linear-algebra

[slow scroll]

(so the third column is correct). Ill leave you with this cuz I gtg, The columns of T_B are exactly [T(v1)_B T(v2)_B T(v3)_B]

We found T(v1)_B = (0, 0, 1), T(v2)_B = (2, 0, 0) (so the second column is right I was dumb before)
and by part a, T(v3)_B = (-1, 1/2, 1)

[violet field]

Ah yes! Sorry I’m trying to work out the math quickly as we speak. Making algebraic mistakes! It’s [ 0 0 1]^t [2 0 0]^t [ -1 1/2 1]^t

And just want to confirm part c is A^-1*[T]B?

[slow scroll]

If you’re using the matrix TB from part b, then it does actually have the form A^-1[T]A since you have to correct for the TB inputs AND outputs in beta. When you first posted you q, I wasn’t looking at it in the context of part b. Sorry if that causes confusion...

[fallow jolt]

Can I get help with c and d?

[quartz compass]

any ideas on how to show they form a basis for R^N?

[gleaming topaz]

I found the determinant of this matrix to be 3a-3 but why do I get a solution when a=1, shouldn't there be no solutions when det=0?

https://gyazo.com/59c666053dd2004a2bc307e603871925

[fallow jolt]

@quartz compass Show they're independent?

That was my first thought, but I'm not sure how to show it

[quartz compass]

yeah

they'll be linearly independent if you can find nonzero scalars that sum to make 0

$c_1 \vec v_1 + \cdots +c_N \vec v_N = \vec 0$

[stoic python]

Merosity:

[quartz compass]

suppose you have it

now take a dot product with one of the vectors with nonzero coefficient and see what happens

[fallow jolt]

Bingo!

Thank you

Would d equal alpha_i? @quartz compass

[quartz compass]

almost

they're orthogonal, but I don't think they said they were necessarily orthonormal did they

[fallow jolt]

it is @quartz compass

Should have mentioned this

[quartz compass]

oh then you're fine

otherwise it'd just by multiplied by the length squared

weird, I wonder what they had in mind, I basically did part d to solve part c, oh well

[fallow jolt]

d seems to be the exact problem as c

but shorter

lol

right?

[quartz compass]

haha yeah idk

I wouldn't worry about it

[fallow jolt]

How would I approach this problem?

I think I got it!

x = V.T*b

e = 0

[fallow jolt]

@quartz compass

Could you help me out with this?

IS V*V.T equal to the 1?

[north sierra]

if you're using gram scimdt for a set of 3 vectors

v1 is easy

but is there a way to check if my work is right for v2 before moving onto v3?

cause like messing up one step will mess up the rest since its recursive

[undone garnet]

any idea

[fallow jolt]

#multivariable-calculus @undone garnet

[dusky epoch]

use the hint

well

do you know the relationship between how many solutions a system has and the det of its coefficient matrix

surely you must have gone over that in class

yes so what must det(B) be then

and det(AB)?

why

yup

so how many sols does ABX = 0 have

so which is it

well there you have it then

you have a solution, more specifically the zero solution

and you can't have only one

so you must have infinitely many

[charred stirrup]

in 2d space, is there a special name for the linear transformation that makes i and j linearly dependent?

[feral mountain]

try singular

@charred stirrup

[charred stirrup]

singular?

[feral mountain]

A matrix is singular iff it has determinant 0.

[gray dust]

for a square matrix with a col of 0s, what do you think?

cool

why don't you try expanding along that col to work out the det

or use something about linear dependence of the matrix's cols

[minor galleon]

Ok guys I need a concept check

So if a matrix is singular

Then one eigenvalue is 0

So the kernel is a subset of eigenspace right?

But if the matrix is non singular

The kernel is NOT a subset of the eigenspace

But can we say that the kernel is the eigenvectors for eigenvalue of 0? Just following from the definition of an eigenvalue

[dusky epoch]

yes. yes we can.

[minor galleon]

Even tho there is technically no eigenvalue of 0?

[dusky epoch]

well if 0 isn't an eigenvalue then your kernel is trivial

[minor galleon]

Wdym by trivial

As in it’s only 0?

[vast torrent]

Right

If no eigenvalue is zero, the kernel is {0}

[half ice]

If an eigenvalue is zero, then its eigenspace = the kernel

[gleaming topaz]

If I want to find a line through 2 lines in R^3, do I just take a point on each line and calculate it from there?

[empty copper]

Sure

[gleaming topaz]

Cool, thanks!

[solemn lotus]

"prove that every square matrix can be expressed as a sum of a symmetric and skew-symmetric matrix"

would appreciate h e l p

[vast torrent]

I didn't know that. Cool.

hmm

wonder if eigendecomposition will help

we're assuming real entries, yes?

[solemn lotus]

yes

[vast torrent]

wikipedia says

[gleaming knot]

Add it with its transpose and subtract it with its transpose

[vast torrent]

the eigenvalues of a skew-symmetric matrix are purely imaginary

icy's method is probably more straightforward

[solemn lotus]

hmm could you elaborate a bit more?

[gleaming knot]

I was actually thinking I gave too big of a hint

Here’s another hint: the same is true for even and odd functions

How does the proof work there?

[solemn lotus]

I think I can do it, I'll try it out

oh wow that was... easier than I thought

matrix + transpose gives one while matrix - transpose gives the other

[vast torrent]

@gleaming knot this way seems obtuse after your hint but can we say that by density we consider diagonalizable matrices and write the real and imaginary parts of the JNF eigenvalues?

[solemn lotus]

thxthx

[gleaming knot]

Well a matrix with real eigenvalues isn’t necessarily symmetric

[vast torrent]

oh right, it's necessary but not sufficient

[gleaming knot]

Yep you can conjugate a matrix by a non symmetric matrix and it doesn’t change its eigenvalues

and conjugation by such doesn’t preserve the symmetric property

[vast torrent]

what do you mean by conjugate a matrix by a non symmetric

[gleaming knot]

Start with M and consider P M P^-1

That’s conjugation

[vast torrent]

More than once I've had to yell at the demon in my head that whispers to make a density argument for lin alg proofs by considering the space of symmetric matrices

It's the devil on my shoulder

tempting me to sin

[gleaming knot]

You can cast it away by telling the devil that the dimension of the space of symmetric matrices is strictly smaller than the dimension of the space of all matrices

[vast torrent]

or I can just pull out an airhorn

doesn't work well on an exam though

the airhorn

hm. what's the dimension of the space of symmetric matrices?

give me aclue, don't tell me the answer

[gleaming knot]

Count how many independent entries you can put in the slots

[vast torrent]

n(n+1)/2 ?

[gleaming knot]

Yep

[vast torrent]

nice

thanks 🙂

I'm studying for my quals in January

[gleaming knot]

😱

[vast torrent]

one day lin alg, one day multi, one day complex analysis

3 hours each

I'm scared

I failed the first time but you can take it more than once

oh let me post one I was having trouble with

practice problems from previous years

Part 1 :

Let $A$ be a $3 \times 3$ matrix with all $0 \le a_{ij} \le 1$

[stoic python]

gfauxpas:

[vast torrent]

Show that $\det A \le 2$ and find such matrices with $\det A = 2$

[stoic python]

gfauxpas:

[vast torrent]

so uh

I figured I'd approach this using a greedy algorithm

on the signs of the cofactor matrix:

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

[stoic python]

gfauxpas:

[vast torrent]

I'd put 1 in a11, 0 in a12,

and 1 in a13

and so on

is that the right start?

[gleaming knot]

Just fix the first row then consider 2x2 minors from then on

Instead of recording

Recursing*

[vast torrent]

okay so we have

$1 \begin{dmatrix} + & - \ - & + \end{dmatrix}$

[stoic python]

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

[vast torrent]

uh

what went wrong

in the latex

$1 \begin{vmatrix} + & - \ - & + \end{vmatrix} + 0 + 1 \begin{vmatrix} + & - \ - & + \end{vmatrix}$

[stoic python]

gfauxpas:

[vast torrent]

yes?

$= 2 \begin{vmatrix} + & - \ - & + \end{vmatrix}$

[stoic python]

gfauxpas:

[vast torrent]

maximized for

$2 \begin{vmatrix} 1 & 0 \ 0 & 1 \end{vmatrix}$

[stoic python]

gfauxpas:

[vast torrent]

so that's an example of a matrix with det(A) = 2

but I didnt really prove you can't get higher

but i wanted to use uihh what's it called

linear programming

not the algorithm, just the theorem that linear programming works

I need help formalizing my thoughts

that to maximize a matrix with entries between 0 and 1, you consider only the "simplices" where entries are 0 and 1

you cna ask Eddy I'm not in a rush

[gleaming topaz]

Given these 2 lines, if I want to find a vector orthogonal to them both can I just use s,t=0&1 and get a vector from that and take the cross product or am I supposed to use their directions vectors? https://gyazo.com/5e73acd8098dddc9b72060ce7ed50ff3

[vast torrent]

why not go one step further with your question. Can you just take s = t = 0?

[gleaming topaz]

You mean cross product of (1,0,-2) x (2,5,-1) ?

[vast torrent]

If taking s = 0 t = 1 would work, then I think taking s = t = 0 would also work. I'm just making your question more extreme

[gleaming topaz]

Oh sorry I might have been unclear

I meant to take both s and t equal to 0 and 1 to get 2 points on the line to get a vector

But not sure if that is right

[vast torrent]

aah

well

because lines through the origin are subspaces

you only need to get a basis vector for each line

and then take the cross product

yes?

[gleaming knot]

Those lines aren’t through the origin though

[vast torrent]

oh, so they're not

[gleaming knot]

Vectors on the line aren’t gonna be parallel to the direction vectors

[vast torrent]

I read it too quickly

[gleaming topaz]

Sorry what do you mean Icy

[vast torrent]

what are direction vectors, unit vectors at the origin?

[gleaming knot]

$\ell_1(t)$ for some random $t$ isn’t gonna be parallel to the direction of $\ell_1$

[stoic python]

Icy001:

[gleaming topaz]

If it's in the line how is it not or am I understanding you wrong?

[gleaming knot]

The vector represented by $\ell_1(t)$ is the vector from the origin to that point

[stoic python]

Icy001:

[gleaming topaz]

Sure, so if t=1 I get (3,2,1)-(1,0,-2) right? Which gives me the vector (2,2,3)?

[gleaming knot]

Oh you did l(t) - l(0) instead of just l(t)

That works

[gleaming topaz]

So say I do this, does the cross product give me the vector which is orthogonal to both the lines?
S=(3,2,1)-(1,0,-2)
T=(3,6,0)-(2,5,-1)
S x T

[gleaming knot]

You can check the orthogonality yourself if you aren’t sure

[gleaming topaz]

Ugh, why didn't I think of that, thanks.

[vast torrent]

:)

Okay back to my problem if you are still willing to help.me icy?

This is only part 1 btw

[gleaming knot]

I can’t spend too long thinking about it but first consider when one of the top entries is 0

[vast torrent]

Mm hmm

[gleaming knot]

Oo I got something

The three 2x2 minors (with appropriate sign) can’t all be positive

Simple algebra inequality manipulation

That solves it

[vast torrent]

are we still only considering entries 0 or 1?

you agree with my reasoning?

[gleaming knot]

No the determinant is cubic in the entries not linear

So you can’t use linear programming

[vast torrent]

no I mean

uh, what do I mean

[gleaming knot]

And the maximum doesn’t necessarily occur at vertices

Of the poly tops

Polytope

[vast torrent]

the answer key reduces it to entries 0 and 1 only but I dont understand its reasoning. the answer key is by students so it might be wrong

[gleaming knot]

Oh hold on

It’s linear in each entry if you fix everythjng else

So fixing 8 of the 9 numbers

The determinant is linear in the last number

[vast torrent]

so you can use linear programming?

[gleaming knot]

So reducing it to 0 or 1 does work

[vast torrent]

okay great, that makes it more tractable 🙂

[gleaming knot]

My way is simpler and doesn’t require reducing it to 0 and 1

[vast torrent]

https://i.imgflip.com/zxoyr.jpg

please*

[gleaming knot]

Prove at most two of the three minors with the appropriate sign can be positive at the same time

You’re basically done then

[vast torrent]

if all 3 were positive you wouldn't be at a maximum?

[gleaming knot]

It’s impossible to have all 3 positive is what I’m saying

[vast torrent]

hmmmm

oh

because if all entries are between 0 and 1

uh

then the product of them is also betwen 0 and 1

no, I dont think Im going the right way

[gleaming knot]

You have to make variables and write down equations lol

Inequalities rather

[vast torrent]

$\begin{vmatrix} a & b & c \ d & e & f \ g & h & i \end{vmatrix} = $

[stoic python]

gfauxpas:

[vast torrent]

$a(ei-hf)-b(di-gf)+c(dh-ge)$

[stoic python]

gfauxpas:

[vast torrent]

[gleaming knot]

yep and I’m talking about the terms ei-hf and so on

I think that’s all the hints i will give

[vast torrent]

I guess I'll come back to this in an hour or so because I'm not seeing it. Thanks for your time icy

[fallow jolt]

Let's say I have an orthogonal matrix

But it isn't square

Call it V

Would V.T * V = I?

I being the identity matrix

[vast torrent]

"orthogonal matrix" is a term used to describe a type of square matrix

[fallow jolt]

Sorry

The columns are orthogonal

And have magnitude 1

[vast torrent]

k

[fallow jolt]

But

Instead of NxN

It’s Nx(N-1)

That’s the trick for this problem

Any ideas?

I think it’ll end up being I

[gleaming knot]

The entries of $V^TV$ have a neat description in terms of the columns of $V$, just derive it from the definition of matrix multiplication

[stoic python]

Icy001:

[gleaming knot]

You’ll find the problem pretty tautological once you do this

[jagged saffron]

elementary matrices represent row operations

they did the row operations for you

now just find the elementary matrix representation

[fallow jolt]

Basically

You can do row operations on the identity matrix

And multiply it

To get the same thing

iirc

Someone correct me if I’m wrong

I remember this being a homework problem for me once

[vast torrent]

@gleaming knot 1 > 1 is the contradiction?

I simplify the inequalities and all the variables end up being 1

assuming all three equalities hold

so I get 1 > 1

[gleaming knot]

I don’t understand very well

[vast torrent]

I have a set of three inequalities

$\begin{cases} ei > hf \ di > gf \ dh > ge \ 0 \le a,b,c,d,e,f,g,h,i \le 1 \end{cases}$

[stoic python]

gfauxpas:

[vast torrent]

yes?

and when I simplify this, am I uspposed to get a = b= c= d = e =f = g = h = i = 1?

[gleaming knot]

You don’t want di > gf

You want the flipped version of that

[vast torrent]

Oooh

[gleaming knot]

Because the coefficient of that is negative so to maximize it need to make it negative

[vast torrent]

Okay so there's no solution to that

Meaning at most 2 minors are positive

[gleaming knot]

Yep

[vast torrent]

So then 2 is a crude upper bound

I need to make it sharp

Oh i do that by my example

Okay great :) part 2 is much harder , ill do it later. Can i @ you when i try it?

Thanks icy

[gleaming knot]

Sure

[rose grotto]

I don't get what to do here

ik how to do

regular simplex

but idk about unbound

[feral mountain]

when you perform the algorithm did you get a non-positive column under an entering basic variable?

@rose grotto

[rose grotto]

what u mean

[feral mountain]

I assume you've tried performing simplex on the program?

[rose grotto]

nvm dw ty

[wind mirage]

http://prntscr.com/q8kmuk

[wind mirage]

yall niggas useless

[quartz compass]

no u

[steady fiber]

maybe if you weren't so useless and posted in the correct channel

could've got answers

y=mx+b is not what #linear-algebra is for

you need #prealg-and-algebra

[rose grotto]

I DONTTT UNDERSTAND AAAA

idk how to tell

if it has no maximum

ty

[rose grotto]

how do I tell if

its unbound

?

<@&286206848099549185>

[rose grotto]

can someone explainnannn

[tranquil trail]

<@&286206848099549185> Is anyone available to help me with my study guide, its involving linear transformations

[scarlet hamlet]

hey

what dimension is this space

i know on M2x2(R), its 4 but idk about complex

would it be the same?

[gleaming knot]

You have to say over what field

For example, $\bC$ is 1-dimensional over $\bC$ (example basis: ${1}$) but it's 2-dimensional over $\bR$ (example basis: ${1,i}$)

[stoic python]

Icy001:

[scarlet hamlet]

oh uh

could you explain the first bit?

C being 1 dimensional over C

i think i get why its 2-dimensional over R because you can express a complex number as a+bi -- so the basis is {1,i}

but im not too sure about the first bit?

@gleaming knot

[gleaming knot]

Every complex number can be expressed as a complex number times 1

So {1} serves as a basis

When we say "over C" it means the field of scalars is C

[scarlet hamlet]

oh

okok i think i get it

[rose grotto]

idk how to tell if it has no maximum

@feral mountain

[viral mason]

i think ur question is more linear programming

[rose grotto]

ya its apart of linear algebra tho

[solemn lotus]

Can i say that det(A+B)=det(A) + det(B)

[gleaming knot]

You can say that but is it true?

$\det(I+I)\stackrel?=\det(I)+\det(I)$

[stoic python]

Icy001:

[solemn lotus]

It is not true then

How would one add two determinants

[gleaming knot]

Do you mean compute the determinant of a sum?

[solemn lotus]

I know that $\begin{vmatrix} a+x & b+y \ c & d \end{vmatrix} = \begin{vmatrix} a & b \ c & d \end{vmatrix} + \begin{vmatrix} x & y \ c & d \end{vmatrix}$

[stoic python]

CoolShot:

[solemn lotus]

i mean the sum of two determinants @gleaming knot

[gleaming knot]

I don't understand, what's stopping you from adding the determinants?

[solemn lotus]

Well I can't exactly add them like matrices, right?

[gleaming knot]

Determinants are 1-dimensional matrices so you can add them

like you'd add two 1-dimensional matrices together

if $\func A V V$ thne $\func{\det A}{\det V}{\det V}$ where $\det V:=\Lambda^n V$ where $n=\dim V$

[stoic python]

Icy001:

[solemn lotus]

$\begin{vmatrix} a & b \ c & d \end{vmatrix} + \begin{vmatrix} p & q \ r & s \end{vmatrix} = \begin{vmatrix} a+p & b+q \ c+r & d+s \end{vmatrix}$?

[stoic python]

CoolShot:

[gleaming knot]

That's just the same as your claim that $\det(A+B)=\det A+\det B$

[stoic python]

Icy001:

[solemn lotus]

yes that is why i am confused

[gleaming knot]

I don't think there's a way to represent the sum of two determinants as a determinant of some formula in terms of the two matrices

[solemn lotus]

i see

oh welp, thanks

[north sierra]

$det\left(A+B\right)\ne det\left(A\right)+det\left(B\right)$

[stoic python]

The-Elite:

[north sierra]

@solemn lotus

but $det\left(AB\right)=det\left(A\right)det\left(B\right)$

[stoic python]

The-Elite:

[dusky epoch]

\det

[north sierra]

oops lol thx!

[wintry steppe]

could anyone give me a hint about this?

[gleaming knot]

Need context for "indices"

and the definition of $T_a^b$

[stoic python]

Icy001:

[wintry steppe]

vector space of tensors

@gleaming knot

[gleaming knot]

I've never used this notation before

[wintry steppe]

wait no

@gleaming knot

[stoic python]

mop:

[wintry steppe]

it's swapped in the book tho

[gleaming knot]

I feel like upper index should represent duals too

[wintry steppe]

it's preference i guess

[gleaming knot]

Ok so it looks like the only thing you are asked to check is that the trace of a sum of two functions is the sum of the traces of the two functions?

[wintry steppe]

huh

[gleaming knot]

Seems like it was defined for you with no issues, the domain and codomain are right, so you just have to check linearity?

[wintry steppe]

i don't really get what the trace is doing here

is it identifying the trace of a tensor with an endomorphism and taking the normal trace of that

[gleaming knot]

Looks like it's taking "componentwise" trace

[wintry steppe]

but then wouldn't that be a map from $T^{k+1}_{l+1}(V) \to T^1_1 (V)$

[stoic python]

mop:

[gleaming knot]

No, it's still a function of the other variables

[wintry steppe]

oh

wait so what's with this

[gleaming knot]

what stopped me from understanding this is m

[wintry steppe]

m

[gleaming knot]

Hm let's consider the simplest case

If $M\in V^*\otimes V$ nad so we write it as $\sum_{k=1}^n \lambda_k\otimes v_k$

[stoic python]

Icy001:

[gleaming knot]

The trace should take $M$ to $\sum_{k=1}^n\lambda_k(v_k)$

[stoic python]

Icy001:

[gleaming knot]

So $F^{m}_m$

[stoic python]

Icy001:

[gleaming knot]

that's gotta be special notation for "evaluation" or something

[wintry steppe]

it's Einstein notation

[gleaming knot]

like combine the indices labeled m with the <V^*, V> canonical pairing

[wintry steppe]

$\sum^n_{m=1} F_{mm} = F^m_m$

[stoic python]

mop:

[gleaming knot]

oh so m is a dummy variable

[wintry steppe]

indeed

[gleaming knot]

what does $F_{11}$ mean for example (taking the first term in the sum)

[stoic python]

Icy001:

[gleaming knot]

Wasn't $F$ an element of $V^*\otimes V$

[stoic python]

Icy001:

[wintry steppe]

$F = F(\varphi^{j_1}, \hdots, \varphi^{j_{l+1}}, E_{i_1}, \hdots E_{i_{k+1}})\varphi^{j_1} \otimes \hdots \otimes \varphi^{j_{l+1}} \otimes E_{i_1} \otimes \hdots \otimes E_{i_{k+1}}$

[stoic python]

mop:

[gleaming knot]

Hm shouldn't it be the dual basis for the simple tensor

[wintry steppe]

wait that's the other way around

but wait what

oh swap the order

[gleaming knot]

Well I understood what you wrote as defining how to understand an element of $(V^)^m\otimes V^n$ as a function on $V^m\otimes (V^)^n$

[stoic python]

Icy001:

[gleaming knot]

Just take the function which when evaluated on $w$ is the component of $w^*$ in its expansion as a tensor

[stoic python]

Icy001:

[wintry steppe]

uuuuh notation

[gleaming knot]

Yep I'm using math notation

[wintry steppe]

indeed

ok wait

if $F \in T^{l+1}_{k+1} (V)$, then $\tr F$ is identified with the trace of an $\Bar{F} \in \text{End}(V)$, since for finite dimensional $V$, $\text{End}(V)$ is iso to $T^1_1 (V)$

[stoic python]

mop:

[gleaming knot]

I feel like the phrase "fixing the rest of the variables" needs to be in there somewhere

[wintry steppe]

brb ree

ye and then you have those two empty slots

[gleaming knot]

ya

[wintry steppe]

so $\tr: T^{l+1}{k+1} (V) \to T^1_1 (V) \to T^{l}{k} (V)$

[gleaming knot]

That doesn't make sense

[wintry steppe]

wait no that's dumb

[gleaming knot]

A map like that would lose so much information

[wintry steppe]

i kinda get the idea

but how would i prove this is well defined

aaaaaa

[gleaming knot]

I don't see any possible issues that could prevent it from being well defined o_O

[wintry steppe]

ok

lemme do latex

$(tr Q)^{j_1, \hdots, j_l }{i_1, \hdots i_k} = Q^{j_1, \hdots, j_l m}{i_1, \hdots i_k, m}$

fuck i need tensor in my preamble

[stoic python]

mop:

[gleaming knot]

Let me express it in math notation

[wintry steppe]

[gleaming knot]

$\sum_{1\leq i,j\leq d}e_i\otimes e_j^\otimes A_{ij}$ represents a general element of $V^{\otimes n}\otimes (V^)^{\otimes m}$

where I picked out a basis $e_i$ of one of the $V$ factors and the dual basis $e_j^$ of one of the $V^$ factors

[stoic python]

Icy001:

Icy001:

[gleaming knot]

ok and $d$ is the dimension of $V$

And trace takes that to $\sum_{1\leq i\leq d} A_{ii}$

[stoic python]

Icy001:

[wintry steppe]

looks like a pumped up matrix

[stoic python]

Icy001:

[gleaming knot]

well $A_{ij}$ is a tensor of rank $(m-1,n-1)$

[stoic python]

Icy001:

[wintry steppe]

wait what

isn't it just a component

[gleaming knot]

I combined the remaining components together

[wintry steppe]

i don't get this part

[gleaming knot]

I wrote $V^{\otimes n}\otimes (V^)^{\otimes m}$ as $V\otimes V^\otimes (V^{\otimes(n-1)}\otimes(V^*)^{\otimes(m-1)})$

[wintry steppe]

i mean

[stoic python]

Icy001:

[gleaming knot]

in order to pick out the factors being traced out

[wintry steppe]

wut

but the vector space tensor prod doesn't commute

[gleaming knot]

I'm writing a tensor product of vector spaces itself

They're isomorphic

like I'm just reordering the factors for convenience

I'm not changing anything

[wintry steppe]

i think my knowledge of tensor products is screwing me here

this is how it was defined

[gleaming knot]

So in the last representation, a generic element is of the form $\sum_{1\leq i,j\leq d}e_i\otimes e_j^\otimes A_{ij}$ for $(e_i)$ a basis of $V$, $(e_i^)$ the dual basis to $(e_i)$, and $A_{ij}\in V^{\otimes(n-1)}\otimes(V^*)^{\otimes(m-1)}$

[stoic python]

Icy001:

[gleaming knot]

Ah yes I view elements of a tensor product as linear combinations of pure tensors

[wintry steppe]

wut

explain plz

[gleaming knot]

If $V$ and $W$ are vector spaces, then $V\otimes W$ is the vector space spanned by symbols $v\otimes w$ for $v,w\in V$ subject to relations $(v_1+v_2)\otimes w=v_1\otimes w+v_2\otimes w$ and so on

[stoic python]

Icy001:

[wintry steppe]

ah

[gleaming knot]

With these relations, a basis of $V\otimes W$ is obtained by tensoring each pair of basis element of $V$ with a basis element of $W$

[wintry steppe]

this makes sense

[stoic python]

Icy001:

[wintry steppe]

ah this does make sense

so they do 'commute' in this case

[gleaming knot]

Well

$V\otimes W\cong W\otimes V$ via the map $v\otimes w\mapsto w\otimes v$

[stoic python]

Icy001:

[wintry steppe]

but them commuting would mean all bilinear maps are symmetric

ok no

wait so then what was the reordering thing above

sorry ee

[gleaming knot]

Commuting means $v\otimes w=w\otimes v$ which isn't true

[stoic python]

Icy001:

[gleaming knot]

like

[wintry steppe]

ye

i got it

they don't commute

[gleaming knot]

I'm just rearranging the function's arguments so to speak

[wintry steppe]

oh i see what you did up there

right i got that

so i get the first part

[gleaming knot]

I wrote that to show that it's self-evidently a linear map

because if I add it to another similar expression but with $B_{ij}$ then it just takes it to the sum of $A_{ii}$ + $B_{ii}$

[stoic python]

Icy001:

[wintry steppe]

well defined means basis independent tho

how would i show that that is basis independent

[gleaming knot]

ooo

[wintry steppe]

i think

[gleaming knot]

Oh I have it

[wintry steppe]

writing it like this may be helpful

but idk

[gleaming knot]

Write a generic element as $\sum_{1\leq i,j\leq n}a_i\otimes b_j\otimes A_{ij}$ for any arbitrary basis $a_i$ and $b_i$ of $V$ and $V^*$

[stoic python]

Icy001:

[gleaming knot]

The formula is that it takes it to $\sum_{1\leq i,j\leq n}\langle a_i,b_j\rangle A_{ij}$

[stoic python]

Icy001:

[gleaming knot]

before, we had a basis and the dual basis so the only surviving terms were the diagonal terms

Hm

Let's take it one step further

Treat $V\otimes V^*$ as $\End(V)$

[stoic python]

Icy001:

[gleaming knot]

We take $f\otimes A\mapsto \tr(f)\otimes A$ and extend by linearity

[stoic python]

Icy001:

[wintry steppe]

what if i just proved $F^{j_1, \hdots, j_l m}{i_1, \hdots, i_k m} = F^{g_1, \hdots, g_l m}{q_1, \hdots, q_k m}$ by means of F's transformation rule

[stoic python]

mop:

[gleaming knot]

Hm you could do that

I was doing it the mathy way which is to provide a description of the map that doesn't refer to any basis at all whatsoever

[wintry steppe]

you do your thing first

[gleaming knot]

I think I already did

$f\otimes A\mapsto \tr(f)\otimes A$, done

[stoic python]

Icy001:

[gleaming knot]

no basis in sight

[wintry steppe]

what

oh

[gleaming knot]

$f\in\End(V)\cong V\otimes V^*$ and $A$ is an arbitrary rank $(n-1,m-1)$ tensor

[stoic python]

Icy001:

[wintry steppe]

àh

that's pretty cool

ty for that

[gleaming knot]

Yey

Dam this brought to explicit knowledge something I had always "known" but never actually known

[wintry steppe]

i need to be better at maff

[gleaming knot]

That $\tr\colon\End(V)\to F$ \textbf{is} the same thing as the canonical pairing $\langle\cdot,\cdot\rangle\colon V\otimes V^*\to F$

[stoic python]

Icy001:

[urban bough]

Is the eigendecomposition of a PD matrix always it’s SVD?

[gleaming knot]

I wanna say yes, SVD of a diagonalizable square matrix is the same thing as the eigendecomposition

[urban bough]

Me too

But I dunno for sure

[vast torrent]

can someone please suggest a video teaching QR factorization? There are a lot of math videos on youtube and I don't know which channels are the best or most liked

[lone quail]

Does anyone know any good books on conics and quadrics in linear algebra?

[charred stirrup]

if two matrices are row equivalent, are they the same matrix? and in turn, have the same RREF?

[dusky epoch]

if two matrices are row equivalent, are they the same matrix?
no not necessarily
and in turn, have the same RREF?
yes

[minor galleon]

Is there a condition that tells us when the inverse of a 3 x3 matrix satisfies the characteristic equation?

[dusky epoch]

????

[north sierra]

lol

what

[vast torrent]

Maybe you're saying that

If Chi(A) is the characteristic polynomial of A

When is Chi(A-¹)=0

That's my best guess

If ChiA=a + bx + cx²+dx³, ChiA-¹=d+cx+bx²+ax³ @minor galleon

[minor galleon]

yeah thats what im saying

[vast torrent]

I dunno how to tell by looking at the mtx

[timber minnow]

you have to ask a question to get an answer :p

[native lodge]

From the given equations, it’s not clear at all what the solution is, or if there is one.

Instead we’ll write the system in matrix form, and tack on the solutions in the augmented matrix

Then we perform elimination

We get to our upper triangular form and now we can see what the solutions are

The formal procedure is called Gaussian elimination

No, upper triangular form means there are zeros below the entries of the main diagonal

[lone quail]

another option would be, without any elimination (this is a more thoeretical approach):

1. find the rank of the matrix made by just the coefficients of the equations
2. find the rank of the matrix made by that matrix + the constants
   3)the difference in rank between the original matrix and the starting space represents the number of solutions

4. this is true if the rank of both matrices is the same

[native lodge]

How would you be able to tell the rank by just looking at the coeff. matrix?

[lone quail]

finding the determinant

if 0, try find a smaller 2x2 matrix with non zero determinant

[native lodge]

How can I tell between rank of 1 or 2? That’s possible for a 3x3

Oh, you said it above

But for the moment, he’s not there yet, lol. So elimination it is.

[lone quail]

yeah, thing is they never taught us elimination xd

[native lodge]

For the first column you’ll have to zero out the all entries below it, since the first entry of column 1 is on the main diagonal.

Yes

There’s no other choice

[charred stirrup]

when you're trying to find the determinant of a matrix, is it advisable to row reduce and make some 0 elements and then apply cofactor expansion?

[vast torrent]

@charred stirrup ri->ri+crj doesn't change the determinant but the other operations do

Replacing a row with a row+constant times another row

Switching 2 rows changes the sign of the determinant

[charred stirrup]

i know interchanging a row makes adds a (-1), but there is more?

[vast torrent]

Multiplying a single row by a constant multiplies the determinant of that constant

Nonzero constant of course

[charred stirrup]

yup

[vast torrent]

Lastly

[charred stirrup]

what is the rule for scaling the whole matrix?

[vast torrent]

Since det(A)=det(A^t), you can do column operations too

Good question. You tell me.

[charred stirrup]

its that project website

i forget

1 sec

[vast torrent]

Nooo

Use your brain

[charred stirrup]

but i document R

lol jokes

[vast torrent]

That is only true for 2x2

Tell me the general case of det(cA)

[charred stirrup]

if we do each row, and there are n rows

c^n is my intuitive guess

[vast torrent]

That's a good guess

The proof would be by induction

[charred stirrup]

omg

u mean that k+1 stuff

?

[vast torrent]

But i wont make you do induction lol

[charred stirrup]

no more

brain hurt

calc 1 sucked

[vast torrent]

Yeah "that k+1 stuff"

Hah, it's not a calc thing

It comes up a lot

Hmm, can you do it without induction? Thinking

[charred stirrup]

the only proof i know other than induction is proof by necessity

[vast torrent]

I don't know that term

[charred stirrup]

if it weren't true someone smarter than me would have noticed it before me

[vast torrent]

-_-

[charred stirrup]

btw, is my c^n guess wrong? I really dont know if im missing a factor somewhere

[vast torrent]

It's right

You can argue geometrically

[charred stirrup]

good

in stats they sneak in a -1 somewhere and it always messes me up

but thank you alot

i still cant find the website

[vast torrent]

The volume of aparallelipiped is scaled in each of its dimensions when you grow or shrink the whole thing

What website

[charred stirrup]

it has all the properties of the determinant for matrix

[vast torrent]

Most of them are probably on proofwiki.org

https://proofwiki.org/wiki/Category:Determinants

[charred stirrup]

my geometric interpretation only goes as far as the basic parallelepiped made from e1, e2, and e3

[vast torrent]

So go to 3d

If i have 1 liter of air in a balloon

And i double the air in the balloon i blow air into the balloon until the diameter is twice as wide

What happens to the volume?

,w volume of a sphere

[stoic python]

Display results online and refine query

Requested by gfauxpas#9497

[vast torrent]

The radius doubles

So the volume grows by 2^3

Because of the a^3

Wait

Aaaahhhh

[charred stirrup]

I thought litres of air is V

[vast torrent]

That's not how balloons work i meant

You blow air into it until its diameter is twice as long

Sorry sorry sorry

Fixed, sorry to confuse you

That's related to why ants can carry things so much heavier than themselves

[charred stirrup]

hey

I think I am realizing what u were trying to teach me earlier

i didnt even know

for the c^n thing

[vast torrent]

Good to know i occassionally teach things successfully

[charred stirrup]

if you were to multiply each row by k, one at a time

it really is the same effect as c^n

[vast torrent]

Yesssss

[charred stirrup]

i feel like there is a mathematician rolling in their grave because i dont know the name of that property

[vast torrent]

Who says it has a name

Idk a name

[charred stirrup]

nice

if someone brilliant like u dont know then nobody needs to

literally feel my brain cell count growin

[quartz compass]

another way to prove it is

det(cA)=det(c*I*A) = det(c*I)*det(A) = c^n det(A)

since det(c*I) is just multiplied by the cs on the diagonal

I guess in some sense you could call factoring out a c from each column or row as one aspect of the determinant being a multilinear operator, being linear in each of the columns, if you are super desperate to name it haha

[vast torrent]

Homogeneity of order n

[north sierra]

if i have a matrix

and i want to say like U1 = the first column of the matrix, U2 = second column of the matrix

how do i write the first column of the matrix... mathematically ?

hmm i could write like am1 maybe

am2

am3

...

[charred stirrup]

@north sierra U1 = [1st entry, 2nd entry, .... , nth entry]

and by entries i mean the entries down the column you want to write out

[dreamy cedar]

@charred stirrup nooo this is Patrick

[wintry steppe]

trying to prove $\Phi: \text{End}(V) \to T^1_1 (V): A \to \Phi A(\omega, X) := \omega(AX)$ is an isomorphism

[stoic python]

mop:

[wintry steppe]

Got the linear part

got to $(A-B)(X) = 0$ when I was trying to prove it was injective

[stoic python]

mop:

[wintry steppe]

Not sure what to do then

Then I guess subjectivity would follow from rank nullity

[desert void]

So here's a really sticker for me

My girlfriend gave me a situation and I'm having a hard time figuring it out

Wait this isn't where I ask

[wintry steppe]

<@&286206848099549185> ^ been >15 mins

but wait is $\dim T^1_1 (V) = \dim \text{End}(V)$

[stoic python]

mop:

[gleaming knot]

Find an inverse

[wintry steppe]

aaaaaaaah

I proved it’s injective

[gleaming knot]

Ok it reduces to constructing a vector from a functional $\func f {V^*} k$

[stoic python]

Icy001:

[gleaming knot]

Which is only possible if $\dim V<\infty$

[stoic python]

Icy001:

[gleaming knot]

so that suggests you have to invoke a basis to construct this vector

well anyway

$\dim(V^*\otimes V)=n^2=\dim(\End(V))$ so you're done

[stoic python]

Icy001:

[wintry steppe]

Wait how

[gleaming knot]

I'm using $n$ as the dimension of $V$

[stoic python]

Icy001:

[wintry steppe]

ye but why $n^2$

[stoic python]

mop:

[gleaming knot]

which side are you unsure of equaling $n^2$?

[stoic python]

Icy001:

[wintry steppe]

both

[gleaming knot]

ok the left side is a result of the fact that dimensions multiply when you take tensor product

[wintry steppe]

Ah yes

[gleaming knot]

the right side is a result of the fact that the space of $n\times n$ matrices has dimension $n^2$

[stoic python]

Icy001:

[wintry steppe]

is $\End(V)$ a vector space?

[stoic python]

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

[gleaming knot]

Yes

whoa that's weird

How do you not have \End in your commands

Do you have a messed up preamble?

[wintry steppe]

over what field

[gleaming knot]

over the same field as the one V is over

[wintry steppe]

idk I’ve never touched my preamble

[gleaming knot]

I only touched mine to add \usepackage[shortlabels]{enumitem}

which I'm sure does not have \End in it

[wintry steppe]

forgive me for sounding nooby but isn’t commutativity of multiplication a vs axiom

wait no it’s not

that’s on the field right

[gleaming knot]

Vector spaces only have scalar multiplication but not multiplication of two vectors

[wintry steppe]

ye got it

silly me

[vast torrent]

@wintry steppe

Are you sure you're a scientist

You seem pretty mathy ;)

Come to the light side

[wintry steppe]

I’m trying lol

So then it’s an iso by rank nullity

[vast torrent]

The dark side is quicker and easier to make money with,but it ultimately will cost you your soul

[wintry steppe]

Indeed

Math is great

[north sierra]

is this a set of vectors ?

[vast torrent]

Your proof is fine

@thin hazel it's correct

[gleaming knot]

Why is $\dim(\bR^{10})$ at least 10? Why can't it be less than 10?

[stoic python]

Icy001:

[gleaming knot]

👀

[vast torrent]

Oh snap

Do you really need to prove R^n has dimension n

At least *

[north sierra]

I would say something like since there are only 6 vectors in the set, it'd be impossible to have 10 pivot rows. thus it wont span R^10

[vast torrent]

Lol

What has he given you

[north sierra]

what the heck

[gleaming knot]

I think you can prove this statement: If $a_1,\dots,a_n$ are linearly independent in $V$ and $b_1,\dots,b_m$ span $V$, then $n\leq m$. Then apply it to this situation by taking $n=10$ and $a_i$ the standard basis vector, and $m=6$ and $b_i$ your chosen 6 vectors

[stoic python]

Icy001:

[vast torrent]

Rank nullity?

Okay so

Hmm

[north sierra]

damn this is tough

[wintry steppe]

rank nullity is op

[vast torrent]

I feel.like you should choose a 10x4 matrix of the 6 vectors chosen

You claim the image of the matrix has dim>=10

Then nullity has a negative dimension

Impossible

?

[north sierra]

👌

[vast torrent]

It's nuking

As mathematicians say

[gleaming knot]

any theorem about matrices seems like it rests on the shoulders of the basic facts about vector spaces

of which this is one

[vast torrent]

Using a super powerful theorem to solve a simple problem is called nuking the problem

[gleaming knot]

\begin{Theorem}If $a_1,\dots,a_n$ are linearly independent in $V$ and $b_1,\dots,b_m$ span $V$, then $n\leq m$.\end{Theorem}

[stoic python]

Icy001:

[gleaming knot]

^ the theorem that leads to this

(you have to prove the theorem of course)

😱

[vast torrent]

But he uses rank nullity

Damn

[gleaming knot]

ok I worked out the proof

[vast torrent]

Of?

[gleaming knot]

This is an exercise in arguing that you can replace the supposed spanning set of 6 vectors one by one with 6 standard basis vectors

[vast torrent]

Did you not like my nuke

[gleaming knot]

at which point you get a contradiction because 6 standard basis vectors obviously don't span $\bR^{10}$

[stoic python]

Icy001:

[gleaming knot]

Your nuke is very likely dependent on proving this

circular proofs are a no go

[vast torrent]

Depends what's written on the notes already

[gleaming knot]

Well I personally wouldn't accept a proof of the fact that every number is divisible by some prime number using the fundamental theorem of arithmetic

[vast torrent]

But presumably you'd know what prime numbers are

[gleaming knot]

The fundamental theorem says every number has a unique decomposition in terms of primes, and therefore even to state it requires you to know that every number has a prime factor

[vast torrent]

Whereas saint "doesn't know" what a vector space is

[gleaming knot]

Well you can still work out the proof I have in mind using $\bR^{10}$ only

[stoic python]

Icy001:

[gleaming knot]

No the replacing preserves the hypothesis that the 6 vectors span $\bR^{10}$

[stoic python]

Icy001:

[gleaming knot]

Start with writing each standard basis vector $e_1,\dots,e_{10}$ as a linear combination of the 6 hypothetical vectors

[stoic python]

Icy001:

[gleaming knot]

No no no

[wintry steppe]

uh can i send a question here

[gleaming knot]

It's a proof by contradiction where you assume that there is a set of 6 vectors which span $\bR^{10}$

[wintry steppe]

1b plx

plz

[stoic python]

Icy001:

[gleaming knot]

Suppose this set exists and then write each of $e_1,\dots,e_{10}$ in terms of this set. Then you say, ok $e_1$ can replace something, $e_2$ can replace something else, and so on up to $e_6$ then you realize that your assumption turned into the statement that ${e_1,\dots,e_6}$ spans $\bR^{10}$ which is obviously false

[stoic python]

Icy001:

[wintry steppe]

ya but idk how to do it with p double prime x

1a was easy

[gleaming knot]

just write out the formula for double prime explicitly

[wintry steppe]

do i write it in polynomial form

[gleaming knot]

yes

[wintry steppe]

so since it is going from P3 to P1

write it as a subscript 3 cubed a subscript 2 squred such and such

and double derivative that

[north sierra]

how would i do 2 b)

[gleaming knot]

Yes 123brocklee exactly

[wintry steppe]

for orthogonl

just find dot product

of u and v

and if that equals to 0

it is going to be mutually orthogonal i believe

[north sierra]

2 b)

[wintry steppe]

oh im retarded

mb

[gleaming knot]

You gonna have to solve some equations

[wintry steppe]

its gonna be in parametric form

free variables

[gleaming knot]

one might suggest "cross product" but you need to write down the equations to prove that the space of such vectors is 1-dimensional

😆

[wintry steppe]

alright ill do 1b and ill brb

and can i check with u guys if its right

[north sierra]

r u guys talkign about my Q?

[gleaming knot]

ya

[north sierra]

oh

so how would i do this?

[gleaming knot]

np!

[north sierra]

i dont think ive learnt cross product

[gleaming knot]

You can write down the condition that a vector $(x,y,z)$ is orthogonal to both of the given vectors with 2 linear equations in $x,y,z$

[stoic python]

Icy001:

[gleaming knot]

just solve them

[wintry steppe]

im assuming i did this right

which came out as not being a linear transformation for 1b

[north sierra]

how to flip

rotate

[wintry steppe]

uh

shit yea mb

ill send a better pic

Or does the constant in the end matter if it doesnt equal

[north sierra]

my prof ha

my prof has this answer

idk how he got it

[odd yarrow]

is anyone a lin alg tutor here?

[gleaming knot]

derivative should be linear

so should second derivative

You probably made a silly mistake somewhere

[north sierra]

what

where are you getting derivative from????

[gleaming knot]

Answering the other guy's questions

[north sierra]

oh

[wintry steppe]

Oh okay

It hust the constant thta doesnt equal

But the 6x equals

[gleaming knot]

The constant should equal too

Unfortunately I can't read your work

[wintry steppe]

Alright will look over again

Thx for help bro

Much appreciated

[north sierra]

what even was your question?

thought that pic was answering my Question lol

im so lost

but i couldn't read it

[wintry steppe]

u guys know how to diagonlize a matrix

[odd yarrow]

https://www.youtube.com/watch?v=U8R54zOTVLw

[gleaming knot]

if you want to blow your mind with an application of this, see (or do yourself) the linear algebraic derivation of the closed form of the Fibonacci sequence

[wintry steppe]

last question

how the fuck do you the find the distance of x^3 plus x^4

for 2a

[gleaming knot]

d is the bottom right entry of the matrix

not the distance function

[wintry steppe]

how do you show that is a kernel of the linear transformation though

because i need to show that kernel is 0 which is a nullspace

[gleaming knot]

The words in your question don't make sense

[wintry steppe]

im just trying to ask, how do you find the kernel of T given that polynomial of P4

[gleaming knot]

find which matrices get sent to the zero polynomial

[wintry steppe]

alright

im taking the class rn and u seem to know more than me

did you already take it or are you taking it rn

[terse mirage]

so pretty much my entire class got the last question on our exam wrong

which is funny because it was basically

sups A is invertible, find the basis for the null space of A

[gleaming knot]

wtf that seems to say more about the teacher

[feral mountain]

did the people who got it wrong put {0}?

[terse mirage]

Essentially yea

the book apparently never went over the fact that null is a basis for a point

[gleaming knot]

oh nvm

I thought by "getting it wrong" I thought they didn't know the answer

[terse mirage]

I mean

this teacher was not a great person to teach intro to linear

so I think a good portion of the students didnt even understand what a basis was

[copper mason]

what's a basis lol

[gleaming knot]

a minimal spanning set, or alternatively, a maximal linearly independent set

[charred stirrup]

is the only elementary row operation that doesnt affect the determinant the operation where you add a multiple of another row?

[dusky epoch]

yes

[charred stirrup]

nice thank you

[gleaming knot]

No

Swapping rows or columns an even distance apart as well

[dusky epoch]

no

any single column swap flips it

no matter what cols they are

[gleaming knot]

I knew that >.<

All transpositions are odd

[charred stirrup]

is this because you make a row/col of 0s, and then if you expand along there you will get a 0 determinant?

[dusky epoch]

well that's one way of establishing it yes

[lone quail]

I was looling at this linear algebra textbook

Why does it say that the set {0,1} is a field with respect to addition and multiplication?

Doesnt a field need an opposite element eg -1?

[dusky epoch]

in the field with two elements, -1 is 1

[lone quail]

Im not sure i get what you mean

[dusky epoch]

0 and 1 shouldn't be taken as representing the real numbers 0 and 1

not in this context

[lone quail]

Ok so what is it representing exactly?

[dusky epoch]

they don't represent anything

[lone quail]

I mean the set

What set is it

[dusky epoch]

it's just a set with two elements

one of which we write with the symbol 0, and the other with the symbol 1

[lone quail]

So the symbol 1 includes its opposite?

[dusky epoch]

the element denoted with 1 IS its own opposite.

because 1 + 1 = 0.

[lone quail]

[dusky epoch]

yes.

the set {0,1} forms a field if it is equipped with the following addition and multiplication operations:
0+0 = 1+1 = 0; 0+1 = 1+0 = 1
0*0 = 1*0 = 0*1 = 0; 1*1 = 1

[lone quail]

Ok cool i got it

Thanks

[charred stirrup]

why are the scalars only between 0 and 1?

[gray dust]

P is the parallelogram spanned by vectors a & b. P has side lengths of ||a|| & ||b||. to fill in all the points in P, you make a set of linear combos of a & b, ie sa+tb where s & t are real scalars. to make sure you don't include points outside P, you limit s & t to between 0 to 1

[charred stirrup]

ooo

and the vectors a and b can be of any length, but the scalar [0,1] ensures it stays within the span(a,b)

[gray dust]

span{a,b} is an infinitely big plane, equivalent to letting s & t take on ALL real values

limiting s & t to between 0 to 1 restricts the linear combos to points inside P

[charred stirrup]

why dont we bother interchanging the 1st column with the 3rd?

[vast torrent]

You don't swap colunns in gaussian elimination

[charred stirrup]

thanks big dad

gonna ace this exam

[vast torrent]

[charred stirrup]

i know D is the diagonal of A's eigenvalues, but what property relates them in this way? A^n makes D^n

[dusky epoch]

consider explicitly multiplying out AA, AAA, AAAA, etc

[charred stirrup]

I guess I can see the matrix getting bigger if our entries are positive

that affects the eigen values by that factor?

[hallow bobcat]

This is simpler than you think
As Ann said, for example:

$A^3 = AAA = (PDP^{-1})(PDP^{-1})(PDP^{-1}) = \ldots$

[stoic python]

Fέliχ:

[dusky epoch]

^

[charred stirrup]

$A^3 = AAA = (PDP^{-1})^3 = \ldots$

[stoic python]

is this the krusty krab: