linear-algebra

[dusky epoch]

do you have the entire problem statement @pine scaffold

[pine scaffold]

i was solving this but now that u say that it is impossible to determine uniquely ig i made a mistake

nvm found my mistake thanks tho

[scenic coral]

I need to find $f_3^-1(M_3)$ Now my issue is that the results are infinite because sin(x) is periodically. But i guess i need to write the results somehow as an interval? How can i write it as an interval while taking care of the fact that its periodically?

[stoic python]

ＬＹＮＩＸ:

[round coral]

@scenic coral you need to find the range of arcsinx where x belongs to [0,1)

the range is simply [0, pi/2)

the fact that it is periodic makes it easy not difficult

[scenic coral]

@round coral but aren't there numbers out of the range of [0,pi/2) for which sin(x) is in [0,1)? For example sin(5/2 pi) is not in [0,pi/2) but is 1 which is in [0,1).

oh nevermind i understand now

Thank you very much

[nocturne jewel]

y'' +a(x)y'+b(x)y = 0, what does it mean for g(x) to be a solution to this?

(This was in a linear algebra problem set so that's why it's going here)

[round coral]

the solution of a differential function it will be like y = some function of x that you have to find

they simply call it g(x)

nothing special

[nocturne jewel]

oh so replace y with g and the thing is satisfied?

[round coral]

yeah after you get the solution just change y to g(x)

[nocturne jewel]

what solution? the solution is g

The question isnt to find what g is, i just wanted to know what a solution to a diff eqn meant

[viscid kernel]

Can a non square matrix have an eigenvalue ?

Or eigenvector

[limber sierra]

not really, no

there is a somewhat analogous notion of "singular values"

which are important because of a concept/algorithm known as "singular value decomposition"

but these arent really eigenvectors

the definition of eigenvector/eigenvalue only works for square matrices.

[red prawn]

@royal ore From the relation on the right, $B = B(AC) = (BA)C$.

[stoic python]

Apopheniac:

[lone current]

Howdy folks - I need to turn a division equation into a multiplication for an Excel function, I'd appreciate any help!

Basically it's 195/1.65

[steady fiber]

that is not linear algebra

[lone current]

oh

[red prawn]

multiply by 195 time 1/1.65 ?

[lone current]

It can't have any division in it 😦

[steady fiber]

again, not linear algebra

pls take the convo to #computing-software

[red prawn]

Well, if you have a matrix with determinant 195 times the inverse of a matrix with determinant 1.65, then the determinant of the product of these matrices is equal to the determinant of the product of a matrix with 195 and a matrix with determinant .606060606060606060606....

[wintry steppe]

help

this isn't linear algebra

#geometry-and-trigonometry

Hey could someone help me understand why sum_i lambda_i x_i + U = 0 implies sum_i lambda_i x_i in U in the following stack: https://math.stackexchange.com/questions/2810480/quotient-space-mathbbf-infty-u-is-infinite-dimensional

surprising someone didn't jump on it and get mad at the guy for just posting a problem without any work

that's by the definition of the quotient space

but their notation is possibly misleading

if $x + U$ is the zero element of $V / U$, then $x + U = 0 + U$, meaning $x - 0 = u$ for some $u \in U$, or $x = u \in U$.

@wintry steppe

[hollow finch]

is there a standard notation for the coordinate map with respect to the standard basis or some other specific basis? u->(u)_B or u->(u)_S

would something like this be confusing? if $B$ is a basis for the $n$-dimensional vector space $V$
$$C_{B}:V\to \bR^n,\quad C_B(u)=(u)_B$$

[stoic python]

nix:

[wintry steppe]

i think this notation is perfectly self explanatory and clear tbh

[hollow finch]

nice thank you

[jaunty wing]

I got a differential equation to solve and idk where to begin, but I think I was not taught it yet

Even tho I wonder if I have to integrate it first

[hollow finch]

you want #multivariable-calculus not #linear-algebra

[jaunty wing]

Oh, alright Nix!

[alpine pulsar]

Can you do this ?

[wintry steppe]

yes

[alpine pulsar]

Can you link me some proof, or suggest search keywords? I need to convince someone and can't find anything online
@wintry steppe

[wintry steppe]

multiplication distributes over addition

that's literally it

[alpine pulsar]

@wintry steppe many blessings

[wintry steppe]

write it out for yourself if you wanna be super convinced

[alpine pulsar]

Oh I am convinced 😄

[wintry steppe]

if you want to be fancy, induct

[alpine pulsar]

I actually did.

[half ice]

Note that
Σ ay = a Σ y
For the same reason

For any a that the sum doesn't depend on ofc

[alpine pulsar]

Yes It makes sense thank you @half ice

[wintry steppe]

thanks @wintry steppe , but how can he state: Let x + U = 0?

or that's an assumption

idk lemme actually read it

this is the only actual solution I've found on the net for dimensions of this type of quotient spaces

I'll use the ideas to solve the other ones

@wintry steppe he's trying to establish linear independence so he says assume that this thing is zero in F^\infty / U

and then he shows that the lambda_i's are zero

so you're right, it's an assumption

thanks a lot @wintry steppe 🙂

So then if I want to prove that the dimension of $l^\infty / l_0$ is $\infty$ where $l_0$ are sequences whose limit is 0, I can take the sequences ${x_n}$ where $x_n \in {0, 1}^\infty$ and is zero everywhere except at indices divisible by $n$.

If the dimension of the quotient space is 2, then assume $a_1 (x_1 + l_0) + a_2 (x_2 + l_0) = 0$

[stoic python]

petarj:

[wintry steppe]

and $a_1 x_1 + a_2 x_2$ will be in $l_0$ only when $a_i = 0$, since the sequence is $(a_1, a_1 + a_2, a_1, a_1 + a_2, ...)$.

[stoic python]

petarj:

[wintry steppe]

then I can do the same with dimension 3, ..., n and n + 1

[stoic python]

petarj:

[opaque lake]

What exactly is a basic solution?

[hollow finch]

@opaque lake can you elaborate? i havent heard that exact terminology and there are a few things i think you might be referring to

[opaque lake]

@hollow finch Hopefully this gives you some context.

[odd kite]

never heard that term before. why not just call them complementary solutions

[hollow finch]

not a fan of calling them "basic" but they are homogeneous solutions

complementary/homogeneous are both fine

if An=0 where A is a matrix, we say that n is a homogeneous solution. basically it becomes zero when multiplied by A (or mapped to zero by the transformation)

[opaque lake]

Ah, okay.

[hollow finch]

in the case of that equation specifically, if v is an eigenvector (or basic eigenvector i guess)
$$(\lambda I-A)v=0$$
$$\lambda Iv-Av=0$$
$$Av=\lambda v$$

[stoic python]

nix:

[hollow finch]

so eigenvectors are just scaled when multiplied by A which is a very special and useful property

we dont call zero an eigenvector though even though its always a homogeneous solution

[opaque lake]

Thanks.

I don't really understand what the purpose of this terminology is.

Wait nevermind. I'm dumb.

I thought multipliction was commutative, so it'd be P^-1•B•P = I•B

But that's not true.

[wintry steppe]

conjugation by P^(-1)

[opaque lake]

@wintry steppe I'm sorry?

[hollow finch]

its called conjugation

when you do pbp^-1

similar matrices basically do the same transformation on a different basis

the main application is diagonalization. the transformation A just scales eigenvectors right? so the transformation of A on a basis of eigenvectors (if you have one) is just a diagonal matrix

and diagonal matrices are very nice and easy to work with

[lament delta]

Hi, I'm looking for a pair of real matrices A and B such that det(A) = det(B) and $$A \not\sim B.$$ Where $$A\sim B$$ if there exists a uni-modular matrix P (a square integer matrix P is uni-modular if det(P) = ±1) such that $$B = P^T A P.$$ Can someone help me?

[stoic python]

feanornumberthree:

[royal ore]

https://media.discordapp.net/attachments/504653099793645568/780812414844534804/image0.jpg

[lament delta]

I've cross-posted this to the questions-theta channel with some of my working.

[wintry sphinx]

BA = CA
(BA)B = (CA)B
B(AB) = C(AB)
B = C

[opaque lake]

Also, is the statement: "The transformation matrix of a linear transformation of itself is an identity matrix" coherent?

[half ice]

"linear transformation of itself"?

Just generally want to avoid "it"

What is "itself" in this context?

[dusky epoch]

@opaque lake no

[cedar oasis]

since two row vectors are given , the normal vector to it will be the cross product of <8, -5, 2> and <0,2,1> , which i get as <-9, -8, 16> so my answer is -1 but in solution they have given it as <9,8,-16> and ans is 1

am i doing anything wrong?

[native rampart]

well,both are correct

But,They specifically asked for the case when a>0

[crystal oracle]

Is the following true?

Suppose 𝔽 is a field, V is a vector space over 𝔽, T is a nilpotent linear operator on V. Then T+I has a square root.

I know this is true for the case 𝔽=ℂ and V finite dimensional, the proof uses ideas similar to the Taylor series of √(1+x). It's listed as the Theorem 8.31 in "Linear Algebra Done Right 3rd edition". I am pretty sure the same proof works for V infinite dimensional. I also think the same proof works for an arbitrary field, but I am not sure.

[native rampart]

I don't think the proof uses the fact that F=C in that case

So, Should be fine for arbitary fields

[crystal oracle]

Alright, thanks for confirmation

[native rampart]

Wait,You will run into problems with finite fields.(let me represent the multiplicative identity of the field as 1. For example
(I+T)^1/2=1+ 1/(1+1) T+.... (For verification,try squaring the term) . The taylor series may not be well defined(for example 1+1=0 in F_2)

[dusky epoch]

let me represent the multiplicative identity as I

youre already representing it as 1

[crystal oracle]

@native rampart indeed, I haven't noticed that. I tried this theorem for ℤ/(7) with one example, got a success and decided that it works for all fields. Guess I was wrong, thanks.

TIL: (1+1+...+1) might not have a multiplicative inverse in some fields, thus proofs that use some kind of division like that might not work.

[wintry steppe]

linear algebra in finite fields

[timber magnet]

need some help here

[odd kite]

so what does that say about ker(A)

that is has the general solution like that

[sly cedar]

can anyone help with this q ?

[native rampart]

Do you think Av in kernel of A for all v according to your condition?

[old flame]

One quick question, could someone explain intuitively what is the adjoint operator $T^{}$, I've seen the definitions and the relation $(Tv,w)=(v,T^{}w)$, but don't really get what's going on

[stoic python]

Otoro:

[pure wasp]

what does each variable mean here ?

[dusky epoch]

i and j are the basis vectors, i.e. unit vectors pointing along the x and along the y axis respectively

[pure wasp]

so which one is which

i = x

?

[dusky epoch]

respectively

yes, i is the one for the x-axis.

[pure wasp]

mint

thank you

[pure wasp]

erm how would i add 3 vectors together

???

[waxen jacinth]

By adding corresponding entries together

[pure wasp]

what does that mean

ive solved what a b and c are in vectors

just need to solve them

[waxen jacinth]

Havent done this before lol

For this Id say ud need to add them graphically

Using the parallelogram rule

Add A and B first, then the resulting vector with C

[pure wasp]

couldnt i do that with vectors as well tho

[waxen jacinth]

Same thing with subtraction, except ud need the triangle rule

You dont have coordinates here, just angles lol

ALthough I wouldnt know

As I said, havent done this be4

[pure wasp]

ive worked the angles out tho

using pythagoras

and trig

[waxen jacinth]

Idk wt ur problem asks u

Idk if its units, angles or whatever

[pure wasp]

niether do i man

rubbish university work

[waxen jacinth]

:p

Im tryna figure out my own rubbish

[waxen jacinth]

Guess u also need magnitude

[pure wasp]

so ive worked out the adjacent and oposite sides

[waxen jacinth]

Did u find the magnitude lol

[pure wasp]

and then used them as my x and y

its at the top of the page there

[waxen jacinth]

I mean the magnitude of the resulting vectors

After addition

[pure wasp]

of the resulting vecors ?

so i got A = [5,11] B = [1.83,9.8] , C = [0,9]

[waxen jacinth]

A+B

Okay, but B should be with negative signs

4th quadrant :T

[pure wasp]

ah yes

good spot

il have to do b again then

[waxen jacinth]

Good luck lul

[acoustic path]

why is the determinant of a matrix 0 when we have infinite homogeneous solutions

like the determinants zero when the vector columns are linearly dependent

[magic acorn]

how would you prove that r2 isnt a linear transformation

[clear arrow]

no

do it yourself

[magic acorn]

im confused on how you would get U

dont just dismiss me for no reason its not like im asking for the answer

[clear arrow]

oh I forgot there was a confinement

you look at zero

[magic acorn]

?

bro fr?

[red prawn]

When the vector columns are linearly dependent, you are no longer guaranteed a unique solution. And if you have more than one solution, you can add and scale those to get infinitely many

[red prawn]

An affine transformation preserves lines, but possibly doesn't fix the origin. To be linear (not just affine) you also must send 0 to 0

[acoustic path]

@red prawn oh so the determinant isnt consistent as well

[magic acorn]

@red prawn but it stated that r and s cant be 0

or is that something different

[red prawn]

They're saying "this may look linear but if r and s are nonzero, then it doesn't fix the origin so it's technically not linear"

err, they're telling you to say something like that, now that i looked closer. lol

[magic acorn]

alright ic ic

[wintry steppe]

x² + y² +16/x + 16/y ≥ 24

Can someone prove that is always true when x,y ≥ 0

[red prawn]

@acoustic path I don't get what "consistent determinant" means. But one of my favorite ways to think of the determinant is volume. When you have determinant zero, it could mean you're squishing a solid (for example) into a flat 2-plane.

[acoustic path]

ye i said it wrong

and thats what youre also doing when you have free variables

@red prawn but how does that prove that the determinant is 0.

[red prawn]

If you have infinitely many solutions, that's telling you that you have lost at least one dimension in the transformation. If you take away a dimension from a solid, it's got no volume

[acoustic path]

but it has area then

[red prawn]

Having area is great if you are only looking at 2 variables to start with. If you start with 3 coordinates though...

[acoustic path]

i get what youre sayin for R^2 vectors and how if theyre dependent then its just a line so thats 0 area

but for R^3 you technically still have area

[red prawn]

for R^3 you technically still have lengths of lines, but do those tell you about the volume of a solid?

[acoustic path]

well nothing

[red prawn]

Another easy way to think about this is remember that the determinant is equal to the product of the eigenvalues. then det is 0 when at least one of the eigenvalues is zero

If an eigenvalue is zero, that means the corresponding eigenvector gets mapped to 0. Then every scalar multiple as well, so you're losing information about an entire line in the process of making the transformation.

[acoustic path]

ill hold on to that thought then

thanks 🙂

[magic acorn]

hey its me again

so for b ii)

im struggling on how you would prove this is an isomorphism

@red prawn

[red prawn]

Is it linear, injective, and surjective?

[magic acorn]

so the 3x3 matrix i got in b i) i have to prove if its linear, injective and surjective?

[red prawn]

They've already told you that it's linear (plus, it's in the form of a matrix, so kinda automatically linear).
You could a) Show surjective and injective, b) show that it has an inverse, or c) use results form linear algebra like calculating the determinant. Each are equivalent methods, some shorter than others.

I think it's easiest to show what it's inverse mapping would be. What's the reverse of adding stuff?

[magic acorn]

subtracting stuff

so if i were to calculate the inverse what would that prove

[red prawn]

If a linear map has an inverse, then it is an isomorphism.

[magic acorn]

alright thank you @red prawn once again

[hollow finch]

is a change of basis an isomorphism ?

[gray dust]

yes & its inverse effects a reverse COB

[icy orchid]

I'm learning about geometric algebra and have been reading through Alan Macdonald's "A Survey of Geometric Algebra and Geometric Calculus" where i've come across this theorem:

i believe I understand the proof, but the paper doesn't go into much detail re: its implications

does this mean that any bivector in G^3, B = B1e12 + B2e23 + B3e31 can be expressed as the outer product of two vectors, or the geometric product of two orthogonal vectors?

[wise flare]

wait

can a matrix be iagonaliable if it has both an imaginary an real part ?

[wintry steppe]

of course

take your favorite diagonal matrix and throw in an i on the diagonal somewhere

[wise flare]

im confuse here

how can u fin if a characteristic polynomail is iagnoaliable without knwing its matrix A

you cant check the im eigenspace

beacuse its lamba I - A

but you not know A

[outer tulip]

wait I missed that it followed on from a previous part, one sec

[icy orchid]

yes, provided they are all in the same vector space which the three basis vectors span

[wise flare]

any help ?

[wintry steppe]

are you asking about diagonalizability over a certain field here

because this doesn't even split over R

[wintry steppe]

do you have a question about this? have you tried to prove this?

[royal ore]

not sure where to start

[wintry steppe]

write down the definition of an isomorphism

[spice storm]

Start by the definition

[wintry steppe]

generally if you don't know where to start on a problem, just write down the relevant definitions in front of you and stare at em a bit

refresh your memory of the things you're being asked to work on

[wise flare]

nope

[outer tulip]

ok previous part doesn't help, will post full q

[wise flare]

just is it igonaliable

[wintry steppe]

not over R, no

because its characteristic polynomial is not splitting

[outer tulip]

Done part i, know how to do the last part of part (ii), not sure how to get a P that does both those things (ik that B represents a positive definite quadratic form iff there's an invertible C with B = C^T C)

[wintry steppe]

I am asked to prove that for a bounded linear operator $T: V \to \mathbb{R}^n$, it is possible to have the direct sum decomposition $V = \text{Ker}T \oplus Z$ where $Z$ is a closed space. I am beginning the proof by taking $Z = (\text{Ker}T)^c \cup {0}$, can I actually do this? I am wondering because from the direct sum decomposition it follows that $\text{Ker}T \cap Z = {0}$.

[stoic python]

petarj:

[wise flare]

what o you mean splitting ?

[wintry steppe]

means you can write it as a product of linear factors. this is a necessary condition for diagonalizability over a field

Then I use $T$ bounded and equivalence of norms to prove $Z$ is closed, but I'm stuck on this. I want to first make sure that I'm constructing $Z$ the right way.

[stoic python]

petarj:

[wise flare]

I don’t agree

It is in linear factor form already

[wintry steppe]

no

[wise flare]

The lambdas are factored out

[wintry steppe]

no

that last factor splits over the complex numbers but not over R

so whatever A is

it's definitely not diagonalizable over R

[wise flare]

Anybody disagrees with some of these ? (Don’t tell me which ones )

T F T T T ? T T T F

[wintry sphinx]

I disagree

[royal ore]

how do i find the inverse of this? it's not 3x3

[wise flare]

Are you sure

[acoustic path]

@royal ore how can you find the inverse if its an odd by even matrix?

[royal ore]

that's what i'm not getting

how do you find the identity matrix

but it's whatever i don't need it anymore LOL

[acoustic path]

identity matrix and inverse arent the same thing

if thats what ur sayin

[royal ore]

oh

ok

still don't get identity matrix that's why

[winter moat]

4x4 identity matrix will work

[wise flare]

Any second opinions ?

[outer tulip]

wolfram alpha agrees https://www.wolframalpha.com/input/?i=[{1%2C-1%2C2%2C4}%2C+{0%2C-1%2C3%2C0}%2C+{2%2C+-2%2C+2%2C+1}]*[{1%2C0%2C0%2C0}%2C+{0%2C1%2C0%2C0}%2C+{0%2C0%2C1%2C0}%2C+{0%2C0%2C0%2C1}]

non-square just means no two-sided inverses/identities so there's no problem with it

[wintry sphinx]

@wise flare yeah pretty sure

[fervent palm]

im really lost with this one ;-; i've been trying to follow along but I'm lost on how the summation turns into a matrix norm:

[acoustic path]

it plugged the values into n

for both matrices

[fervent palm]

Hmm, i think i figured it out, I missed out on the definitions of the holder norms...
2norm(X) = (summ(X_i)^2)^1/2

[mossy trail]

Hello I was hoping one of you could help me. I was given two vectors in R^4 and I was also given the magnitude of their sum and difference. It wants me to compute the dot product and the largest and smallest value that each vector could be. I'm just having some trouble solving it. Any help would be appreciated.

[mossy trail]

@wintry steppe Expanding out the norm into a dot product seems to lead me to a unfavorable situation as the given vectors are not orthogonal, so when I expanded out the norm of the sum I got.

u^2 + 2uv + v^2 = 2.

My instructor recommended using the triangle inequality and the reverse triangle inequality to solve the problem. I'm just having problems applying it to the problem.

[wise flare]

any help ?

[wintry steppe]

Please help with this

[wise flare]

Anybody disagrees with some of these ?
T F T T T ? T T T F

[wintry steppe]

seems fine

[wise flare]

an i got false for F

@wintry steppe

[wintry steppe]

mmhm

[wise flare]

you ont agree ?

[wintry sphinx]

I still disagree with the first one

if you include the zero vector in U, you're done

[wintry steppe]

ugh

right

it's always the degenerate cases

[spiral garden]

so a vector can be represented as a matrix with the first row being the change in x and the second row being the change in y for a 2 by 1 matrix right?

also, what's the difference between a matrix with parenthesis and a matrix with brackets?

[half ice]

I don't know anything about this "change in x - change in y" business

There's no difference between the two

[spiral garden]

what

well one way to represent a vector u for example is u=(change in x, change in y)

I'm just wondering how that works with a matrix

[limber sierra]

also, what's the difference between a matrix with parenthesis and a matrix with brackets?
nothing

some sources prefer the former, some prefer the latter

its just an aesthetic choice (and also parentheses tend to be faster to draw on a whiteboard)

though do note that if it has bars on the side, like $\begin{vmatrix}a&b\c&d\end{vmatrix}$, this usually denotes the determinant $\det\begin{pmatrix}a&b\c&d\end{pmatrix}$

[stoic python]

Namington:

[limber sierra]

rather than the matrix itself

no difference between brackets/parentheses though.

so a vector can be represented as a matrix with the first row being the change in x and the second row being the change in y for a 2 by 1 matrix right?
it's unclear exactly what this means; if you mean in the sense of matrix multiplication:
[
\begin{pmatrix}a&b\c&d\end{pmatrix}\begin{pmatrix}x\y\end{pmatrix}
]

then this product actually evaluates to:

[stoic python]

Namington:

[limber sierra]

[\begin{pmatrix}ax&by\cx&dy\end{pmatrix}]

[stoic python]

Namington:

[limber sierra]

so if anything, the columns correspond to this information, not the rows

if you mean using the matrix to directly represent a vector in R^2... why do you think a matrix would have a nice visual intuition as a vector? a matrix represents a linear transformation, not a vector.

(unless that matrix is itself a vector, of course, such as a 2x1 matrix)

(it's also worth noting that your visual intuition for a vector falls apart outside of R^2, or when working with nonstandard inner products, but that's probably not a concern for the time being)

(it does, however, suggest that this visual interpretation is "less fundamental" in some ways than the definition of a vector - and this would be a true observation. there's no reason the first entry of a vector should correspond to a horizontal axis, for example - we just accept that by convention/the standard basis.)

[slate fox]

how can i show dim(W1+(W2+W3))= dimW1+dimW2+dimW3 if and only if W1+(W2+W3) is a direct sum

i got the forward direction but the backwards one is trippin me up

all Ws are subspaces of finite dimensional space

actually now that i think of it mmy forward direction doesnt really work

<@&286206848099549185> anyone?

i managed to show if theyre a direct sum then the equation holds

[strange dove]

Could someone help with these

[slate fox]

channel is kinda occupied :/

[slate fox]

https://tenor.com/view/kitties-cuteness-overload-halp-help-help-me-gif-16189870

[slate fox]

i think if the equation holds then W2 n W3 is {0} @wintry steppe

[slate fox]

im honestly not sure anymore and ive come to realise i dont realise the q doesnt really make that much sense to me @wintry steppe

i mean how can W1+(W2+W3) be a direct sum of itself?

[spring tide]

Does anyone have any experience with graph theory and Feidler value (also known as algebraic connectivity)?

[wintry steppe]

How do I show that a vector space has a unique additive identity?

I think I am going in circles here

[dusky epoch]

suppose it has two, show they're equal to each other

[wintry steppe]

Yeah that's what I've been trying

It's harder than it sounds lol

[dusky epoch]

suppose it has two identities 0 and 0'

0 + 0' = ?

on the one hand since 0 is an identity 0 + 0' = 0'

[wintry steppe]

Yeah 0 and 0'

[dusky epoch]

on the other hand since 0' is an identity 0 + 0' = 0

0 = 0 + 0' = 0'

by the transitive property of equality you have 0 = 0'

[wintry steppe]

[dusky epoch]

it's big brain i know

[slate fox]

like, the question says ''show that W1+(W2+W3) is a direct sum, ie W1+(W2+W3)=W1 direct sum sign (W2 ditect sum sign W3)"

word for word

[ebon cloak]

can an eigenfunction be any function that holds true for Af=λf?

or is it like a specific function

[odd kite]

any function

[ionic grotto]

anyone doing slope?

and y-intercept?

[wintry steppe]

wrong channel, read the pinned message

[lethal sequoia]

Hey can anyone help me with the pneumonoultramicroscopicsilicovolcanoconiosis theory?

Please I really don’t understand

[half ice]

Please avoid unnecessary use of helper pings. Ask a question, wait 15 minutes, ect.

[lethal sequoia]

I understand why they call u not cool

Jk

I’ll wait

Sorry

My bad

[limber sierra]

are you trying to be funny

[summer wagon]

Hi could someone help me figure out this question?

[tacit storm]

Arte there only two bases vector for this subspace

with dimension 2

[silver tree]

If I have found the eigenvectors and eigenvalues for 2 matrices, and they are identical (i.e. they are similar), How do I find the regular matrix M, that satisfies the following:

B = M^-1 * A * M

Where B and A are the similar matrices?

[wintry steppe]

@tacit storm seems like it but im too lazy to actually compute it

cause like you have two independent variables in the first two entries

and the last one is not

[tacit storm]

ok thanks bro @wintry steppe

[wintry steppe]

just draw the subspace

one way to see it without computing is to notice that it contains (1,0,1) and (0,2,3) which are linearly independent, but it doesn't contain (0,1,1), so it can't have full dimension

another way to do it would be to write it as the image of a rank-2 matrix but that's probably overkill

[tacit storm]

yeah thank you

[wintry steppe]

Prove that -(-v) = v for every v in V (where V is a vector space.)

Proof. -(-v) - v = -[(-1)v] + (-1)v = [(-1)v](-1 + 1) = (-1)v = 0, so by transitivity, -(-v) - v = 0 so -(-v) = v.

Suppose a ∈ F, v ∈ V, and av = 0. Prove that a = 0 or v = 0.

Proof. Suppose neither a nor v is 0. If that is the case, then av = 1av ≠ 0, which implies that one of a or v must be zero.

Suppose v, w ∈ V. Explain why there exists a unique element x ∈ V such that v + 3x = w.

Proof. For v, w ∈ V, we have that v - w = u for some u in V. Note that v - w = u = 1/3 u + 1/3 u + 1/3 u. Letting u = -1x = -3x, for some x in V, we have that v - w = -3x so v + 3x = w.

Can someone check these proofs and tell me if they are good or what I should change?

[native rampart]

For second,why does av=1av imply a or v must be zero

[wintry steppe]

Well, because av ≠ 0 if neither a nor v are zero, so if av = 0, then one of them must be zero

Since I looked at the case when neither are zero

[native rampart]

How do you av cannot be zero when a and v are both nonzero

[wintry steppe]

I don't know

[native rampart]

Let's say a and v are both nonzero

av=0

Multiply both sides with inverse of a

a'av=a'0
1v=0
v=0
Contradiction

[wintry steppe]

oh ok

Is there a way to fix my proof though using the contradiction?

[native rampart]

This is the contradiction

[wintry steppe]

Okay. What about the other proofs?

[native rampart]

1st is fine, although you could have just used uniqueness of additive inverse.(because a vector space is a abelian group over a field)

[wintry steppe]

Okay. What about the third?

[pure wasp]

Does anyone know if any of these are possible and if not why

[cunning arch]

Think about your matrix multiplication rules @pure wasp

For matrix multiplication, the number of columns in the first matrix must be equal to the number of rows in the second matrix. The result matrix has the number of rows of the first and the number of columns of the second matrix.

So for AB, # of columns in A is 2, # of rows in B is 2, so yes it's possible

[pure wasp]

Thank you :))

i was very confused

[cunning arch]

Yup, np :) Once you multiply AB out, your resultant matrix should be 2 rows and 3 columns

[pure wasp]

so for example what should i multiply in the first one ??

[cunning arch]

$\begin{pmatrix}3&1\ -1&3\end{pmatrix}\begin{pmatrix}1&4&0\ 2&7&-1\end{pmatrix} = \begin{pmatrix}3\cdot :1+1\cdot :2&3\cdot :4+1\cdot :7&3\cdot :0+1\cdot \left(-1\right)\ \left(-1\right)\cdot :1+3\cdot :2&\left(-1\right)\cdot :4+3\cdot :7&\left(-1\right)\cdot :0+3\left(-1\right)\end{pmatrix}$

Rip, you get the idea haha. It's rows of A * Columns of B

So first, we have the first row of A * first column of B = 31 + 12

Then first row A * second column B, etc...

[stoic python]

strawberrypocky:

[pure wasp]

is the dot a dot thingy or a multiplication

[cunning arch]

Just regular multiplication lol

[pure wasp]

okidoki thank you

[ocean sequoia]

$x \in N(B) \cap R(V_{k+1})$

why is that true?

[stoic python]

brzig:

[ocean sequoia]

is it just because there are only k independent vectors in the image of B so if its not one of those it must be in the null?

am I missing something obvious?

[wintry steppe]

If I have $$A = {(x,0,0) \in F^3 : x \in F}$$ and $$B = { (0,y,0) \in F^3 : y \in F }$$ How can I compute $A + B?$

[stoic python]

LINEAR_ALGEBRA_GUY:

[ocean sequoia]

unless I misunderstand something it will just be C = A + B = (x,y,0)

you can add vectors

@wintry steppe

[wintry steppe]

Ok but what about if

$A = { (x,x,y,y) \in F^4 : x,y \in F }$ and $B = { (x,x,x,y) \in F^4 : x,y \in F }$

[stoic python]

LINEAR_ALGEBRA_GUY:

[wintry steppe]

Then what is $A+B$?

[stoic python]

LINEAR_ALGEBRA_GUY:

[ocean sequoia]

same thing just add em

[wintry steppe]

Yeah but now there's a collision

[ocean sequoia]

I dont think it matters because they both subspaces of F^4

maybe someone else could weigh in

just specificy that its for all x,y in F

[half ice]

What's a "collision"? Haha

[wintry steppe]

There is an x in one slot

And a y in the other

[half ice]

Just add each of the components as you would

Note that x + y is in F, as that's how F do.

[whole solstice]

@half ice

let us begin

[half ice]

Aight so it is pretty tedious looking haha

[acoustic path]

teach us

[half ice]

But basically you just want to compute

• T(ax) and aT(x)
• T(x + y) and T(x) + T(y)

[whole solstice]

ok sure

[half ice]

Wait is P(R) just any polynomial? Eew

[whole solstice]

P(R) denotes all polynomials with coefficients in R

[half ice]

😫

[whole solstice]

what's wrong with that notation?

[half ice]

The notation is fine haha, but dealing with that is eh

[whole solstice]

ok let's computer T(x + y) first

[half ice]

So yeah I used x and y to represent vectors, which are polynomials in this case

[whole solstice]

T(x + y) = (3(x+y)(4) + 5(x+y)'(6) + b(x+y)(1)(x+y)(2), int from -1 to 2 of .............)

it's long

and i don't know latex that well

plus

i'm in class

[half ice]

I guess we could define the functions X and Y to make this a little easier to think about

[whole solstice]

ctually can we wait till after this class

[half ice]

T(X + Y)
= 3X(4) + 3Y(4) + 5X'(6) + 5Y'(6)...

And go on like that haha

Note that I did just use the linearity of the derivative there

[dusky epoch]

are we allowed to use the fact that the sum of two linear transformations is itself linear here

[robust pond]

so this seems like B would just be A inverse

but A is singular

granted the second works if B is the 0 matrix but not the other way around

[dusky epoch]

AB needs to act like the identity on the colspace of A

[robust pond]

eigenvalue 1

[dusky epoch]

[robust pond]

idk linear algebra

lemme see

that doesnt make sense does it lol

ann can i have another hint

[wise fractal]

does NN need linear alg?

[dusky epoch]

NN?

[robust pond]

neural networks?

i dont really see what "AB needs to act like the identity on the colspace of A" means

tbh i dont need to do this problem but im curious because no one else in my class has figured it out 🤔

teacher is about to just give us all credit for it but i wanna see

[thin bloom]

Network Neutrality?

[robust pond]

[wise fractal]

nueral networks for AIs

im a freshman in high school considering skipping precalc so that I can take linear algebra in senior year and understand formulas for Neural networks

[thin bloom]

Pls don't do that

[wise fractal]

y?

[dusky epoch]

i dont really see what "AB needs to act like the identity on the colspace of A" means
pay it no mind then

[thin bloom]

Who knows if NN would be that tech which gets nowhere

[dusky epoch]

anyway uhh

[robust pond]

[wise fractal]

still precalc doesn't seem that interesting

I might as well just take it over the summer

[robust pond]

i mean i dont get like

[thin bloom]

Network neurality is more important subject tbh

[dusky epoch]

linalg is necessarsy for neural networks yes

[robust pond]

are you saying that it will act differently on the column space than the matrix or

[dusky epoch]

but i would say the stuff they teach you in precalc is good to know too

[thin bloom]

^

[robust pond]

i know thats a dumb question

[thin bloom]

It's not like, precalc is useless or sth

(Unless you already know it, but then you won't be asking this I think)

[wise fractal]

yea but its like I'm already kinda doing the textbook itself

and I'm not really struggling

i'm just taking it over summer not to learn the course but to just get the credit

I plan to just learn it throughout this year

[thin bloom]

Hm do you already know about much of precalc?

(Tho tbh I'd strongly discourage doing sth for NN, there have been lots of things like that which eventually got forgotten)

[wise fractal]

35% of it

it's really not that hard

its just algebra 2

but on steriods

as they say

[odd kite]

yes it's mostly a waste of time if you paid attention in algebra 2

[wise fractal]

I still don't get factorials lol

its the most confusion I'm having

[shrewd mortar]

wdym

[wise fractal]

like let give u a problem

[shrewd mortar]

n! = n(n-1)....(3)(2)(1)

[wise fractal]

like this problem

I'm trying to understand it

[wintry steppe]

wrong channel...?

unless this is one of those non-LA looking problems with a clever LA solution

[wise fractal]

tf

[wintry steppe]

like the fib sequence

[odd kite]

no, they were just talking about whether to skip precal and do linear algebra instead

[wintry steppe]

oh

y'all were having a convo

ok

mb

[odd kite]

prob. does belong in another channel though if you really want to discuss how to solve it

[robust pond]

ive asked everywhere else i have 😅

google isnt helping, idk what im supposed to see in this problem

[thin bloom]

That problem looks like an easy one

I guess the class would be useful

[robust pond]

class?

im talking about

I know its probably easy my teacher seems to think so too 😅

my teacher gave the hint of try the zero or identity matrix

idk what thats supposed to mean

[thin bloom]

@robust pond o sry I was talking abt the one of prachod

[robust pond]

oh

[thin bloom]

I think the teacher mean to try identity for B

Who knows it might work

[wise fractal]

lmao prachod is my friend irl

hes a literal chod

[robust pond]

the identity matrix for B

i mean its not possible

wouldnt that mean that A^2=A

[thin bloom]

Yep

Try it

[native rampart]

That doesn't work

[robust pond]

it doesnt work lol

[native rampart]

Because it implies A=I

[thin bloom]

?

[robust pond]

yea theres only a handful of cases where that works

[native rampart]

2nd equation

[thin bloom]

A^2 = A is possible without A being identity

[robust pond]

ABA=A and BAB=B

[thin bloom]

Oh right

I misread the question I guess

[robust pond]

unless im not being uhh

imaginative

[thin bloom]

You know what

[robust pond]

which im probably not

im p bad an this subject

[thin bloom]

What if

AB is identity

[native rampart]

A is not invertible

[robust pond]

singular

[thin bloom]

O

Hmm

[robust pond]

so uhh

i mean i suck at linear algebra so i know this is wrong

well nvm

i know its wrong

[thin bloom]

Perhaps you can find pseudo identity then I guess

[robust pond]

i dont see how you get identity type behavior if an inverse doesnt exist

i was thinking maybe some kinda dumb answer like constants

[thin bloom]

Iirc there are pseudoinverse and stuffs

[robust pond]

but looks like dim B has to be dim A

pseudoinverse

[thin bloom]

Also could try like diagonalization

[robust pond]

hmm

[native rampart]

Diagonalization is horrible

[robust pond]

let me try that

okay nvm

bwahaha

they eigenvalues looked shitty

[thin bloom]

So that is why Ann talked about "identity on the colspace of A"

[robust pond]

ill be honest i have no idea what that means

[thin bloom]

Why

You should have learned abt it

[native rampart]

Well,The matrix is diagonalizable,though

[robust pond]

i mean

[thin bloom]

O wait is it diagonalizable

[robust pond]

we learned about the column space

and we learned about identities

idk what it means for something to behave like an identity on the column space of a matrix

[native rampart]

Char is x(x^2-2x-32)

[robust pond]

without just being the identity

i dont see how that could be a thing

[thin bloom]

Hmm, ye that would make it diagonalizable

[thin bloom]

Well, like consider this case: if column space is like a plane made by e1 and e2

Matrix would be identity on the columbspace if it fixes e1 and e2

[robust pond]

oh

then uhh

[thin bloom]

?

Having problems?

[robust pond]

im trying something

damn it was wrong

the column space is a plane

bases are (5 -4 1) and (-3 -2 -1)

are you saying any like

well lmc

[thin bloom]

Perhaps simplify the basis first so you could move them more easily

[robust pond]

yea that thought was wrong too but i already checked it was wrong

move?

i mean you can fix the negatives idk what youd do thatd make it prettier

wait does B have these as eigenvectors?

is that the trick

[thin bloom]

AB def have those as eigens I think?

[robust pond]

AB?

i gotta repost tired of scrolling

I get B having colspace bases as eigenvectors

[thin bloom]

Hmm wait a second, I'm a bit confused rn

Oh, actually if B has those as eigenvectors

That is nicer

[robust pond]

seems like a nightmare

[thin bloom]

Why

[robust pond]

trying to match an eigenvector to a matrix i dont know anything about

[thin bloom]

I mean, you could get more conditions out of like BAB=B

[robust pond]

idk tbh im not really seeing what B having these eigenvectors means

[thin bloom]

Some vectors in Diagonalization?

[robust pond]

idk ill have to review diagonalization

i dont think weve used it in this class

[thin bloom]

? The class did not deal with diagonalization?

[robust pond]

we see it in the final week

i think

tbh well probably end up skipping it because the whole class is behind

which means ill have to take this class again at uni even though everyone is probably going to get >90%

let me try diagonalize

[thin bloom]

Oh then it might be solvable without diagonalization

[robust pond]

idk my teacher gives us shit we havent seen all the time

[thin bloom]

(AB-I)A = O, B(AB-I) = O Hmmm

[robust pond]

yea i was lookin at that too

but i didnt see it being useful

[timber magnet]

anyone has some idea on b)

[robust pond]

columns sum to 1?

try something like

[timber magnet]

that only proves stochastic right?

[robust pond]

a 1x5 vector

x_1 = x_2 = ... = x_5

just a guess

err,

column vector

let me double check it

nope not even close

[timber magnet]

😆

[robust pond]

fwiw

here

idk linear algebra

[timber magnet]

Came across this theorem and tried finding the stationary distribution, but i got all terms to be 0 z, so probably this way won't work

[robust pond]

but heres where my guess comes from at least

but idk i didnt play around with anything higher than 2x2

[timber magnet]

oh let me have a read

[thin bloom]

Eigenvalues and vectors are gud

[timber magnet]

welcome for help anyone 😄

[robust pond]

oh wait

i typed in the matrix really wrong

@timber magnet

yea

just use a row vector

x_1 = x_2 = ... = x_2

youll get an eigenvector of x_1 = x_2 = ... = x_n

let the row vector be <a, a, ..., a>

since each column in P totals to one, the product c_1 a, c_2 a, ...., c_n a will have c_1 + c_2 + ... + c_n = (1) a

then 1 is an eigenvalue of all such P

[odd kite]

@distant merlin not a linear algebra question. Also it's an expression, and we don't solve expressions

[distant merlin]

whatever that problem is, do you think it's solvable

like can u solve for x

[odd kite]

try asking in #precalculus maybe

[distant merlin]

ok

thx

[robust pond]

lol my teacher asked me if i got an answer to that problem that the site isnt accepting

motherfucker you made the problem

[thin bloom]

? Wait what

[thin bloom]

What happened

[robust pond]

nothing

he told us not to do that problem

[shrewd mortar]

fun little putnam problem from 1985:

[stoic python]

Archsys:

[shrewd mortar]

okay i bothered to make the formatting marginally better

$\textbf{Putnam 1985 B6.}$ Prove that if $G$ is a finite subgroup of $\text{GL}n(\bR)$ such that $\sum{M \in G} \text{tr}(M) = 0$, we must have $\sum_{M \in G} M = \mathbf 0$.

[stoic python]

Archsys:

[thin bloom]

Lmao the teacher

[dreamy iron]

Any one know who the instructor is and what institute he’s out of?

https://youtu.be/KiXM4nCkQ2w

I’d like to see course materials and maybe his course notes

He’s got a fun take in that he’s motivating category theory and general homomorphisms using linear transformations.

[acoustic path]

thx for the vid

[trim galleon]

when are points not coplanar?

[acoustic path]

if their vectors are not parallel

and they dont intersect

[dreamy iron]

How many points?

Two points are always coplanar.

[trim galleon]

4 points

[dreamy iron]

Okay. That’s the toughie.

[acoustic path]

just said it

[dreamy iron]

Can you compute a cross product?

[trim galleon]

a what

i just put the 4 points in a matrix and do rref

[dreamy iron]

Okay

[trim galleon]

so when do u know if its coplanar or not

how*

nvm found

In an equal space, the following planes are given.
Calculate k and m so that those planes are parallel. Can they coincide?

k and 'm are elements of real numbers

23

[trim galleon]

<@&286206848099549185>

[fair vector]

Can anyone explain the underlined statement?
I am trying to show if A is an upper triangular invertible matrix then all diagonal entries of A are non zero

[round coral]

what can't you understand in it?

one thing you have to remember that you have to keep your matrix invertible

that is the null space must be zero

@fair vector

[fair vector]

Hmm so I understood that if a nn is 0 forces b to be zero but then i didn’t understand why it is impossible to find an x ...

[cedar oasis]

how is the eigen value of projection matrix zero?

[steady fiber]

the projection matrix maps down a dimension

[cedar oasis]

since a 3d vector is projected as a 2d vector, the dimension of transformation matrix is 2X3. how to calculate eigen value for rectangular matrix

[steady fiber]

which means at least one dimension is lost

any vector that belongs solely to that dimension

would be lost

during the projection

in that example, any purely vertical vector would be projected to a point

so the eigenvalue is 0 for any eigenvector that only has a z component

and you can pretend it's 3x3, not 2x3

except the last row is always just 0

as in, instead of thinking it is (x, y, z) -> (x', y')

think of it as (x, y, z) -> (x', y', 0)

[cedar oasis]

how can the last row of A be always zero? we can have something like 1 ,-1 ,0 right?

[steady fiber]

why can't the last row be 0

there's no rule against it

[cedar oasis]

i understand this one, but can you tell me how you calculated that the eigen value is zero?

[steady fiber]

it's a 3x3 matrix

you can calculate the eigenvalue however you normally calculate them

if your original matrix is
[a b c
d e f]
just make a new matrix
[a b c
d e f
0 0 0]

and use that

it's square, and does the exact same projection

[cedar oasis]

why can't i take the matrix as 1,2,3 4,5,6 1,-1,0

[steady fiber]

because that doesn't work

the new row has to be 0s

so it changes nothing

[jagged gulch]

@cedar oasis it's true that this mapping sends a 3d space to a 2d space, but the transformation isn't a 2x3 matrix because that 2d space is a subset of 3d space

[cedar oasis]

kindly excuse me now, i am getting a call

[red prawn]

Projections are idempotent

[jagged gulch]

you're sending R^3 to the xy plane, but it's the xy plane treated as a subspace of R^3

we know this because the output is a vector in R^3

so, as @steady fiber said, the transformation will be represented with a square matrix

[sharp shell]

Trying to solve this system with a parameter with gauss jordan

But I'm stuck

If u use the 1 in the second row everything gets way to complicated, but I don't see what else to do

[cedar oasis]

Thank u @jagged gulch @steady fiber I understood it now

[hollow finch]

Get reduced row echelon form before introducing a parameter

[red prawn]

I think it can be a little tidier to aim for unreduced echelon form, waiting until the end to multiply last two rows by appropriate amounts to get 1's in the diagonal. That way, there's not as many fractions to write. Also writing each expression in its most factored form, ie m^2 - m = m(m-1), I think this helps here

that's a good idea too 🙂

[hollow finch]

Assuming you mean rref except the pivots aren't 1 (but still all the same) then I totally agree

[red prawn]

like unreduced ref until the end, then reduce

[hollow finch]

It makes the homogeneous part of the solution much easier but you have to be careful about particular solutions since you need to divide at the end

[red prawn]

ohhhh right

[hollow finch]

The benefits of rref are still mostly there. That being the pivot variables are purely in terms of the parameter ones. So no solving is really necessary.

[red prawn]

Is there any trick coming from the matrix being symmetric?

[hollow finch]

Nothing comes to mind but someone else might know. The main benefits of symmetric matrices are the convenient properties of their eigenvalues and eigenvectors (imo at least)

If a symmetric matrix had a nontrivial null space then every vector in the null space will be orthogonal to all the eigenvectors with a nonzero eigenvalue I guess

So if you have some eigenvectors to start you may be able to use gram schmidt or a cross product maybe but that would be very situational

Not to mention probably a lot more difficult than just row reducing

[silver tree]

I have a quick question. How many eigenvalue (with algebraic multiplicity) will a n x n matrix have? Will it always have n eigenvalues or?

[gray dust]

considering eigenvalues in C and counting potential duplicates, always n

[karmic ginkgo]

just at a glance, is this a valid/solvable question? it confuses me as it says R^3 -> R^3 but the matrix has 4 rows

[peak pelican]

how do i calculate how many months i will be in position A before either going to F or B? i have calculated that N which is (I-Q)^-1 is [7.5 5.0; 1.25 2.5]

[ocean sequoia]

is the rh saying the l1 norm times the infinity norm?

[dusky epoch]

yes

[ocean sequoia]

am I underthinking this or couldnt I just square both sides to show that $v_1^2 +...+V_n^2 =< (v_1+v_n)^2?$

[stoic python]

brzig:

[dusky epoch]

wat.

[ocean sequoia]

ill take that as Im an idiot lol

isnt the l2 norm the eucildean norm?

[stoic python]

brzig:

[ocean sequoia]

$\sqrt{v_1^2 +...+v_n^2} \le |v_1| + ... + |v_n| * |v_{i\inf}|$

[stoic python]

brzig:

[ocean sequoia]

so we square both sides and get $v_1^2 +...+v_n^2 =< (v_1+v_n)^2* |v{i\inf}| = v_1^2 +...+v_n^2 \le v_1^2 + v_n^2 +v_1*v_2 + ...+v_{n-1}*v_n *|v{i\inf}|$

[stoic python]

brzig:

[ocean sequoia]

cause if we expand that binomial wont that show its bigger

$x^2 + y^2 \le x^2 + 2xy + y^2$ if x,y\ge0

[stoic python]

brzig:
Compile Error! Click the reaction for details. (You may edit your message)

[slow scroll]

you're kind of on the right track

[ocean sequoia]

any hints?

i know that misses the x infinity step but that works for the l2 <= l1 i think...

[slow scroll]

first of all, are we talking about Rn here, or spaces of sequences?

[ocean sequoia]

Rn

and all v > 0 i think because they are norms

[slow scroll]

ok, so you have sum x_i^2 = sum x_i x_i.
You can introduce the infty-norm right away

since x_i <= ||x||_\infty for all i....

[ocean sequoia]

because the infinity norm simply "selects" the largest x correct?

sorry norms are fairly new so its kinda odd working with them for me

[slow scroll]

yea, that's right

[ocean sequoia]

ok cool

[slow scroll]

do u see what to do from here?

[ocean sequoia]

i think so

I just distribute that over and because its larger than l2 it stays larger

wait

hang on

fuck lol

im try some more on my own

[cobalt marlin]

Is it impossible to find an eigenbasis if the characteristic polynomial has a repeated root?

Ah, I see. So if I have a 2x2 matrix whose characteristic polynomial has a repeated root, how do I use the one eigenvalue I found to determine the eigenbasis?

That is, after I find the eigenvector.

Right, so I have the matrix:

$\begin{bmatrix}
3 & -1\
13 & 4
\end{bmatrix}$

[stoic python]

Tristan

[cobalt marlin]

And found the eigenvalue to be 4, with the eigenvector to be:

[1
1 ]

Sorry idk the LaTeX lol

I'm still confused how I use that to find the eigenspace?

I took the determinant of the matrix minus the identity matrix*lambda and solved the characteristic polynomial

${\lambda}^2 - 8{\lambda} + 16 = 0$

[stoic python]

Tristan

[cobalt marlin]

Hmm okay, I'll re-check my work

Weird, I still get the same thing

I took the determinant of:

$\begin{bmatrix}
5-\lambda & -1
1 & 3-\lambda
\end{bmatrix}$

[stoic python]

Tristan

[cobalt marlin]

Oops

I messed up the LaTeX

Like I subtracted lambda from the 5 and from the 3

Oh oops I wrote down the wrong matrix earlier my bad lol

It should have been:

$\begin{bmatrix}
5 & -1\
1 & 3
\end{bmatrix}$

[stoic python]

Tristan

[cobalt marlin]

So I plugged the eigenvalue into:

$\begin{bmatrix}
5-\lambda & -1\
1 & 3-\lambda
\end{bmatrix}$

[stoic python]

Tristan

[cobalt marlin]

Which gives me

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

[stoic python]

Tristan

[cobalt marlin]

And then I multiplied that by

$\begin{bmatrix}
x \
y
\end{bmatrix}$

[stoic python]

Tristan

[cobalt marlin]

$\begin{bmatrix}
1 & -1\
1 & -1
\end{bmatrix}
\cdot
\begin{bmatrix}
x \
y
\end{bmatrix}$

[stoic python]

Tristan

[cobalt marlin]

Right

So is it literally just:

\begin{bmatrix}
1 \
1
\end{bmatrix}

?

[stoic python]

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

[cobalt marlin]

So it's still considered a valid eigenbasis even though it doesn't have a non-zero determinant (or any determinant at all for that matter)?

Because my understanding was that you need to span all of 2d space in order for it to be a basis

[stoic python]

Salicina

[cobalt marlin]

eigenspace == eigenbasis right?

Just to be sure

Oh so eigenspace is the matrix whose columns are the eigenbasis vectors?

So in this case my eigenbasis is not
[1
1]?

Doesn't the eigenbasis not exist since the determinant of the matrix is 0?

Okay, so what is the eigenbasis and what is the eigenspace in this case?

[stoic python]

Salicina

Salicina

[cobalt marlin]

Oh gotcha. So the eigenspace is just the line spanned by the vectors that satisfy that ^

So it's the set of the infinitely many vectors whose span is on that line

👍

Okay so that makes sense that [1 1] is a part of that set

Okay, so the eigenbasis still "exists" then. Would I write it as:

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

?

[stoic python]

Tristan

[wise flare]

i have a question

[cobalt marlin]

Okay so it's just like ([1 1]) then

[wise flare]

im looking to find if a given vector is in the span of the matrix

[cobalt marlin]

Alright, thanks!

[wise flare]

but the matrix augmentaion gives me an identity matrix

what does that mean

any help ?

like do you need to be independent to be in the span of a matrixc ?

[wintry sphinx]

finding whether b is in the span of the columns of A is the same as determining whether there exist solutions to Ax = b

[wise flare]

@wintry sphinx hey

are you good with matlab ? i have a question

[wintry sphinx]

no I can't use matlab

[cedar oasis]

https://youtu.be/oafswPUBICg

3:10 why does he say that dimension of eigen space corresponding to eigen value 3 is two?

if we take the span of [1 2 0] and [0 0 1] won't it cover the entire 3d space?

no its not possible 😅

i immediately added the two vectors and thought [1 2 1] looks three dimensional, did not consider that it is still on a plane in 3d space

thanks @warm briar

[ebon cloak]

https://prnt.sc/vs3qaf

is this written incorrectly?

vk and vj are 3d points

and H is supposed to be a real number

Uk and Uj are the distance between s' and vk and s' and vj respectively

[round coral]

Sorry, I had a small question to ask. When we make a matrix of a linear map and do gaussian elimination in it. Then the rref form of the form of matrix, is it similar to the matrix of the original map

meaning are they the same linear map in different bases

[native rampart]

No,Take any non identity invertible matrix

RREF is I, but PAP^-1 is not I unless A=I

[round coral]

Yeah, that makes sense

Thank you

[timber magnet]

anyone can help for C)

[hollow finch]

what have you tried

[thin bloom]

Seems like it won't be that hard even

[timber magnet]

tried using REF, but will get all terms to be 0

[shrewd wharf]

idk if it belongs here but
https://lupucezar.files.wordpress.com/2010/10/worksheet_linear_algebra__undergraduate_competitions_.pdf
I am stuck on problem 3
Can I get a little hint

thanks

[red prawn]

If AB = BA then A and B are simultaneously diagonalizable over the complex numbers. The determinant is the product of the eigenvalues of a matrix. Try working with eigenvalues?

[errant cedar]

Give an injective but not a surjective linear mapping P (F) → P (F).

do you guys have i idea

[wintry steppe]

try something that increases the degree

so that you can't get constant polynomials

that's one way to lose surjectivity

[errant cedar]

but its not the lowest degree

isnt it

Tp(x) = x^2p(x) is injective but is it surjectiv or not idk

Tp(x) = x^2*p(x)

T : P (F) → P (F)

[round coral]

Could anyone please tell me if X and Y are two sets , what does X⊔Y mean ?

[quartz compass]

means disjoint union

which doesn't actually mean anything other than union

and that the sets X and Y have empty intersection beforehand

it's just a way of emphasizing it

[round coral]

Oh! I see so it is the union of disjoint sets

[quartz compass]

yup

[hollow finch]

does there exist a polynomial q such that T(q(x))=1?

[round coral]

@errant cedar the map you defined is not surjective, so your work is done

[hollow finch]

x^2(q(x))=1

can q be a polynomial?

[errant cedar]

Yes q is polynomial

[hollow finch]

how

is 1/x^2 a polynomial?

[round coral]

it is not surjective Fatih, you can clearly see that there is no one degree polynomial in the image

@hollow finch no that isn't

[hollow finch]

i know that you know it im asking Fatih

[errant cedar]

😄 no itsnt

it isn't

[hollow finch]

right

so since there is no input to get that output, the transformation is not surjective

[errant cedar]

Aah yes 🙂 thanks i have a few more questions

Give an example of two linear mappings K, L: F ^ 5 → F ^ 2, which have the same kernel kern (K) = kern (L) but so that K is not a scalar multiple of L.

[wintry steppe]

hello kanishk