linear-algebra

[main badger]

That means T(u+v) = T(u) + T(v) and T(cu) = cT(u)

[winter harbor]

Just compute the product LU and verify yourself if this gives A

[wintry steppe]

Where'd I go wrong. Answer in the back of the book given on the right

It's the setup, not my computation:

[sullen hazel]

i don't know where to begin with this

the only thing i'm able to do is distribute the a,b, and c and having it equal to t(x) but i'm lost on solving for the individual variables

[wintry steppe]

t(x) is also equal to -20x^2 + 8

use gaussian elimination to find a, b, and c

make into a matrix equation

you can determine if a vector b is a linear combination of some column vectors by solving the matrix equation Ax = b where A is the matrix of column vectors a1, a2, ...

and you can "vectorify" the polynomials by taking the coefficient of increasing degree terms as each component

ex: 1 + 3x + 23x^2 could be written as the vector (1, 3, 23) (where its understood that the nth entry has degree n-1)

so you can boil that problem down to a 3x3 matrix * (a, b, c) = (8, 0, -20)

[sullen hazel]

okay that makes sense now including the 0 coefficients on the variables missing

ty to the both of you

[wintry steppe]

When can we view polynomials as vectors?

[dusky epoch]

what?

[wintry steppe]

Typo

[dusky epoch]

ok, then the answer to your question appears to be "When we have a vector space whose elements happen to be polynomials."

[wintry steppe]

How can polynomials be elements of a vector space?

[dusky epoch]

very easily

for example, take the set $V$ consisting of all polynomial functions from $\bR$ to $\bR$ with degree at most 4, with addition and scaling operations defined in the obvious manner.

[stoic python]

Ann

[dusky epoch]

this is a vector space.

we can even calculate its dimension (5) and find a basis for it (one possible basis would be {1, x, x^2, x^3, x^4})

@wintry steppe does this answer your question?

i have more examples of vector spaces where the vectors are polynomials.

[wintry steppe]

Sort of

[dusky epoch]

like, just for the simplest stuff --- there's a whole family of \textbf{named} vector spaces, typically denoted $\mathcal{P}_n(\bR)$, which are defined as consisting of polynomials of degree at most $n$ for some fixed $n$. what i showed there is $\mathcal{P}_4(\bR)$.

[stoic python]

Ann

[dusky epoch]

but it seems you still have your doubts?

if you do have some doubts, i'd like you to voice them now.

[wintry steppe]

I think I understand

[sand tundra]

Hi, how to prove that a positive defined symmetric matrice is invertible and that its inverse is also symmetric? Thanks in advance

[lavish jewel]

positive definite has all eigenvalues > 0 by definition

[sand tundra]

Thanks! how to prove that its inverse is also symmetric?

[stoic python]

c squared

[sand tundra]

Great thanks!!

[sage shale]

Hi, I see people here asking questions; is it okay to ask a question in here?

Or do I need to relegate it to the Math Help

[teal grotto]

if ur question has to do with lin alg then go ahead and ask

[sage shale]

thanks ^_^

So a question similar to this was given to me last week on a quiz, and the prof never mentioned it to me, ignored my request to know the answer and all. I know it's going to be on my final next week; and I'm unsure as to how to get to any answer, just the process of getting there even

Sorry for the bad handwriting, I'm writing on a track pad

[teal grotto]

i think there should be a way to write B as the product of some invertible matrices with A.

this is sort of like row reduction, just transform A into B with some elementary matrices

[dusky epoch]

^

[teal grotto]

the you use the fact that det(EF) = det(E) det(F)

[sage shale]

Sorry. I forgot to kinda write the question. It was given the determinant of A, what would the determinant of B be equal to

They don't give us our quizzes or the answers to the quiz

[dusky epoch]

what c squared said still stands

[sage shale]

so perform a row reduction?

[teal grotto]

it’s similar to the idea of row reduction. it is not however row reduction.

[sage shale]

Do you have a resource I could look at to better understand it?

[stable kindle]

look

find C such that AC = B

then det(A)det(C) = det(B)

C will be very simple

[sage shale]

Right, but I don't have a C, right? Or are you talking about the variable inside the matrix?

OH

[stable kindle]

no

find C

[sage shale]

I'm going to ask a student to see if they have a copy of the question asked.

[lavish jewel]

notice that stuff is happing to the matrix A row-wise in order to yield B

e.g. the whole row 2 is scaled by 5

this type of operation can be written as a matrix multiplying from the left

so it'd rather be CA = B

e.g. if C = [[1,0,0],[0,5,0],[0,1,0]], you'd get the desired effect of keeping the first row the same, and the second row scaled by 5

what would the 3rd row need to be?

then find det C

[sage shale]

Gotcha

Thank you

Everyone, thanks a lot 😄

[dusky epoch]

will this not be CA=B instead?

oh

[stable kindle]

does it matter?

[dusky epoch]

im late

but yes it does

[stable kindle]

wait damn

yes

i see

[dusky epoch]

you will not find C s.t. AC=B here

[bold sun]

Hey can someone explain this to me please?

i get hpw you expand the barackets

but what is A' and B'???

like its diffrernt to A and B but i dont get it is it like a transpose of a or b??

[stable kindle]

they're just distinct matrices

you could call them A, B, C, D if you wanted

it's just that A and A' have the same dimensions, and B and B' have the same dimensions

but otherwise they can be completely different

[bold sun]

oooh okk lol i was tryna find the answer everywhere but couldntt

thanksss

that actually makes sense

[lavish jewel]

yeah just meme notation

[golden reef]

If i have 3 matrices, A B and C
If i Know what B^TB is can i substitute it in this equation?
can i do something like this:
C * B^T * B * A
let D matrix = B^TB
C * D * A

does this break any rules or is this a valid equation?

[winter harbor]

It depends, you are substituiting B^t B into what????

What exactly do you want to do?

I mean, sure you can let D = B^tB and compute C*D*A

[golden reef]

because I know what B^TB is equal but both B^T and B by themselves are not very simple matrices so it is much harder for me to do C*B^T than to do C * D

[lavish jewel]

matrix multiplication is associative

you could put parenthesis anywhere in that expression

including C (B^T B) A, as you wanted

[golden reef]

so like A(BC) = (AB)C

[lavish jewel]

mhm

[golden reef]

or (AB)(CD) = A(BC)D?

[lavish jewel]

mhm

[golden reef]

yeah this helps so much, this way its so much more simpler

[sage shale]

I realized where I was being dumb earlier

When you're given something like
[a b c]
[d e f] det = 4
[g h i]

[a b c ]
[3d 3e 3f ] =?
[3g+3d 3h+3e 3i+3f]

I'm supposed to manipulate the top one to make it look like the bottom, and what I do to the elements of the matrix, I do to the determinant as well, correct?

so i'd be r2 = r2*3, then the determinant = 12

[winter harbor]

Yup

Exactly

[west flame]

abyine help

[winter harbor]

That's not linear algebra.

[torpid socket]

Can someone help me with c and d or at least explain how to solve them also a and b are correct right? I would have to use the aximos to solve them

[lavish jewel]

i'm not sure this is the right channel but, you can just test all the properties of a field one by one

[torpid socket]

Its in my linear algebra class ;-;

I’m still in my first week that’s probably why

[lavish jewel]

for example Z is not a field because multiplicative inverses are rational

[torpid socket]

Oh

Thank you ed!

[lavish jewel]

so think about each of the properties separately and see if you can come up with counterexamples or explain why the do work

[torpid socket]

Wait isn’t the multiplicative inverse 1

[lavish jewel]

for 1, sure

or did you mean an element of the field times its multiplicative inverse is 1 (or rather the multiplicative identity)?

[torpid socket]

Yeah

That

[lavish jewel]

sure

now take 2

2 * x = 1 makes x be the mult inverse of 2

is x an element of Z?

[torpid socket]

Oh I understand

Well thank you ❤️

[wintry steppe]

Is this reasonable?
\Given invertible matrix with real entries $A$ prove that $A^ku_0=\lambda_0^ku_0$ for any integer $k$
\Begin by recalling that any invertible real matrix is at the very least diagonalizable such that:
[A^k=PD^kP^{-1}]
Furthermore recall that the matrices $P$ are generated through the eigenvectors $u_j$ such that:
[P=[u_0,u_1,...u_n]\iff P^{-1}=\begin{bmatrix}v_0\v_1\...\v_n\end{bmatrix},]
[ u_j v_j=e_j,\ v_ju_j=1,\ u_jv_i=\vec{0},\ v_iu_j=0,\ i\not=j ]
for the $j$th unit vectors $e_j$ where the $j$th position is the only non-zero value, which is equal to 1
Take $A^ku_0\Rightarrow PD^kP^{-1}u_0$:
[(PD^k)(P^{-1}u_0)=([\lambda_0^ku_0,\lambda_1^ku_1,...\lambda_n^ku_n])(\begin{bmatrix}v_0\v_1\...\v_n\end{bmatrix}u_0)]
However by above we get that:
[([\lambda_0^ku_0,\lambda_1^ku_1,...\lambda_n^ku_n])(\begin{bmatrix}1\0\...\0\end{bmatrix})=\lambda_0^ku_0]

[stoic python]

Manzareh

[winter harbor]

Wait, not every real invertible matrix is diagonalizable...

Consider the following matrix

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

[wintry steppe]

woops you're right

[stoic python]

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

[winter harbor]

And I suppose you are misinterpreting the question a little bit

I suppose the question is concerned with the following

[wintry steppe]

don't worry, i've already done the question in a 'valid' way, i just wanted to see if this was correct and you've reminded me why it wouldn't be correct for every invertible matrix

[winter harbor]

Suppose that you have a real matrix A, such that $\lambda_{0}$ is an eigenvalue of $A$ and $u_{0}$ is an eigenvector associated to $\lambda_{0}$. Then, $\forall k \in \mathbb{N}$ we have $A^{k} u_{0} = \lambda_{0}^{k} u_{0}$

[stoic python]

MisterSystem

[winter harbor]

This is true

And in fact, the only eigenvalues of A^k would be the k-th powers of the eigenvalues of A

This is known as the spectral mapping theorem (for polynomials)

You can prove this using Jordan Canonical Form.

[lavish jewel]

you forgot to say that u_0 was an eigenvect earlier

[winter harbor]

Me?

[lavish jewel]

or is this not mentioned anywhere in the original prob?

no, the other person

[winter harbor]

Oh, alright.

[wintry steppe]

that is how the question is stated, that you're given A invertible with real entries and lambda 0 and u0 as eigenvalue and eigenvector

so wait

[lavish jewel]

ok, then yeah

[wintry steppe]

This would be valid then?

[lavish jewel]

i mean, you don't even need anything fancy

just associate from the right

A^k u_0 = AAAA...Au_0

[winter harbor]

You can even prove this by induction

Yeah, exactly

[lavish jewel]

now make parentheses from right to left

and repeatedly exploit A u_0 = lambda_0 u_0

[winter harbor]

With Jordan Canonical Form you can also see that the only eigenvalues of A^k are in fact the powers of the eigenvalues of A

[wintry steppe]

that is quite a bit easier actually, the way i did it before was pretty messy

[winter harbor]

Which is kinda nice

[wintry steppe]

regardless, so with the conditions shown, the thing i brought up earlier would be valid? It would just be invalid if there wasn't an assumed eigenvalue and eigenvector?

[lavish jewel]

for diagonalizable matrices and an eigenvector u_0, sure

[wintry steppe]

Well hold on, if I substitute PDP^{-1} with the SVD factorization USV It DOES work then

right?

[lavish jewel]

only if the left and right singular vectors are mutually orthonormal

i.e. when the SVD equals the EVD 😛

[wintry steppe]

but it's not hard to construct a matrix like that when you're given the eigenvectors and eigenvalues, no?

[lavish jewel]

it's in general not true tho

unless ur mat is symmetric

[wintry steppe]

hmm, and here i thought i had an umbral calculus moment

[pliant atlas]

can some on solve it ?

[lavish jewel]

looks like wrong channel

[torpid socket]

Sorry about the quality but can someone explain how to find the inverse of square root 2 like they are asking us in question 3 I’m gonna be honest I don’t understand the Q square root 2 which is why I’m asking 😅

[winter harbor]

The inverse of √2 in the field Q[√2] would be √2/2

Why?

Well

If we think about it we have "intuitively" that the inverse of √2 should be 1/√2

Now we can multiply this one out by 1

Which is the same as multiplying by √2/√2

[torpid socket]

Alright perfect

[winter harbor]

Notice that we have to do this rationalization

Otherwise

We don't even know what "1/√2" means in Q[√2]

[torpid socket]

I thought it was harder lmao

[winter harbor]

√2/2 makes sense in Q[√2] because it is a rational multiple of √2

[winter harbor]

Np

In any case

These questions feel more like intro to number theory

Or introductory algebra

Rather than linear algebra

[torpid socket]

It is

[winter harbor]

I know that your professor is prolly introducing you to fields

[torpid socket]

Mhm mhm

[winter harbor]

And introducing them via linear algebra makes perfect sense

But this is exactly what linear algebra is concerned with

[torpid socket]

Oh yeah we wont see fields in the exams right ?

Like I checked some linear algebra and calculus 1 exams and I didn’t see any fields

Only in discreet math

[winter harbor]

It depends on the way your professor handles the course man

For instance

I would say fields rather than R or C wouldn't come up in an engineering oriented course

[torpid socket]

R or C ?

😳

[winter harbor]

Maybe in a CS oriented course they would talk about finite fields.

But I wouldn't bet it

[winter harbor]

The reals or the complex numbers

[torpid socket]

Oh yeah we came across finite fields

Oh

[winter harbor]

Most intro to linear algebra courses care mostly about doing things over R or C.

The nice thing is that

Most stuff generalizes to any field almost trivially

Except stuff that explicitly uses some properties of R, say it being an ordered field

Which comes up some times, say with Silvester's Law of Inertia

Or the fact that C is algebraically closed

Say with Jordan Canonical Form

These are particular properties of these fields

And we explicitly use them to prove these results

But again

We don't need C explicitly in order to prove Jordan Canonical Form, only the fact that it is algebraically closed

So we might as well talk about algebraically closed fields instead

[torpid socket]

Got it

Thanks for the explanation

[winter harbor]

Yeah, this is what happens with most linear algebra. Most courses deal with R or C for pedagogical reasons

But most stuff can be generalized to arbitrary fields, modulo some other hypothesis.

[winter harbor]

They are quite nice

[meager steppe]

Hi could somebody explain how to prove that u dot v is equal to u norm times v norm times cosine theta given the cauchy shwarz inequality?

[torpid socket]

Super nice

:)

Tbh I feel like medicine would have been easier than engineering ngl

[winter harbor]

Oh, you do engineering

[lavish jewel]

given only cauchy schwarz, that's not enough info. you'd rather need to do some shenanigans projecting onto the canonical basis

[winter harbor]

Uh, idk how to prove it via Cauchy Schwarz tbh. But it is possible to do via the parallelogram law/polarization identities.

Then you apply the law of cosines

I remember doing it this way

[meager steppe]

Well it just says the inequality and then it says that theta is defined as follows

Ignore the other writing

[lavish jewel]

can you show the full question?

[meager steppe]

It’s not a question it’s just telling me that

Like it’s part of the notes

And it was the very first part of the notes too

[lavish jewel]

and you want a proof that the sum of the products of the vector elements is the same as the product of the norms times the cos of the angle?

[torpid socket]

Oh yeah I have another question what does
Field R/ field Q mean

Only irrational numbers ?

[torpid socket]

Sadly

Started last week

[winter harbor]

R \ Q is the set of irrational numbers, yes.

[stable kindle]

mostly irrational

there should be like

0 in there

[meager steppe]

that was my original question. But just based off of the notes, how do you get from that inequality to the inverse cosine expression

[torpid socket]

Can rational numbers be in it

[winter harbor]

Oh, it should be R \ Q btw

[torpid socket]

Yeah

[winter harbor]

R / Q would be interpreted as R quotient by Q.

Which is a totally different things

[torpid socket]

Oh

[stable kindle]

yeah that's what i was thinking

[torpid socket]

Thank you for mentioning it

[lavish jewel]

ok mumu, so it goes something like this

[torpid socket]

Damn

[lavish jewel]

the dot product of p and q is geometrically given as the length of q times the length of the bottom leg of that triangle

[stable kindle]

is that a stick figure in the bottom right

that's q??

wild

[torpid socket]

Sorry cause my question are too stupid but can someone explain d 😅

Like the circle and the plus inside

[winter harbor]

You are defining a new operation on R.

[lavish jewel]

it's telling you to forget what you know about addition

addition is a whole new thing

[torpid socket]

Hmm

[winter harbor]

And you are interpreting this as the new operation sum. And this new operation of sum + standard multiplication makes R into a field.

[torpid socket]

Oh I see

[winter harbor]

In fact, you have to verify if this is indeed true or false.

[torpid socket]

Thanks!

[lavish jewel]

@meager steppe look at this image, the vectors p and q have associated lengths $\Vert p \Vert$ and $\Vert q \Vert$

[stoic python]

Edd

[lavish jewel]

and from them, the dot product is geometrically defined as the length of the bottom leg times the length of q

you can do some basic trig to find that $\cos(\theta) = \text{leg} / \Vert p \Vert$

[stoic python]

Edd

[lavish jewel]

and so $\text{leg} = \Vert p \Vert \cos(\theta)$

[stoic python]

Edd

[lavish jewel]

and finally, $p \cdot q = \Vert p \Vert \Vert q \Vert \cos(\theta)$

so far so good

[stoic python]

Edd

[lavish jewel]

now we wanna get from here to the other definition

we note that a vector $p = [p_1 p_2 p_3 \dots]$ can be decomposed as $p = p_1 e_1 + p_2 e_2 + \dots$

[stoic python]

Edd

[lavish jewel]

where the e_i are the canonical basis, which is orthonormal

e.g. [1,0,0], [0,1,0], [0,0,1]

geometrically, these vectors are orthogonal to each other

this means e_i dot e_j = 0 if i is different from j

now we want to take p dot q again

$p \cdot q = p \cdot \sum_i^N q_i e_i$

p does not depend on i and we can bring it into the sum

so that

[stoic python]

Edd

[lavish jewel]

$p \cdot q = p \cdot \sum_i^N q_i e_i = \sum_i^N q_i (p \cdot e_i)$, where we moved the q_i to the left as it's just a scalar

[stoic python]

Edd

$p \cdot q = p \cdot \sum_i^N q_i e_i =  \sum_i^N q_i (p \cdot e_i)$, where we moved the q_i to the left as it's just a scalar
```Compilation error:```! Missing $ inserted.
<inserted text> 
                $
l.55 ... q_i (p \cdot e_i)$, where we moved the q_
                                                  i to the left as it's just...
I've inserted a begin-math/end-math symbol since I think
you left one out. Proceed, with fingers crossed.```

[lavish jewel]

i don't see the mistake :x

anywho

it still looks ok

now, we expand p in the same cursed way

[slow scroll]

It doesn’t like that q_i isn’t in math mode I think

[lavish jewel]

$\sum_i^N q_i (p \cdot e_i) = \sum_i^N q_i \left( \sum_j^N (p_j e_j ) \cdot e_i \right)$

and now we exploit the orthonormality of the basis

[stoic python]

Edd

[lavish jewel]

due to orthonormality of the e_i, the inner sum is equal to p_i, since all the other terms are multiplied by 0

leaving us with $\sum_i^N q_i p_i = p \cdot q$

[stoic python]

Edd

[lavish jewel]

which we knew to be equal to $\Vert p \Vert \Vert q \Vert \cos(\theta)$ from the geometric def using a triangle

[stoic python]

Edd

[lavish jewel]

and finally, from this same equation, just solve for theta to get the expression with the inverse cosine

that should've covered all your questions, i think @meager steppe

[sick sandal]

is this free now :p

[lavish jewel]

[sick sandal]

im trying to prove this and i just want some leads to go off :/
1st of all how do i define the space L(V, W)

[wintry steppe]

it's just the set of linear maps from V to W

the idea is that, if you choose a basis of V and a basis of W, then you can associate a unique matrix to each operator T by which matrix multiplication in coordinates is just application of T

so you should try to prove that L(V, W) is isomorphic to M(m by n, field)

[sick sandal]

hoffman kunze didnt define isomorphism yet

what do you mean by that

[wintry steppe]

read a different book then

/s

[lavish jewel]

lol

[wintry steppe]

which statement did you want me to clarify

[sick sandal]

^

[wintry steppe]

give me a moment

you don't actually have to prove that it's isomorphic to some matrix space, it's just more intuitive that way imo

[lavish jewel]

decompose the vectors in some basis and then play around with sums

[wintry steppe]

my bathroom is leaking i need a minute

[lavish jewel]

[sick sandal]

tyt ile look and try to start the proof

[lavish jewel]

here you go

https://math.stackexchange.com/questions/1201784/proving-isomorphism-between-linear-maps-and-matrices

[slow scroll]

Is this some kind of euphemism or is your bathroom like literally leaking

[wintry steppe]

water was dripping from the ceiling and is all over the floor

when im done cleaning it ill type some linear algebra

[lavish jewel]

from the ceiling? shit

[sick sandal]

nvm i figured it out
took a hint from the proof but worked out

[lavish jewel]

proof by reading the proof

[sage shale]

Sorry, I promise this is the last thing about this type of solution. So when I have something like 3h+4e, and my original matrix is just h. Do I multiply my determinant by 3 and add 4

[lavish jewel]

in your problem, the last row is a scaling + addition of another row

[empty siren]

solving the image above using this
𝑥[𝑛] = 𝑢[𝑛 + 2], 𝑎𝑡 − 5 ≤ 𝑛 ≤ 10
can someone help me with this

<@&286206848099549185>

[wintry steppe]

Let $V,W$ have bases $\beta={v_1,\dots,v_n}$ and $\gamma={w_1,\dots,w_m}$. Every vector $v \in V$ can be uniquely expressed in the form $$v = c_1v_1 + \cdots + c_nv_n$$ for some $c_1,\dots,c_n\in F$, so we get a coordinate vector $$[v]\beta = \begin{pmatrix}c_1 \ \vdots \ c_n\end{pmatrix} \in F^n,$$ and similarly a $[w]\gamma \in F^m$ for $w\in W$. Also, for each $i = 1, \dots, n$, $$Tv_i= \sum_{j=1}^m a_{ji}w_j$$ for some scalars $a_{ij} \in F$. Then we have
\begin{align*}
Tv &= T\left(\sum_{i=1}^n c_iv_i \right) \
&= \sum_{i=1}^n c_iTv_i \
&= \sum_{i=1}^n \sum_{j=1}^m c_ia_{ji}w_j \
&= \sum_{j=1}^m \left(\sum_{i=1}^n c_ia_{ji}\right)w_j.
\end{align*}
If we take coordinate vectors with respect to the basis $\gamma$, then we obtain
$$
[Tv]\gamma = \begin{pmatrix}
c_1a{11} + \cdots + c_na_{1n} \
\vdots \
c_1a_{m1} + \cdots + c_na_{mn}
\end{pmatrix} = \underbrace{\begin{pmatrix}
a_{11} & \cdots & a_{1n} \
\vdots & \ddots & \vdots \
a_{m1} & \cdots & a_{mn}
\end{pmatrix}}{:=[T]\beta^\gamma}[v]\beta.
$$
So $T$ in the coordinates given by $\beta$ and $\gamma$ corresponds to exactly multiplication by this matrix $[T]\beta^\gamma$. The map $$L(V,W) \to M(m\times n, F), \qquad T \mapsto [T]\beta^\gamma$$ is an isomorphism of vector spaces (an invertible linear map). Showing linearity is just a computation. For injectivity, if this matrix is zero, then it's clear that $T$ is zero from $[Tv]\gamma = [T]\beta^\gamma[v]\beta = 0$. For surjectivity, define $T$ by the equation $[Tv]\gamma = [T]\beta^\gamma[v]_\beta$.

nuked

[stoic python]

TTerra

[wintry steppe]

[lavish jewel]

such read

much wow

[wintry steppe]

the indices might have some typos

im used to using einstein notation whereever possible

[empty siren]

can you help me with this?

[lavish jewel]

TTerra be like

the sum is there, don't you see it?

the sum:

[wintry steppe]

#point-set-topology message

[lavish jewel]

😌

[lavish jewel]

what's u here

[empty siren]

its this

this is a guide

[lavish jewel]

aight

[empty siren]

i just dont get how to solve it

[lavish jewel]

just substitute the stuff you know

$E_x = \sum_{n=-5}^{10} \vert x[n] \vert ^2 = \sum_{n=-5}^{10} \vert u[n +2] \vert^2$

[stoic python]

Edd

[lavish jewel]

the abs isn't needed and you can do a change of vars

[empty siren]

can you guide me

since im just trying to find the answer since im coding this into matlab to see if the answer is similar

[lavish jewel]

bruh just do it

the unit step func is 1, so you can remove the abs and the square

and if you like you can substitute w = n + 2

$E_x = \sum_{w = -3} ^{12} u[w]$

[stoic python]

Edd

[lavish jewel]

idk, 13?

[empty siren]

im trying to solve it on paper

[sick sandal]

ile understand this approach better when i learn isomorphism :3

[wintry steppe]

i taught you what an isomorphism is right there 😉

it's something you should know the moment you learn what a vector space is

[sick sandal]

[wintry steppe]

it's just a formal way of saying that, after choosing bases, linear operators and matrices are the same

[quaint sage]

Possible dumb question:

If I have a plane, and all the info defining it (i.e 3 points on it, equation, normal vector) how do I determine the change of basis matrix to transform a point into a point relative to the plane (if that makes sense?)

I think if I take the normal and normalize it, does that by itself (or perhaps along with the centroid of the plane) define a CoB onto the plane?

Hopefully that makes sense..

[lavish jewel]

well

you have a point in 3D space, for example, in the canonical basis

and you want to represent it presumably in the bases of 2 vectors on the plane and the plane's normal

so put those 3 vectors in a matrix

as its columns

and you get that known point = matrix * coordinates in the new basis

you know the point and the matrix, not the coords

so you can do matrix^-1 * known point = coords you want

and that's all

a nice choice is two mutually orthogonal vectors on the plane, since you get an orthonormal basis and the inverse is simply the transpose

[sick sandal]

holy fking shit
it all makes sense now
all transformations are determined by what they do to a the basis of V as it maps to W and all of them can be repsented as a matrix between 2 finite dimensional spaces

[lavish jewel]

add and multiply too stronk

[quaint sage]

@lavish jewel Thanks!

[lavish jewel]

👍

[teal grotto]

bruh who even cares at this point 😭

[sage shale]

How do you find 4 unknown coefficients with 4 (x,y) points?

I've been given a curve, and I need the points to match

[teal grotto]

can u post the entire problem?

[sage shale]

It's a written problem

I have points on a graph and I need to find coefficients

basically I have a polynomial and I need to find certain points on the graph

[shrewd vine]

Should a Simplex / Big M question get asked here? Even though it has nothing to do with Early University?

[meager steppe]

Thank you. This wasn’t even on my test but I understand it now

[wintry steppe]

👍

yup

the matrix encodes exactly what the transformation does to your basis of V

you got it

[lethal yoke]

yo im not sure which type of math this is

so ima ask here but you guys lemme know if im wrong

so this is a real-world thing, not a hw question or anything

im building an aquarium and the sheets of glass come in panels 48x60", so i want to know what most optimal layout it to get the most gallons

i think mayb its calc?

idk

from one panel. ik that cubic inch/231 is gal

[north anvil]

Single-Variable Calculus: Optimization

[bold sun]

hi i really dont get this

how do i find k??

basically if i multiplied it correctly i got [ k^2 +2k k+2 -1]= 0

but i dunno what to do after like make each induvidual eqal zero or what?

[pallid rampart]

You should get a single number instead of vector as the answer

[bold sun]

huh how?

[pallid rampart]

Well

The dimension of the left multiplication will give you a row vector of length 3

Then multiplied with the column vector will give you a scalar

[bold sun]

sorry i dont really get that?

[pallid rampart]

$\begin{bmatrix}k&1&1\end{bmatrix}\begin{bmatrix}1&1&0\1&0&2\0&2&-3\end{bmatrix}$ should be a row vector with three numbers

[stoic python]

Whoever

[pallid rampart]

Right?

[bold sun]

so a 1 by 3 matrix?

[pallid rampart]

Yes

[bold sun]

ok yh

[pallid rampart]

Now if you multiply this 1 by 3 matrix by the 3 by 1 matrix, you get a single number

[bold sun]

ooh okk

lemme try that again then gimme a min

if u dont mind

ooh okk i think i got it.... k squared plus 2k+1= 0

so k is -1?

[winter harbor]

Damn TTerra, that's unironically cursed

Diff geo does this to your brain

[bold sun]

okk i substituted it back in and it worked so thank you!!

[lethal yoke]

ok cool thanks

oops

[north anvil]

All good

[shrewd vine]

I'm having a hard converting "What I want to optimise" in a properly Big M expressed problem. It's not for school or work.
Here is what I got:
ProductA takes 3 seconds to make and require 2 ProductB and 3 ProductC.
It takes 5 seconds to make a batch of 2 ProductB.
It takes 2 seconds to make a batch of 1 ProductC.
I want to make at least 2 ProductA per seconds.
No left-over byproducts of ProductB and ProductC. (aka perfect ratios)

Optimization function: How many machines X, Y and Z, (making A, B and C respectively) should I have if I want to minimize the number of machines used.

The way I see it is I could express it in this way:
Max: P = -x -y -z
Constraints:

ProductB used and created = 0: -2/3x + 2/5y = 0
ProductC used and created = 0: -x + 1/2z = 0
At least 2 ProductA /seconds: 1/3x >= 2

But it doesn't seem like I can solve that. What am I expressing the wrong way? Is the Big M method even appropriate for that?

[shrewd vine]

<@&286206848099549185>

[wintry steppe]

Where'd I go wrong? Answer in the back of the book given on the right.

[earnest sigil]

$A^T*A$ is incrct

[stoic python]

lil_giant

[sturdy portal]

What's the canonical notation and name for the following mathematical object

[torn leaf]

Hii

Is the B1 correct?

Ping me too

[raven lotus]

Give an example of a 2 x 2 matrix with no inverse ( singular).
Give an example of a matrix which is its own inverse ( that is , where A−1=A)

[dusk drum]

$\begin{bmatrix}6 & 4\3 & 2\end{bmatrix} $ has no inverse.

[stoic python]

Euclid31415

[dusk drum]

${\begin{bmatrix}0 & 1\1 & 0\end{bmatrix}}^{-1}= \begin{bmatrix}0 & 1\1 & 0\end{bmatrix}$

[stoic python]

Euclid31415

[void cloud]

anyone know how to approach this?

https://cdn.discordapp.com/attachments/611298383637905438/900579917887078400/unknown.png

what i have so far

https://cdn.discordapp.com/attachments/611298383637905438/900580105791885332/image0.jpg

[torn leaf]

anyone?

[winter harbor]

By rotation group I suppose you mean $\text{SO}(3)$, right?

[stoic python]

MisterSystem

[void cloud]

yes the set of real orthogonal 3x3 matrices with det = 1

[winter harbor]

SO(n) for any n is by definition the set of n×n matrices A for which AA^t = A^t A = I and det(A) = 1.

Ok

So

[void cloud]

is how rotation groups are defined

[winter harbor]

First of all

Do you know that the determinant satisfies $\text{det}(AB) = \text{det}(A) \text{det}(B)$.

[stoic python]

MisterSystem

[void cloud]

yes

the property

[winter harbor]

I.e, the determinant of the product of two matrices is the product of the determinants

[void cloud]

i'm thinking det(A) = 1 and det(B) = 1

[winter harbor]

Yup

[void cloud]

because i am assuming both A and B are part of the rotation group

[winter harbor]

So if we have C = AB, then det(C) = det(AB) = det(A) det(B) = 1 * 1 = 1

So C has determinant 1

Now

We need to check that CC^t = C^t C = I

For this

[void cloud]

yeah that proves it's orthogonal

[winter harbor]

We are going to use some properties of the transpose

Are you familiar with the fact that (AB)^t = B^t A^t ?

[void cloud]

yes

[winter harbor]

Nice

[void cloud]

my struggle is applying those properties

for this problem

[winter harbor]

So CC^t = (AB)(AB)^t = (AB)(B^tA^t) = A(BB^t) A^t

Now do we see what we have to do?

B is, by hypothesis, an orthogonal matrix

So BB^t = I

And we have CC^t = A(BB^t)A^t = A(I)A^t = AA^t

In the same way, AA^t = I

So CC^t = AA^t = I

And we are done with the first part

Now we need to check C^t C = I

And I will leave this to you

[void cloud]

alright, thanks!

[winter harbor]

No, it is not.

[torn leaf]

Then how to do

[winter harbor]

There's standard approaches to this kind of problem in general

Like, finding square roots of semidefinite matrices with real coefficients

You don't need anything fancy like this

But think about it

What could possibly be the simplest form matrix B could have?

For me, it seems that a matrix with equal entries would do the job

So I will try to find a matrix with equal entries

Whose square is A

Say B has all entries equal to a certain number "a"

Then, if we compute B^2

We will find that we want 2 a^2 = 2, so a^2 = 1 and this implies a=1 or a=-1.

And you can try this out by yourself

Let B_1 be the matrix whose entries are all equal to 1

And let B_2 be the matrix whose entries are all equal to -1

B_1 ^ 2 = A

And B_2 ^ 2 = A too

[void cloud]

@winter harbor i think i got it

[winter harbor]

Hmmm

We would want to show (AB)(AB)^t = I too

[void cloud]

ahhh

[winter harbor]

Yeah

[void cloud]

well

couldn't i just say C^T C = C C^T ?

which = I

since that's its definition

[winter harbor]

No

Because C and its transpose might not commute

[void cloud]

oh

[winter harbor]

Well

I am going to say we don't need exactly to prove CC^t = I too

But for a totally different reason

Namely because if a square matrix has a left inverse, then it has a right inverse

[void cloud]

yeah

[winter harbor]

And vice-versa

And these coincide

[void cloud]

and 3x3 matrices are square, right

[winter harbor]

This is also a corollary of the more general fact about monoids, i.e if in a monoid every element has a left inverse, then they also have a right inverse and these coincide and etc...

You definitely don't need this too

But just commenting

In any case

I would say that it would be good if you explicitly showed CC^t = I

Instead of relying on any of these facts

[void cloud]

i'm wondering if i can just cite the textbook

[winter harbor]

Because you want to practice dealing with these properties.

[void cloud]

rather than doing separate proof

[winter harbor]

Nah, just do it.

It's not that long of a proof

It's a 1 liner.

[torn leaf]

So it’s 1?

[winter harbor]

$$
B_{1}

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

[stoic python]

MisterSystem

[winter harbor]

And you can take B_{2} = - B_{1} ofc

[zinc timber]

$A, B \in M_n(\mathbb{C})$ then if $B$ is invertible, then there exists a scalar $c$ s.t. $|A+cB|=0$

[stoic python]

Ryuzaki

[zinc timber]

any other method other than using the continuity of det,

[loud fog]

description of ker(T).```

i've algebraically determined that dim(ker(T)) = 2, but don't really understand why

any help is appreciated 🙂

[zinc timber]

what do u need? a proof of dim=2 or geometrical interp

[loud fog]

geometrical interpretation

well and an explaanation as to why it's 2 i g

[zinc timber]

rank=2 means it's a plane in R³

it gets mapped to 0 means the entire plane gets squished to the point 0

[zinc timber]

because R³ has a dimension 3 and R has 1, so a vague analogy is that if u have a bag that can only fit one object but u have 3, u can fit one of them and 2 will be left outside,

[loud fog]

hm

[lavish jewel]

you can do this by picking a basis and writing the functional as a matrix

the matrix would be a 3 x 1 row vector

if you then look for the null space, you get 2 basis vectors for it

and then for an inhomogeneous problem c = v^T x (where c is a constant in the field and v^T is the 3x1 matrix)

then having found a basis of the null space, say vectors a and b, v^T a = 0 and v^T b = 0, you find that if there is a particular solution c = v^T p, then c = v^T (p + s a + t b), where s and t are scalars also in the field

due to linearly and a and b being in the null space

you can then geometrically interpret this

s a + t b is just a linear combination, and if a and b are linearly independent, then this is a plane

so this tells you that for every particular solution p, there is a whole plane containing it that also yields the same value c

[torn leaf]

Got it thank you

Which answer is wrong

[loud fog]

the steps you took make sense; could you walk me through an example? say T(x, y, z) = x + y + z; what would a basis for ker(T) look like then

[lavish jewel]

so, x + y + z can be written as [1,1,1] [x,y,z]^T, yeah?

a row and a column vector

so we look for these x,y,z that yield zero

off the top of my head, one of them is 1, 0, -1

another could be 1, -2, 1

and i chose those so that they are mutually ortho... i think?

so now if you find one particular solution for x + y + z = c, for a fixed c

then you can construct infinitely many solutions by adding scaled versions of [1,0,-1] and [1,-2,1]

for example

let c = 3

then we want [1,1,1] v = 3, where v = [x,y,z]^T

and the claim would be we can arbitrarily add scaled versions of the other vectors and still get 0

let's test it out

,w {1,1,1} dot {1,1,1}

[stoic python]

Results provided by WolframAlpha

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

Requested by Edd#5182

[lavish jewel]

simple enough

now

,w {1,1,1} dot ({1,1,1} + 5*{1,0,-1} + 38.2*{1,-2,1})

[stoic python]

Results provided by WolframAlpha

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

Requested by Edd#5182

[lavish jewel]

i just made up some scalars

[loud fog]

wait i'm utterly mindgamed

[lavish jewel]

this is a consequence of the linear functional being, well, linear

you could replace that 5 and 38.2 by anything you want

and it'll still work

cuz [1,1,1] dot [1,0,-1] = 0

[loud fog]

oh

yeah ofc

bc theyr'e solutions to the

[lavish jewel]

and [1,1,1] dot (x + [1,0,-1]) = [1,1,1] dot x + [1,1,1] dot [1,-2,1] = [1,1,1] dot x + 0

that's easy to see algebraically, simply with linearity

and then you notice that it's a linear combination of vectors in the null space

and since we could build 2 linearly independent ones, they span a plane

which tells you there's a whole plane that your functional maps to 0

[torn leaf]

.

[loud fog]

i see now; i think i was tunnel visioning on the idea that dim(ker(T)) = 3 bc it's elements are contained in R^3

[lavish jewel]

the kernet is a subspace, it need not be equal to the whole space

and it can be trivial too, i.e. contain only the 0 vector

[loud fog]

👍 needed an example to fully convince myself of that fact

thanks a bunch

[lavish jewel]

might wanna look at the rank nullity theorem

[torn leaf]

Anyone?

[zinc timber]

just multiply

[torn leaf]

I did but something is wrong

The system says 8/10

Can u please check it

[zinc timber]

send, someone will ig

[loud fog]

this is what i used to algebraically determine that dim(ker(T)) was indeed equal to 2, but my brain refused to accept this fact since i thought i had a basis for ker(T) that had 3 elements

[zinc timber]

send the product you calculated, AxB and BxA

[torn leaf]

I did it on white board and erased already

[dusky epoch]

well if you don't show or even keep your work then how do you expect to learn from your mistakes?

okay, now that i have done the multiplication

it looks like you misread part D

it asks for the second column of AB but you wrote the second row

[torn leaf]

Got it

Thank you

[stoic python]

MrFlaze20MS

[wintry steppe]

If $F^{S}$ denotes the set of functions from S to F where F is a field, why does considering $F^{S}$ to be a vector space over F require S to be a non empty set?.

[stoic python]

unchained

[stable kindle]

look at the axiomatic conditions for something to be a vector space

[wintry steppe]

Wait a additive element must exist so therefore they must be a element of S such that f(x) = 0. is that it?

[stable kindle]

yes

[wintry steppe]

thanks.

[acoustic path]

Hi there, so I've solved this linear system of equations, but the question is, "how many solutions are there?", how do I figure that out? (a^2-5a+8≠0, a^3-9a^2+23a-15≠0)

[stable kindle]

that's a solution?

[acoustic path]

no

its the matrix

[stable kindle]

you said you solved it

but you haven't solved it

[acoustic path]

like ive got the x values

but the fact that i have a variable "a" confuses me

[marble lance]

a is just some arbitrarily fixed number

It doesn't vary though. So for a given a, you have found one solution.

[stable kindle]

yeah

[acoustic path]

wouldnt it be infinite solutions then cause a can be any number

[stable kindle]

assuming a fulfils the requirements on the right

[stable kindle]

yep

but for any given a, you have one solution, assuming it satisfies the conditions on the right

[acoustic path]

ty, as a last question then, for some reason this calculator wants to put in the zeros?

whats this for?

[wintry steppe]

Suppose a $\in$ F, v $\in V$, and av = 0. Prove a = 0 or v = 0.

[stoic python]

unchained

[wintry steppe]

Should I treat this in a case like?. For example I prove it by assuming a or b is equal to show then show the equality holds. I feel like it something else.

[lavish jewel]

well, av is a vector, right?

[wintry steppe]

yeah

[lavish jewel]

do you know anything about V?

[wintry steppe]

shit sorry I copied my problem, ok V is a vector space over a field F.

[lavish jewel]

i guess that doesn't matter much. for two vectors to be equal, they must be everywhere equal

so since av is a vector, it must equal the zero vector

then you use the definition of scalar mult

[wintry steppe]

ok thanks.

[lavish jewel]

after that you can do case work

e.g. if a is nonzero, you can divide both sides of the eq and you get v = 0

and go from there

[winter harbor]

Yup, just look at the spectrum of both A and B.

Since B is invertible, its spectrum does not contain 0.

And then you do some more memes.

And it is done.

[zinc timber]

[winter harbor]

The idea here is using the fact that the determinant is the product of the eigenvalues of a matrix.

There's another way to do it

Using the fact that C is algebraically closed

Notice that since B is invertible, then we have:
$$
\text{det}(A+cB) = 0 \iff \text{det}(AB^{-1}+c I) = 0
$$

[stoic python]

MisterSystem

[winter harbor]

But C is algebraically closed

And so the characteristic polynomial of AB^-1 has a root for some c

And we are done

[zinc timber]

yeah nice

I was using continuity, which kind of has the same spirit but it requires more work

this one's nice

[winter harbor]

Ofc you would have to write down a full argument tho

But this idea is nice, yeah.

[zinc timber]

I agree

[dusky epoch]

bad tex

[lofty topaz]

Hi guys. Please can you check if my solution of homework is correct ?

[primal schooner]

help plz >_<

[winter harbor]

That's not linear algebra tho

Check out #prealg-and-algebra or #precalculus

[winter harbor]

You definitely did something, but it is incomplete.

In order to prove something is a commutative/abelian group you have to check out 4 axioms

Namely, that the operation is associative, there's an Identity element, every element has an inverse and moreover the operation is commutative.

You didn't check, for example, if R × R {0,0} has an identity

And if each element in here has an inverse

You need to do more work

Moreover

Be more explicitly about what you are trying to check.

It is always good.

Like "Now we are checking the commutative property, and now the associative property and etc..."

[lofty topaz]

Ok, thanks for response. I will try to make an improvements

[fluid lintel]

Can someone help me with this? I have to find for which values of k the matrix is diagonalizable
I tried something but it's very much a mess, I found the eigenvalues which should be 3, k+3 and k-3
but i have trouble with the eigenvectors, especially with those that depend on the value k

[stoic python]

Gaarco

[winter harbor]

I think it is better to think about this in terms of the minimal polynomial.

You have the characteristic polynomial

We know that the minimal polynomial has the same irreducible factors as the characteristic polynomial

And a matrix is diagonalizable iff its minimal polynomial splits into distinct linear factors / has no repeated roots.

So think about what the minimal polynomial of this matrix should be

In terms of k

And for which values of k it has no repeated roots.

You can also think about this in terms of the Jordan blocks of its jordan canonical form.

[fluid lintel]

Thanks, that helped 🙂

[stable kindle]

not sure where this goes but

are there any solutions to
$A^2 = \begin{pmatrix}
cos(a) & sin(a)\
sin(a) & -cos(a)
\end{pmatrix}$ if A can have complex entries

[stoic python]

Kaisheng21

[stable kindle]

my guess rn is no

[lavish jewel]

can't you rotate twice by half the angle?

[stable kindle]

no

the matrix is slightly different

determinant -1, if you see

[lavish jewel]

ah the second col is neg

hmm

you can probably SVD it

[stable kindle]

idk that method

[lavish jewel]

for a fixed value of a

[stable kindle]

nah, i want the general case

trivial solutions for, say, a = 0 or a = pi/2

[lavish jewel]

you can diagonalize both AA^T and A^TA and use that to build the svd

just really annoying to do in the general case rather than numerically

[stable kindle]

yeah again i don't actually know what that method is

what i've tried is

let H be the usual half-angle rotation matrix, and let M be such that M^2 = (1, 0; 0, -1), as then (H^2)(M^2) is the original matrix we're trying to square root, as (1, 0; 0, -1) will flip the second column's signs

then, if (H^2)(M^2) = (HM)^2, we're done.

the above is true if HM = MH. however, i brute-forced it a bit by literally multiplying entries and i think there's no M where M both commutes with H and can be squared to give (1, 0; 0 -1)

[lavish jewel]

right

[stable kindle]

checking that last part rn just in case

[lavish jewel]

i'd say if the matrix isn't diagonalizable in general, it doesn't have a square root

so if you can come up with any case where it isn't, then that's that

otherwise idk

[stable kindle]

hm

right, yeah, my checking shows it works in the specific case where sin(a) = 0

[lavish jewel]

right, since diagonal matrices commute

[stable kindle]

yeah

it's just not an elegant proof

[lavish jewel]

it doesn't generalize to all a

[stable kindle]

yeah i know

to be clear

i'm saying it only works if sin(a) = 0

the condition seems to be that isin(a) = sin(a)

but the way i got there wasn't nice

i just want a nice proof 😔

[lavish jewel]

lemme think

ah

how do you like EVDs instead of SVDs

[stable kindle]

??

[lavish jewel]

eigenvalue decomposition?

[stable kindle]

oh

i mean, if i have to

... not experienced with the implications of eigenvalues tho

[lavish jewel]

then that's it

it's real symmetric -> it's diagonalizable by an orthonormal matrix

i.e. it has a square root for any a

[stable kindle]

...

sounds plausible, i didn't know that

[lavish jewel]

but if you don't like EVDs and SVDs, big oof

[stable kindle]

i mean i can find eigenvalues and eigenvectors

i just don't really know how they fit into the grander scope of LA, if ykwim

how they connect, how to use them in proofs properly...

[lavish jewel]

diagonalization is one of the most powerful results of linalg

if you do an eigenvalue decomposition of a matrix, you express it as some QDQ^-1, where D is diagonal

and Q may or may not have a nice structure

[stable kindle]

right yes

i vaguely know this

[lavish jewel]

the important thing to notice is that a matrix inverse is equivalent to a change of basis transformation

for example

[tranquil steeple]

>> [V,E]=eig([cos(a) sin(a);sin(a) -cos(a)]);
>> V

V =

[cos(a)/sin(a) + (cos(a)^2 + sin(a)^2)^(1/2)/sin(a), cos(a)/sin(a) - (cos(a)^2 + sin(a)^2)^(1/2)/sin(a)]
[                                                 1,                                                  1]

>> diag(E)

ans =

 (cos(a)^2 + sin(a)^2)^(1/2)
-(cos(a)^2 + sin(a)^2)^(1/2)

[stable kindle]

no i think i see that

[lavish jewel]

i give you a vector y, and i want you to represent it in a new basis

[stable kindle]

you just swap the two pairs of sets of basis vectors?

[lavish jewel]

put the vectors of the basis as columns of a matrix A

and we say that y = Ax

[stable kindle]

yes

[lavish jewel]

x is the coordinates of y in the basis A

so then A^-1 y = x

[stable kindle]

yeah

[lavish jewel]

if A is invertible, which it should be if its columns are a basis, cuz they have to be linearly independent

now let's look at an eigenvalue decomp

let's say A = QDQ^-1

and we multiply A by some vector x

[stable kindle]

Q is the eigenvectors, D has the eigenvalues?

iirc

[lavish jewel]

yes

so Ax = QDQ^-1 x

and we associate from the right

step one: change x into the eigenbasis by doing Q^-1 x

step two: take the transformed x into the eigenbasis and apply a transformation

but since we're in the eigenbasis of the matrix, the linear transformation is simply a scaling of coordinates

this is given by the diag matrix D

step three: take the scaled coordinates in the eigenbasis and transform back into the original basis by multiplying the coordinates with the matrix Q

if a matrix is diagonalizable and you know its eigenvectors, you can instead just see how each eigenvector is scaled, and then add them up (take their linear combination)

this can be done for any kind of diagonalizable linear operator, not just matrices

pops up all the time in differential equations, for example

[stable kindle]

hmm

[lavish jewel]

where one uses the fourier method to solve the diff eq for one complex exponential at a time instead of looking at crazy signals

then you add them up

anyway

if A is symmetric, it's diagonalizable

so A = QDQ^-1

[stable kindle]

oh wait

[lavish jewel]

and the square root of A?

Q D^1/2 Q^-1

[stable kindle]

the square root is QD^0.5 Q^-1?

yes

[lavish jewel]

yep

[stable kindle]

ok wild

[lavish jewel]

symmetric matrices behave super nicely

[stable kindle]

yeah

ok let's go for that

ty

[lavish jewel]

the proof that symmetric matrices (with distinct eigvals) are diagonalizable is also pretty straightforward

[stable kindle]

fuck

these eigenvectors are awful

i got eigenvalues 1 and -1

and the x-value of the eigenvector for eigenvalue = 1 is (cos(a)+1)/0??

not sure if i've done something wrong or this is intended lol

[signal mulch]

Help with b pls!

[lavish jewel]

divided by 0? wth

[stable kindle]

idk, i'm bad

[left snow]

what is the matrix?

[stable kindle]

$\begin{pmatrix}
cos(a) & sin(a)\
sin(a) & -cos(a)
\end{pmatrix}$

[stoic python]

Kaisheng21

[stable kindle]

i get eigenvalues are 1 and -1

[dusky epoch]

\cos

\sin

[left snow]

,,
\begin{pmatrix}
\cos a & \sin a \
\sin a & -\cos a \end{pmatrix} \begin{pmatrix}x \ y\end{pmatrix} = \begin{pmatrix}x \ y\end{pmatrix}

[stoic python]

ricey

[left snow]

,align
x \cos a + y \sin a &= x \
x \sin a - y \cos a &= y

[stoic python]

ricey

[winter harbor]

This is note necessarily the case!

Consider the following matrices

[stable kindle]

ok i found something slightly better

and non-awful

[winter harbor]

$A

\begin{bmatrix}
0 & -1 \
0 & 0
\end{bmatrix}
$
\
\
And
$
B

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

[stoic python]

MisterSystem

[winter harbor]

Notice that A is nilpotent

You can compute A^2 and see it is 0

[stable kindle]

you're nilpotent

[signal mulch]

But how would that prove that B-A is invertible?

[winter harbor]

But we have
$
B - A

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

[winter harbor]

It is a counterexample...

[stoic python]

MisterSystem

[winter harbor]

Notice A is nilpotent

B is invertible

But B-A doesn't have full rank

So it can't be invertible

[signal mulch]

could u explain what nilpotent means? Sryy we havent learned that yet

[winter harbor]

A nilpotent matrix just means that there exists some natural number k for which A^k = 0

So the matrix of your example is nilpotent

[winter harbor]

In particular, if A is an n×n matrix

We have that if A is nilpotent, then A^n = 0

[signal mulch]

so since a^31 is 0; Matrix A is nilpotent

[winter harbor]

Yup

[signal mulch]

and if it nilpotent does it also mean that it is not invertible?

[stable kindle]

consider determinants

[winter harbor]

Yeah, a nilpotent matrix can't be invertible because it has non trivial kernel

By definition

The image of A^k-1 is in the kernel of A (we also have to make sure we pick k to be the smallest natural number for which A^k = 0, so we can guarantee A^k-1 ≠ 0)

But yeah, looking at determinants may be easier.

[signal mulch]

yea is it possible to prove that Matrix A is not invertible using determiants?

[winter harbor]

Yeah, a matrix is invertible iff its determinant is non zero.

The idea is that determinants encode multiplicative information about matrices.

[signal mulch]

yes I was trying to figure out how to prove that det(A) = 0

to show that its not invertible

but I wasnt sure how to prove that

[lavish jewel]

you can help yourself with det(AB) = det(A)det(B)

[stable kindle]

wait

[winter harbor]

det(A^k) = det(A)^k

[stable kindle]

i've had a brainwave

[lavish jewel]

you mean det(A)^k

[winter harbor]

oh shit

Brainfart

[hard vault]

yo is it okay if I ask a question or should I wait? I just want something verified real quick

[lavish jewel]

FartSystem

[winter harbor]

Thanks lmao

[lavish jewel]

probably better to wait til this question is done

[torn hornet]

It’s true in characteric 1

[signal mulch]

soo det(A^k) = det(A)^k right?

[lavish jewel]

yes

[winter harbor]

Yeah

[stable kindle]

what did they say originally

[gray dust]

kdetA

[stable kindle]

lmao

[winter harbor]

That was cursed

[signal mulch]

so now I essentailly need to prove that a invertible matrix - noninvertible matrix is not invertible

[stable kindle]

bruh

[winter harbor]

No

[lavish jewel]

oof

[winter harbor]

You don't need to do any of this

Just consider the counter example I just gave

[lavish jewel]

they asked you to prove the opposite, so it suffices to give one counter example to show it's not true in general

[signal mulch]

no they said that if B-A is invertible then prove that it is invertible... if its not then prove that it is not invertible using a counter example

[stable kindle]

and it's not

bc we found a counterexample

[lavish jewel]

here

they say to prove it or show a counter ex

[signal mulch]

yes

[stable kindle]

man this is much easier

than chugging through it myself

[lavish jewel]

note that erik gave you the answer like 1 hr ago

[lavish jewel]

here

[bold pagoda]

hey, I answered the question on the other channel

[winter harbor]

[bold pagoda]

but if it helps, the steps are these

[stable kindle]

oh i'm a fool

[bold pagoda]

figure out what properties A needs to have so A^31 = 0

figure out what properties B needs to have to B is invertable

figure out if B-A can become non invertable

[stable kindle]

i mean tbf

what they wrote was borderline incomprehensible

[bold pagoda]

if I've given too much away I can delet

[hard vault]

okay so if A and B are n x n matricies then AB = I implies that A is invertible, right? This comes from the fact that if you have a linear transformation from R^n to R^n that is surjective then it must be invective. Is this right or am I having a brain fart?

so that I don't need to check that BA = I too

[lavish jewel]

technically you need to show that BA = I

[hard vault]

yeah I thought so too

[winter harbor]

That is correct

[stable kindle]

it works

[winter harbor]

You can also give an algebraic proof too

For instance

[bold pagoda]

is there a case where A and B are nn and AB = I but B*A =/= I?

[stable kindle]

if it's not surjective, then determinant is 0 so it can't multiply to give I

[hard vault]

okay good I thought that I was on to something very wqeird lmao

[stable kindle]

ez

[lavish jewel]

for rectangular matrices, but not square, yea

[winter harbor]

If you have a monoid where every element has a left inverse, then every element has also a right inverse and these coincide.

[hard vault]

oof okay I see

[stable kindle]

well this isn't a monoid where every element has a left inverse, surely

[hard vault]

Okay great, thanks to both of you!

[stable kindle]

don't think so

wait

hmm

[winter harbor]

https://proofwiki.org/wiki/Left_Inverse_and_Right_Inverse_is_Inverse

[winter harbor]

You also don't need any of this and only use properties of matrices too lol.

In any case, a totally algebraic proof is possible.

[lavish jewel]

that was actually simple enough that i understood it without knowing what a monoid is

but yeah, same as saying a matrix A has left and right inverses L and R

so LA = I

LAR = R

[gray dust]

noone needs to know monoids

[lavish jewel]

associate AR = I

then L = R

[gray dust]

not enough structure

[winter harbor]

Yup, the proof goes the same.

[winter harbor]

I mean, Eckmann-Hilton is nice :(

Probably the nicest result about monoids tho

Or maybe I am just too damn ignorant

[lavish jewel]

eckmann sounds like a reaction to a nasty food

[gray dust]

that's like the best of the worst

[lavish jewel]

minmax moment

[winter harbor]

Guys

Don't talk don't about monoids like this ...

[stable kindle]

monoids are cool

i think if you take like groups or vector spaces or whatever

with the operation of the cartesian product/direct sum/whatever you call it

they're a monoid

[winter harbor]

https://groupprops.subwiki.org/wiki/Equality_of_left_and_right_inverses_in_monoid#:~:text=In a monoid%2C if an,thus a two-sided inverse.

[stable kindle]

which is just really neat