329161, 5.5249745158002, 5.649]

you see the issue xd

[reef latch]

right so get the bottom left vertex, get the bottom right

[sinful valve]

can i like plug these into a linear equation though

these are just x, y pairing first array is x second is y

[reef latch]

subtract bottom right from bottom left, theres your horizontal basis

[sinful valve]

these 2 you mean

[quaint steppe]

sorry been busy. So, i can use this for the first proof?

[sinful valve]

in my book it says you can just line up the coordinates in an equation though will that not work

[reef latch]

yes and presumably the F starts out completely straight?

[sinful valve]

so i think my x and y of the transformed shape are joined on right

[reef latch]

thats weird, whats with the big matrix

I guess I'm not familiar with whatever this is :)

[reef latch]

oh wait, it's a transformation that includes translations?

[sinful valve]

it includes anything

shears /rotations

[reef latch]

I see

[sinful valve]

so in that recent slide they have the column with some single vertex

on LHS with the trasnformed point

and they are solving teh matrix to get to it i think

where the matrix is the unknowns

in my example i think i would just have a fat column vector

[reef latch]

I guess the thing with the basis vectors wont work then

[sinful valve]

with [x1,y1, x2, y2...., xn, yn]

ill see if there is an online tool to solve this exact issue

[reef latch]

hmm, I feel like 3 points should be enough to characterize any affine transformation

[still lodge]

are there any named theorems besides cayley hamilton that i should know as an undergrad

[reef latch]

I hear Pythagoras Theorem is all the rage

[sinful valve]

3 points how though

it may be a homography acc my bad

[reef latch]

well just think about it

like, idk, put three points on a graph

[sinful valve]

yeah ofc only 3 for affine tbf true true

dunno why i said not

[reef latch]

2 is enough for linear transform

[sinful valve]

so i dont need to pass all my vertices in

i can just solve teh matrix from one

[reef latch]

but like, you need to distinguish between a translation and a series of linear transforms

hence the third point

[sinful valve]

yeah I get that

uh just need to find out how to compute these things

by that then are you saying from my F

I need to get any 3 points of x and y

[reef latch]

probably!

[sinful valve]

to solve the affine matrix

ight ill see

[winter harbor]

You mean, in linear algebra?

Stuff you need to know (In my opinion)

Cayley-Hamilton Theorem

Steinitz Exchange Lemma and the Dimension Theorem (Existence of a basis and Uniqueness of dimension)

Spectral Theorem for Normal Operators (and its corollaries such as Spectral Theorem for Hermitian (Self-Adjoint) and real symmetric operators).

Rank-Nullity Theorem (First Isomorphism Theorem)

Lémme de Noyaux (a.k.a primary decomposition theorem)

Jordan Canonical Form Decomposition

Jordan-Chevalley Decomposition

Schur's Theorem (Schur's Decomposition)

SVD decomposition theorem

Rational Canonical Form and Smith Normal Form (maybe?)

Cholesky decomposition theorem

Cayley-Hamilton's Theorem

Stuff you maybe need to know (I guess).

Gershgorin Circle Theorem

Sylvester's Law of Inertia

Spectral Mapping Theorem

Witt's Extension Theorem and Witt's Decomposition Theorem

Hadamard's Inequality

Cartan-Dieudonné theorem

Idk, these are prolly all the linear algebra theorems I know that have a name I guess.

The "stuff you need to know" is prolly shit people go through in most intro to linear algebra classes I guess.

[quaint steppe]

I'm sorry if I'm constantly asking if my proof is correct. I don't have anyone to ask to. And i don't have a solution manual to check. So yeah Is my proof on this exercise correct?

I'm an undergrad and really new to college maths

[winter harbor]

This is not right

I wi give you a hint

Since dim(V) > 1

[fresh canopy]

So for this problem, I know I can just set up "Ax=b" and multiply the right vector by the inverse to solve for "x", which would be the coordinates being requested. But is that the ideal way to solve it? Should i be using QR factorization for this problem?

[winter harbor]

Choose a basis $e_{1}, \cdots, e_{n} \in V$ of $V$ and let $T,U \in \mathcal{L}(V)$ be such that:
\begin{align*}
T(e_{1}) = 1
\
\
T(e_{i}) = 0 ; \forall i > 1
\end{align*}
and
\begin{align*}
U(e_{1}) = 0
\
\
U(e_{i}) = 1 ; \forall i > 1
\end{align*}
Then, notice that
$$
(T+U)(e_{i}) = 1 ; \forall i \geq 1
$$
And thus, $T+U = \text{Id}_{V}$ is the Identity on $V$, which is invertible.
\
\
Therefore, we have that the set of non invertible operators is not closed under addition and as such is not a subspace of $\mathcal{L}(V)$.

Notice that we are explicitly using the fact that dim(V) > 1 here in order to construct T and V.

Moreover

You can fill in the details yourself if you want

But T and U are well defined

Because we especified how the act on a basis of V

And we can use this to extend T and U to to whole of V by linearity like this.

And uniquely as such

I should be more precise on why your argument does not apply.

It is NOT true that every linear operator T : V -> V between finite dimensional vector spaces is invertible.

You could take, for example, the 0 linear operator.

But that are more interesting examples

Such as the ones I have constructed.

Nilpotent operators are another important class of operators which are not invertible.

[stoic python]

MISTERSYSTEM

[winter harbor]

What is true tho

[quaint steppe]

ohhhh so what does this mean?

[winter harbor]

That's what I was about to say

As a consequence of the rank-nullity theorem

If an operator T : V -> V is injective, then it is invertible.

Or if it is surjective, then it is invertible.

But not every linear operator between finite dimensional vector spaces satisfy these criteria.

For example, 0 : V -> V which takes a vector v in V and does 0(v) = 0 is not injective nor surjective.

[quaint steppe]

ohhhhhh

thank you very much

im taking notes

[vital pagoda]

one of the things which define a normed space is that $| x | = 0 \iff x = 0$ where $x \in X$ where $X$ is a vector space

but i dont understand why that's needed

is there an example of something which satisfies the other two norm axioms but doesn't satisfy this one?

[stoic python]

Joy!

[winter harbor]

Yup, there are.

[vital pagoda]

isn't it obvious from the definition of the modulus that this will be the case though?

[gray dust]

take the constant 0 map

[still lodge]

random unrelated interjection - i asked about the studying role before but what is/how do i get permastudying (i saw it on system)

[gray dust]

ask for it. want it now?

[vital pagoda]

smh

ok yeah the constant 0 map does indeed satisfy the other two while violating positive definiteness

im still not fully comfortable with the idea of a norm tbh, does it get more natural once you learn topology?

[winter harbor]

Let $(X, \mathcal{B}, \mu)$ be a measure space, where $X$ is a set, $\mathcal{B}$ is a sigma algebra on $X$ and $\mu : \mathcal{B} \rightarrow \mathbb{R}{\geq 0} \cup {+ \infty}$ is a measure on $X$.
\
\
Let $1 \leq p < + \infty$. We define the real vector space:
$$
L^{p}(\mu) = \left{f : X \rightarrow \mathbb{R} : \int\limits{X} |f|^{p} , d \mu < + \infty \right}
$$
We can prove that for any $f \in L^{p}(\mu)$ we have that the function defined as:
$$
|f |{L^{p}} := \left(\int\limits{X} |f|^{p} , d \mu \right)^{\dfrac{1}{p}}
$$
Satisfies all the other product of a norm. I.e, the triangle inequality (here known as Minkowski inequality). It is positive homogeneous and we have that $| f |{L^{p}} \geq 0$ for every $L^{p}$ function.
\
\
But, we have function with $|f |{L^{p}} = 0$, but $f \neq 0$. Basically, if $f$ is 0 almost everywhere (i.e 0 everywhere up to a set of null measure), then $|f|_{L^{p}} = 0$ even if $f \neq 0$.

Oof

[vital pagoda]

damn yeah buddy I'm still a few months away from learning about measure spaces

[winter harbor]

This example comes from functional analysis.

Yeah

I couldn't think of a good example on finite dimensional vector spaces

The 0 one is prolly the easiest

[stoic python]

MISTERSYSTEM

[vital pagoda]

whats interesting is that what inspired my question is the proofwiki proof of the p norm on the p sequence space being a vector space norm

[winter harbor]

Ohhh

Well

That's actually weird

[vital pagoda]

ahahahaha

[winter harbor]

Btw

This is called a pseudo-norm

[vital pagoda]

bruh just when i thought the rabbit hole doesnt get any deeper

[winter harbor]

I can't think of any non zero pseudo norm which is not a norm.

Btw

The thing with L^p spaces is that we usually identity functions that agree almost everywhere.

[vital pagoda]

yeah i looked at the definition of the pseudo norm and immediately II . II_0 popped in my mind

[stable kindle]

proofwiki is kinda weird

[winter harbor]

So that we in fact have a norm.

And not a pseudonorm

This is usually a technical detail

[lofty topaz]

is this a vector subspace ?

[still lodge]

what are the requirements for something to be a subspace

[lofty topaz]

1 include zero vector
2 closure under vector addition

3 closure under scalar multiplicatioon

[stable kindle]

stupid q

aren't 2 and 3 sufficient to give 1

0 is a scalar

am i wrong?

[winter harbor]

These 3 conditions are equivalent to

1 it is a non empty subset
2 closure under vector addition
3 closure under scalar multiplication

[still lodge]

i mean youre right but no need to be condescending, it can be helpful to check if 0 vector is included as a way to eliminate the other 2 sometimes

[stable kindle]

?

[winter harbor]

We need the subset to be NON EMPTY in order for it to be a subspace.

[stable kindle]

no i was saying i was about to ask a stupid question

i wasn't saying your q was stupid

[still lodge]

oh

[winter harbor]

The empty set satisfies the axioms 2 and 3

But not 1

[stable kindle]

right

[still lodge]

then yeah

[lofty topaz]

but are these of two vector subspaces ? I'm struggling with a notation of proof

[still lodge]

i mean you seem to have two sets there

so it depends on the particular instructions...?

[lofty topaz]

I need to prove for every set if it is a subspace

[reef latch]

iirc a subspace is closed under linear transformation

[sinful valve]

i dont think the method is working tbf

this one i solved the values

but somethign is off

im not even sure if this is even possible

[nocturne jewel]

subspace is closed under addition and scaling

to which similar properties are linearity

[sinful valve]

is there any other way to find the parameters of an affine trasnformation

given the start and end shape

like we have this F right and i computed the transformation for a using that method

but the matrix didnt move it correctly not sure why

what is the standard method of getting the affine matrix uesd in a transformation

[glad acorn]

I have no idea how to work on this question

[reef latch]

seems like it should be orthogonal to the other two basis vectors

[stable kindle]

this seems false

let a basis for V be v1, ..., vn

oh wait no it works

[glad acorn]

It dosen't work. For example, <-6/5,-2/5,1,0> is the vector orthogonal to the other but it's not the answer. okay I've figured out the method.

[tribal fossil]

-> Prove that every m × n matrix over the field F is row-equivalent to a row-reduced matrix.
-> Let e be an elementary row operation and I be the m × m identity matrix. Prove that for every m × n matrix A, e(I)A = e(A).

Hey someone help me prove these two theorems??? pls

[still lodge]

can someone help me understand why a matrix (let's say B = {1 1}{0 1}) acts as a change of basis from B to E where E is the standard basis pls

ive been trying to get this into my head since fucking freshman year and it still doesnt click

just a discussion would be v nice

[zinc timber]

@glad acorn has it been resolved?

[zinc timber]

obviously

[still lodge]

yeah i was gonna say, isnt that like... the definition of those things...?

[tribal fossil]

i mean my teacher asked for a neat proof

[zinc timber]

given a matrix, u apply row operations blah blah and u get a row reduced matrix

so original matrix~the row reduced matrix we have just calculated

[still lodge]

ig you could do some visual stuff?

[zinc timber]

do I need to?

[still lodge]

also pls 👉 👈

[zinc timber]

I would have but I'm on my phone now, hard to type, I'll try once I get back to my pc

[still lodge]

<3

any chance on an ETA? i'll be here for a while but will probably close discord for a while

[zinc timber]

ok i'll whatever I can for now

say in 2D you have 2 LI vectors (1, 1) and (1,0). say I want to use them as my new coordinate vector. so any (x,y) vector in 2D can be written as a•v1+b•v2

i.e. the vectors (1,0) and (0,1) which was our old basis can be represented in terms of the new basis v1, v2

so let the the new coordinates of e1, e2 be

$a v_1 + b v_2 = e_1 \
c v_1 + d v_2 =e_2$

[stoic python]

Ryuzaki

[zinc timber]

Now this can be written in the compact form
$\mqty(v_1 & v_2 ) \mqty(a & b \ c & d) =\mqty( \imat{2})$

[stoic python]

Ryuzaki

[zinc timber]

now you can solve for a,b,c,d with it

similarly given we know the coordinate wrt one basis we can transform them into another just by multiplying the inverse of the basis transformation matrix

[still lodge]

so ig it's bc of how that matrix of the new basis vectors (or rather each basis vector) acts upon our basis E...?

[quaint steppe]

Hi again, practicing another exercise in the textbook. Is my proof correct? Can it be proved better?

[zinc timber]

no

you are supposed to show a invertible lt exists given S is injective

but here u assumed the existence

[quaint steppe]

all rightt

[zinc timber]

hint: || extend to a basis ||

[teal grotto]

if S is injective, then take a basis {u1,…,uk} of U. then extend this to a basis {u1,…,uk,v1,…,vr} of V (where of course k = dim U and k + r = dim V). try to define a new function T determined solely by the new basis of V

[quaint steppe]

Thank you. I kinda get it already. I also saw the proof online

[viscid lily]

I have constructed the matrix with importance scores, but dont know how to solve to get the page rankings

Can anyone guide me?

[zinc timber]

[cobalt tartan]

Wanted to doubel check something: Given an arbitrary inner product, does that change how matrix mulitplication works? I.e, say that my inner product is <u, v>, then is matrix multiplication now applying <u, v>, or is it still applying the "standard" dot product?

[zinc timber]

didn't understand what your question is exactly

[cobalt tartan]

So say that you define an inner product over R3, $<x, y> = x_1y_1 + 2x_2y_2 + 4x_3y_3$

[stoic python]

Liria ^(;,;)^

[zinc timber]

so you mean a inner product induced by a matrix?

like $\innerproduct{u}{v} = v^TQu$

[stoic python]

Ryuzaki

[cobalt tartan]

Then normally, when you do matrix multiplication of $AB$, you're applying the standard dot product between the rows of $A$ and the columns of $B$

[stoic python]

Liria ^(;,;)^

[zinc timber]

for that you must have a positive definite matrix Q

[cobalt tartan]

Then in this IPS over $R^3$, when we multiply $A$ and $B$, do we apply the standard dot product, or do we apply this inner product?

[stoic python]

Liria ^(;,;)^

[cobalt tartan]

Yes, I believe that is my question

As this is the definition of inner product in my textbook

[zinc timber]

actually no, matrix multiplication is universal

[nocturne jewel]

IP acts on the vectors, so idk why you're worried about its effect on matrices

when matrices arent the vectors

[cobalt tartan]

Ok, so it just follows standard matrix multiplication

Ok

[lavish jewel]

probably cuz they noted the result of matrix mult is an array of dot products

[cobalt tartan]

Yea

[lavish jewel]

but when you need to interpret it as such, it'll be apparent

[cobalt tartan]

And so matrix multiplication, even though it looks like a series of dot products, isn't really so?

[lavish jewel]

it won't just be the composition of 2 linear transformations, but explicitly written as a matrix whose elements are inner products

[zinc timber]

it's a composition of LT's (in their matrix form). The basis is already taken into account in the matrix representation so you don't need to apply it again while multiplying

[cobalt tartan]

ok

[lavish jewel]

for a nice symmetric pos semi def matrix M, if you want to compose something whose elements are inner products w.r.t. that matrix, you'd get a matrix whose elements are of the form u^T M v, and then you could factor it as U^T M V or something

but then that no longer has the interpretation of composition of linear transformations

so, not a "matrix" in the usual sense

[zinc timber]

yeah but when you try to multiply with the adjoint, then it'll be different. but I'll not go into the rabbit hole for now

[wintry steppe]

Hey Liria if you don't mind me asking what book is that?

[cobalt tartan]

Some prof at my school wrote a textbook specifically for our linear algebra courses

[wintry steppe]

I see thanks.

[keen mirage]

help

how do i show dim(S(V))=n+n(n-1)/2

<@&286206848099549185>

[zinc timber]

For a symmetric bilinear form (which is a symmetric matrix) we must have $a_{ij} = a_{ji}$ given $i\neq j$

[stoic python]

Ryuzaki

[zinc timber]

means you are only free to choose arbitrary value for matrix where j>= i

[keen mirage]

ahhh yeah

[zinc timber]

that answers your question?

[still lodge]

is it accurate to say that the alg multiplicity of an eigenvalue is the degree of the part of the polynomial in which said lambda sits (ik this is worded weird) and the geometric multiplicity is how many vectors make up it's nullspace..?

[keen mirage]

yeah

symmetric bilinear form correspond to symmetric matrix

[zinc timber]

though your terminology is not accurate, I get what you are trying to say. Yes it is the case

[still lodge]

i can do problems with it i just wanna make sure the wording in my head is right so merci

*or at least the idea behind it

and a matrix is defective is the alg multiplicity of any of its eigenvalues is greater than the corresponding geometric multiplicity

[zinc timber]

If
\begin{align}
p(\lambda) = (\lambda-\alpha_1)^{k_1}(\lambda-\alpha_2)^{k_2}\cdots(\lambda-\alpha_n)^{k_n}
\end{align}
then $k_i$ is the algebraic multiplicity of the eigen value $\alpha_k$ and the geometric multiplicity is $\dim \text{ null } (T-\lambda_iI)$

[stoic python]

Ryuzaki

[still lodge]

yep that is what i meant, epic

is there a faster way to do this than rational root thm

that's easy enough to apply but tedious ofc

[zinc timber]

yes

,w roots x^3-10x^2+33x-36

[stoic python]

Results provided by WolframAlpha

Click here to refine your query online
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Requested by Ryuzaki#5323

[zinc timber]

[stable kindle]

yeah i was thinking 3 from the start

[still lodge]

yeah i'll just pull out WA on my exam no biggie

[stable kindle]

just stare at it until you see something that looks good

[zinc timber]

,w formula for cubic equation

[stable kindle]

god that looks like shit wtf

[still lodge]

because it is shit

[zinc timber]

yeah or memorize this ig

lol

[stable kindle]

no like even shittier than usual

anyway here's how you actually solve cubics

basically

try 0, 1, -1, 2, -2, 3, -3

and if none of them work then you're doomed

[still lodge]

yall got any way to intuit how many roots it might have

[zinc timber]

= degree

[stable kindle]

it will always have at least one root of the form 0, 1, -1, 2, -2, 3, -3, by Kai's Conjecture

then just do the long division trivially

and you can tell pretty easily if it'll have more

[still lodge]

bleh fine

[quaint steppe]

What does this fancy R means?

[zinc timber]

range of T2

[quaint steppe]

all right thank you

[still lodge]

ive asked this before but... what's the dif between invariant factors and elementary divisors

ik the latter depends on our choice of minimal polynomial

are invariant factors just each distinct elementary divisor of every possible minimal polynomial (in cases where you dont know the minimal polynomial ofc)

[granite kraken]

I solved this question using adjunct matrices but my professor said to solve it without using adjunct matrices. Does anyone see a possible solutions start or tips to this problem?

[lavish jewel]

if A is diagonalizable, the matrix exponentiation is pretty simple, one need only exponentiate the diagonal matrix of eigenvalues

then you can factor out the matrices of eigenvectors to the left and right, leaving you with a sum of diagonal matrices of the form of the polynomial

[ionic laurel]

Hi I have a question

If you have two bases B and C

would the change of coordinates matrix from B to C be the inverse of C to B and vise versa?

B to C is the inverse of C to B
C to B is the inverse of B to C

[lavish jewel]

yes

[ionic laurel]

👍

My professor wrote it as

[lavish jewel]

in general, you put the basis vectors as columns of a matrix M

[ionic laurel]

the inverse of B to C is is C to B

but I did not know if the converse would be true

thank you nonetheless

[lavish jewel]

then if x are the coordinates in that basis, we have that v = Mx, so that v has coordinates x in the basis M

if you do it backwards, then you get that M^-1 v = x, and now v is the coordinate vector for x in the basis of the columns of M^-1

[raven parrot]

Is there an easy way to LU Factorize a nxm matrix? The method I know and use is to RREF to get U, but in the process, get values for L.

[lavish jewel]

by hand, that's about the easiest

[swift minnow]

Hi i need help

If A,B are squared matrix, and AxB is unsymmetrical so A and B are transposed

[marble lance]

To disprove it, you just need to give a counterexample

[swift minnow]

do you have any example? i have tried

i didn't find any

is it a proof or disproof

[marble lance]

@swift minnow Sure

Consider any symmetric matrix

Then take one of the entries not on the diagonal

And +1 to it

And then take the same entry and -1 to it

The two matrices you get won't be symmetric

But their sum will be

[mystic dagger]

is this conclusion right?

like, h(x)=-h(x) breaks the definition of function right

[stable kindle]

nope

let h(x) = 0

[mystic dagger]

it would be the x-axis right?

and the only function that obeys h(x)=-h(x)

yes bc 0=-0

oh i got it

thank u

[swift minnow]

can you give me an example?

[marble lance]

I told you how to construct an example just now

[swift minnow]

so is it a disprove?

[marble lance]

Create an example the way I said

Then that serves as a counterexample, yes

[swift minnow]

i asked you the wrong question at start and i guess you answered it , my apologoy.
i edited the question

i typed the wrong question

[marble lance]

Is the new question: If A and B are square matrices such that AB is not symmetric, then A = B^T?

Or what do you mean by A and B are transposed?

I don't get it

[swift minnow]

it means that a x b = b x a

that's what i mean by transposed

a x b = anti symmetric

[marble lance]

I don't think you are using terminology correctly

Can you send a screenshot of your actual question?

[swift minnow]

Ye thing is that’s in a different language 😅

[wintry steppe]

Hello I have a possible LA question in #computing-software 😄 any help is appreciated

[granite kraken]

Anyone got any clue how to start this problem,

[zinc timber]

let p=algebraic multiplicity of the eigen value

[finite plume]

Where have I done wrong? The correct answer is sqrt(186)/2.

[dusky epoch]

your P1P2 and P1P3 are wrong

they should be (3, -1, 5) and (-1, 0, 4)

[finite plume]

Wouldnt it be (2, -1, 5) then? On P1P2.

[dusky epoch]

oh oops, typo on my end

you're right

[granite kraken]

Can anyone help?

[dim epoch]

ryuzaki gave you a hint

[umbral void]

If I have an $n\times n$-matrix with integers on the diagonal and 1s on the "immediate parallels" of the diagonal, can I say something about its signature?
To be more precise, I mean matrices like
[ \begin{pmatrix} a_1 & 1 & 0 \ 1 & a_2 & 1 \ 0 & 1 & a_3 \end{pmatrix} ]
and by signature, I mean $#$ positive eigenvalues $- #$ negative eigenvalues

[stoic python]

expectTheUnexpected

[umbral void]

i.e. this

[zinc timber]

can immediately think of Gresgorion thm, other than that, I'll have to think

[winter harbor]

You mean Gershgorin Circle Theorem?

[zinc timber]

yeah but that won't be of much help I think

there's also one convoluted method usigng strum sequence

[umbral void]

Ok, I guess I'll just not write a general formula then would have been nice, but I suspect that it's hard

[zinc timber]

If you wanna know I can just give a ss of the theorem

all yr b_i will be 1

[umbral void]

oh no, recursion

[umbral void]

was to be expeceted

[zinc timber]

then you can put v=0 and check for sign changes

[umbral void]

oof, gotta think

thank 👍

[zinc timber]

I'wd like to know anything better myself