linear-algebra

[stoic python]

Display results online and refine query

Requested by davidL#1578

[lone quail]

if A and B are both similar and diagonalizable, so there exists matrixes P and Q so that:
(P^-1)AP=D and (Q^-1)BQ=D.
by transitivity, the textbook says, that the basis by which f is represented by B is formed by the columns of P(Q^-1), why is that?
For reference (in the image below) A is actually the representative matrix of f while B is the matrix at the bottom of the image
https://cdn.discordapp.com/attachments/540211747613704221/650392600208998430/image0.jpg

Can anyone help pls? sorry for repost

[timber minnow]

@lone quail what confuses you

[lone quail]

why is B is formed by the columns of P(Q^-1)?

[lone quail]

@timber minnow

[lone quail]

anyone has any idea?

(if the question is confusing please do tell)

[timber minnow]

@lone quail can you be more specific

I understand the question but I don't understand where you're confused if that makes any sense haha

[lone quail]

i dont understand why B is formed by the columns of P(Q^-1)

where does that come from

@timber minnow

[timber minnow]

so if B is similar to A

then B = P-1 A P

(P^-1)AP = D = (Q^-1)BQ

[lone quail]

yes

now what?

hmm why?

is it? i never studied such property

[timber minnow]

whoops sry

got different similarity mixed up, fixed

[lone quail]

so any ideas?

[timber minnow]

yes

(P^-1)AP = (Q^-1)BQ

-->(Q^-1)BQ = (Q^-1)((P^-1)AP)Q by similarity

[lone quail]

im not sure you can move it that way

[timber minnow]

I didn't do any operations

I just substituted by definition of similarity

because we established that because B is similar to A, then B=P(-1)AP

[lone quail]

but you substituted A

[timber minnow]

I replaced B with P^(-1)AP

because that's how it's defined

[lone quail]

you replaced A with P^(-1)AP

if you try to invert the first equation

swap the sides

thats what you did

ok cool but its a different matrix P right?

so shouldt we call it S or something

because the matrix such that (P^-1)AP=D is different than the matrix P in (P^-1)AP=B

right?

@timber minnow

[timber minnow]

hm you might be right, I'm still here, was rereading the textbook statement

Ok

so we don't need a new matrix S

instead of thinking of it like A is similar to B

let's think of it like both A and B are similar to D

since both satisfy similarity: (P^-1)AP=D and (Q^-1)BQ=D

Q * (Q^-1)BQ = Q * (P^-1)AP

(BQ)^-1 = (Q * (P^-1)AP)^-1

(Q^-1)(B^-1) = (Q * (P^-1)AP)^-1

Q * (Q^-1)(B^-1) = Q * (Q * (P^-1)AP)^-1

(B^-1)^-1 = (Q * (Q * (P^-1)AP)^-1)^-1

B = Q * (P^-1)AP Q^-1

[lone quail]

@timber minnow what did you do from the first to the second step

[timber minnow]

First step I multiplied through by Q

second step I took inverse of both sides

[lone quail]

Yes but dont you have a Q on the left side

Where did it go

[timber minnow]

that's why I multiplied by Q

[lone quail]

Ah ok i get it

[timber minnow]

(Q * (Q^-1))BQ = Q * (P^-1)AP

[lone quail]

So we got a different answer?

[timber minnow]

I'm not done haha

thinking of how to tie this up

last step

[lone quail]

Oh ok thanks

[timber minnow]

davidL

B = Q * (P^-1)AP Q^-1

Dec 1, 2019, 10:53 PM

last step is just to show the term Q * (P^-1)A cancels to identity

then we will have B = P Q^-1

[lone quail]

We could rearrange it this way

I think

And then maybe view PQ^-1 as a change of base matrix?

[timber minnow]

yeah definitely, I just thought that was your question haha

if you're ok with it being change of basis then we are done

[lone quail]

I was just trying to understand the algebric steps to get here, thanks a lot

You’ve bee a great help

[timber minnow]

for sure, sorry it took me a minute haha

haven't slept today lol

[vast torrent]

Why is

$x^\dagger H x$

[stoic python]

gfauxpas:

[vast torrent]

Real if H is hermitian?

Is that true?

[feral mountain]

take dagger of the whole thing

[vast torrent]

Get the whole thing ba- ooohhh

Nice

[cloud glen]

I don't understand what E_lambda(A) is

[wintry steppe]

all its saying is that the algebraic multiplicity is equal to geometric multiplicity

[cloud glen]

oh ok ty

[dusky epoch]

E_λ(A) = ker(A - λI), most likely.

[cloud glen]

im comfused what it means by eigenspace dimension equal to multiplicity

i don't know what multipllicity of lambda means

you're saying that eigenspace = nullspace/kernel of eigenvalues?

[dusky epoch]

multiplicity of λ means multiplicity of λ as a root of the charpoly of A

[cloud glen]

ohh so like (x-2)^2 = 0, the eigenvalue 2 is multiplicity of lambda i think

[tame root]

how do I find eigenvectors from an eigenvalue

for example, A = [.5 .4 ; -.104 1.1]

eigenvalues are 1.02 and .58

[A - 1.02I | 0] = [-.52 .4 0; -.104 .08 0]

[cloud glen]

u plug into identity matrix subtract by A

then solve system as homogenous

[tame root]

but that gives [.52 .4 0 ; 0 0 0]

so wouldn't that make the eigenvector [.52 ; .4]

the book has [10 ; 13]

which is obviously not a multiple

[cloud glen]

eigenvector can be multiple

yeah

they multiply by 25

they take eigenvalue of 25

25[.52, .4] = [10, 13]

@tame root

[tame root]

but thats backwards isnt it

why doesn't that matter?

[cloud glen]

u got result eigenvector of .52 and .4

but eigenvector result comes in form

t[.52, .4]

plug t = 25

and u get same answer as book?

[tame root]

and you get [13 10]

not [10 13]

[cloud glen]

oh sorry made mistake

not sure

if u can give full question i can try

[cloud glen]

they dont give u matrix A?

[tame root]

that is A

the coefficient matrix for that formula

[cloud glen]

sorry im confused why they chose p = .104

[tame root]

for arbitrary reasons

because they said so

[vast torrent]

do you have a better suggestion Misha?

because if you don't, maybe you shouldn't criticize

0.104 is a good boy

he tries hard

[cloud glen]

im trying to help

not be rude

sorry

[tame root]

I figured it out

it does reduce to [.52 .4 0 ; 0 0 0]

which is $.52 x_1 = -.4x_2$

[stoic python]

Migillope:

[cloud glen]

nice job

[stoic python]

Migillope:

[tame root]

thanks

[vast torrent]

@cloud glen i was just kidding

[cloud glen]

ohh okay did not understand, no problem

[cloud glen]

how do they know that theta is pi/4

(3,3) is coordinate but i dont understand how they how it results in pi/4

1 because 3/3?

oh wow thank you very much

wait sorry follow up question

what if coordinate (2,0)

how come answer would be pi/2

and not any period of pi/2

since tan undefined at pi/2, pi, 3pi/2, and 2pi

for example z = 2i

to find theta i need to find (0,2)

but 2/0 = undefined?

or am i mistaken

arctan(2/0) nonexisitant

using ur approach from earlier

https://www.varsitytutors.com/assets/vt-hotmath-legacy/hotmath_help/topics/tangent-of-an-angle/tan4.gif

oh

makes more sense, i though it was like unit circle (0,2) => (cos, sin)

then i need to find tan => sin/cos

[eternal jasper]

hey could anyone help me start this?

don't really know where to begin

would be really appreciated

[empty copper]

Translate (4, 3) to origin, rotate, and then translate back

[eternal jasper]

@empty copper because I'm solving for any real numbers what would the calculations look like?

[empty copper]

Ok, let's start off simple

If we have a vector (x, y), and we translate our coordinate system to a new system such that (4, 3) in the old system is the origin in the new system, then what are the new coordinates of (x, y)?

I'll give you a hint: it's (x - 4, y - 3)

[eternal jasper]

@empty copper I tried something and my numbers were similar to that, do u think this is right?

[empty copper]

You did it correctly, except for the very last vector you wrote down: the first entry should be 7 instead of 3 (as you wrote down correctly in the line before)

[eternal jasper]

oh whoops... was looking at the example of a similar solution ....

but thank u

[empty copper]

🦈 🙏

[eternal jasper]

this one looks way harder

so for A is it safe to assume there's a translation of 4, 4?

as the origin is moved to 4,4

not too sure how to work backwords

[eternal jasper]

hey im really having trouble figuring this out @empty copper do you have any ideas on where to start?

[slow scroll]

Since T is affine and T(0) = (4,4), T(x) = A(x) + (4,4) where A is some linear transformation. You're given some outputs of T, and you can use that to find what A does to a basis for R2

[eternal jasper]

so maybe like

would i take 8,4 and subtract the translation ?

getting 4, 0

then divide by 2 ?

idk

[slow scroll]

yea ur on the right track

T((2,2)) = A((2,2)) + (4,4) = (8,4) so
A((2,2)) = (4, 0) and since A is a linear transformation,
A((1,1)) = (2,0)

[eternal jasper]

using a program i got this

[slow scroll]

a program?

[eternal jasper]

didnt write it btw

[empty copper]

nice

[eternal jasper]

im not sure how to calculate the answer tho

like i can get a few steps

[slow scroll]

Are you able to get that
A((1,1)) = (2,0)
A((1,3)) = (0, -4)?

where A is the linear part of T?

Once you do that, All you need to know is what A(1,0) and A(0,1) is to write a matrix for A

[eternal jasper]

yeah that makes sense but how u go from that to the final product is where im lost

oh

how would u find 1,0

[slow scroll]

well, the standard way is to use change of basis. However, in this case it is not hard to see that A((1,3)) - A((1,1)) = A((1,3) - (1,1)) = A((0, 2)) = (0, -4) - (2, 0) = (-2, -4). Therefore, A((0, 1)) = (-1, -2). Ill let you do the other one.

Also, the answer that the code gave you seems to be the transpose of the correct matrix. weird.

[eternal jasper]

okay thanks ill try the other one now

@slow scroll hm i ended up getting (2, 1.333)

for (1,0)

[slow scroll]

ok so how did you get that?

[eternal jasper]

i multiplied (2,2) = (4,0) by 3 to get (6,6) = (12,0) then subtracted 2 * (1,3), which got (6, 0) = (12, 0) - (0, -8), which gets (12, 8) and divided by 6 is (2, 1.333)

omg

i missed the (2, 6) subtracted for the x

it would be (4, 0) = (12, 8) which is (1, 0) = (3, 2)

thats also what the program got huh

anyway thank u for all the help that was extremely good

[slow scroll]

@eternal jasper note that this is NOT what the program got. What we got was $A = \begin{pmatrix} 3 & -1 \ 2 & -2 \end{pmatrix}$

[eternal jasper]

oh

hm

[stoic python]

kxrider:

[slow scroll]

By the definition of matrix multiplication, A(1,0) = first column of A and A(0,1) = second column of A. Thats how you construct the matrix

[eternal jasper]

so would the full answer be the thing i wrote down or the matrix u posted?

[slow scroll]

The full answer is whatever T(x,y) = A(x,y) + (4,4) is

[hexed crow]

Just a general question. Can vectors be greater than three-dimensional? I normally see $\begin{bmatrix}1 \ 2 \ 3 \end{bmatrix}$. Can vectors $\begin{bmatrix}1 \ 2 \ 3 \4\.\.\.\n\end{bmatrix}$

[stoic python]

tryingtolearn:

[slow scroll]

sure

[hexed crow]

thanks @slow scroll

[dreamy acorn]

errr this might be the right place to ask....
is there a way i can represent a certain variable can not exceed a value of 115 inside the formula it is in? (i realize i could do something like Z<=115)
I want to show that Z in this equation can not exceed 115.
(Y + (X * 0.2) + W + (35 + Z)) * 1.05

is there some special super/sub script i can use?

[dusky epoch]

this isn't an equation.

[wintry steppe]

i have a vector and 2 bases, how am i supposed to find the change of basis matrix A -> B?

[half ice]

Is any of these two basis very simple?

@wintry steppe

[wintry steppe]

yes

[half ice]

For something like
1,0,0

You can easily put this into a pretend matrix and find what a column should be

[wintry steppe]

A is the basis of R2, B is {(1, 1), (1, -1)}

both vectors ofc

[half ice]

Sick. Your change of basis is
1 1
1 -1

Try putting in one of the basis vectors of R² to see why it works

[wintry steppe]

you mean multiplying by that matrix?

[half ice]

Ya ya. I mean try carrying out
[1 1] [1]
[1 -1] [0]

And the other one with 0,1

[wintry steppe]

it works but i dont quite get why lol

[half ice]

(1,0), when multiplied by a matrix, just gives the first column of that matrix

[wintry steppe]

sure

[half ice]

So, the first column should be (1,1) since that's in the other basis

[gleaming topaz]

Sorry I suck at formatting, but you know that the basis of R2 = {(1,0),(0,1)}, right? So you want x1 * (1,0) + x2 *(0,1) = (1, 1), (1, -1) right

Where everything I've written are column vectors

[wintry steppe]

$ v =
\begin{bmatrix}
2 \
3
\end{bmatrix},
\boldsymbol{A} =
\begin{bmatrix}
1 & 0 \
0 & 1
\end{bmatrix} ,\boldsymbol{B} =
\begin{bmatrix}
1 & 1 \
1 & -1
\end{bmatrix} $

well technically 2 vectors in each matrix

[gleaming topaz]

So you get an augmented matrix of A | B = x

[stoic python]

yami:

[wintry steppe]

yeah i got the coordinate vectors wrt A and B

[gleaming topaz]

Are you asked to find v's coordinates in A and B?

[wintry steppe]

yeah thats the first question, i have that already

v_A is trivial, v_B looks like this

,tex $ [v]_B =
\begin{bmatrix}
\tfrac52 \ \
\tfrac{-1}2
\end{bmatrix} $

[stoic python]

yami:

[wintry steppe]

ugh

[gleaming topaz]

Wait sorry, what is the question even?

[wintry steppe]

i have to determine the change of basis matrix A -> B, i dont quite understand why it has to be the B matrix

i get that B's (1, 0) vector would look differently from A's (1, 0) vector

but isnt there a more.. mathematical way to determine what the A->B matrix is? lol

[gray dust]

there's very little work to be done here since the columns of A are the standard basis vectors of R^2

[half ice]

Yes, you can solve the system of equations.

[wintry steppe]

now that sounds mathematical

[half ice]

[a b] [1] = [1]
[c d] [0] [1]

And the same set up with the other

But.. uh...

[wintry steppe]

so my thing would look like

[half ice]

Is gon be pretty simple to solve this

[wintry steppe]

the abcd matrix is B yes?

okay so for the sake of an example lets do this then

[gleaming topaz]

Well you have Ax = B right, where A multiplied by something gives you your B matrix and that something x is the change of basis matrix from A->B

Which gives you A | B = x

[half ice]

No, the abcd is the change of basis matrix

[wintry steppe]

fml lol

[half ice]

Basically, we literally calculate a matrix that takes one basis to the other

[wintry steppe]

$ v = \begin{bmatrix}
1 \ 0 \ -1
\end{bmatrix},
\boldsymbol{A} = \left {
\begin{bmatrix}
1 & 0 & 0 \
0 & 1 & 0 \
0 & 0 & 1
\end{bmatrix} \left }, \boldsymbol{B} = \left {
\begin{bmatrix}
1 & 0 & 0 \
1 & 1 & 0 \
1 & 1 & 1
\end{bmatrix} \right } $

[stoic python]

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

[wintry steppe]

how would we do this

[gleaming topaz]

Are you asked to calculate v_b or what?

[wintry steppe]

same task

find v_A, v_B wrt bases A and B, find the change of basis matrix A->B, and with that change of basis matrix determine v_B from v_A

[gleaming topaz]

Do you need help with all of them or just the change of basis matrix one?

[wintry steppe]

seems reasonable to assume that v_A * ch.of.b matrix would result in v_B

since its just a transformation in the end no?

the way i found v_B was augmenting the basis B with the given vector and then bring it to RREF, the vector column is what my v_B is

[gleaming topaz]

Yeah that's right

B | v = x and x = v_b

Do you understand why you did that? If so, it doesn't work any different for the change of basis matrix

[wintry steppe]

uhhhh

no. lol

i think im lacking the deepest fundamentals of what it even means to bring something to RREF

[gleaming topaz]

Well, you have Bx = v, so you need to multiply B by some vector/matrix to get v right?

[wintry steppe]

sure

[gleaming topaz]

So x in this case would be v_b since B * v_b = v

[wintry steppe]

wait, isnt v a vector wrt basis A?

[gleaming topaz]

Well since A is the three standard basis vectors in R3 all vectors are the same wrt basis A

[wintry steppe]

yeah

[gleaming topaz]

Since 1 * (1,0,0) + 0*(0,1,0) + -1(0,0,1) = (1,0,-1)

So if you have Bx = v, how would you solve for b then

where x = (x1,x2,x3) or also v_b

[wintry steppe]

i want to say a system of equations but only because this is linear algebra and it seems like this is what its all about

[gleaming topaz]

So basically you have x1 * (1,1,1) + x2 * (0,1,1) + x3 * (0,0,1) = (1,0,-1) right

So you want to know is what you need to multiply the first, second and third vector in B by to arrive at (1,0,-1)

[wintry steppe]

oh, yes

or how to scale them i guess

[gleaming topaz]

So this brings us back to Bx = v where x is (x1,x2,x3) as a column vector and when you put B|v = x and get B to RREF then it gives you x right

[wintry steppe]

what u just said may have been trivial to u but i think this actually gave me new insights lol, or at least i feel like i understand it somewhat better

yes

[gleaming topaz]

And we agreed on that (x1,x2,x3) are how we need to scale our three vectors in B to get v right

so that becomes v_b

v's coordinates expressed wrt basis B

[wintry steppe]

yes

[gleaming topaz]

Do you understand this, if so change of basis matrixes isn't any more difficult

[wintry steppe]

i understood it now (i think)

[gleaming topaz]

Because we're asking us if we have A, what do we need to multiply it by to get B?

so Ax = B which gives you A | B = x and when as A is in RREF you get your change of basis matrix A -> B

[wintry steppe]

oh then whatever is on the | B side is my change of basis matrix A->B then?

so its like finding the inverse of a matrix except you augment with another matrix instead of the identity matrix?

[gleaming topaz]

Yes when you row reduce A into RREF the other side in your augmented matrix will be your A->B matrix

[wintry steppe]

and now it makes sense why when A = I the change of basis matrix A->B is just B

lol

[gleaming topaz]

Yeah, I hope that helped

[wintry steppe]

that was very good

and in order to find v_A from v_B i just need to get the inverse of B because its just an inverse transformation, yes?

[gleaming topaz]

Not sure how you meant but to find v_a from v_b using change of basis matrices you need to multipy v_B with the change of basis matrix B->A

Or you multiply v_b by B to get v in terms of the standard basis vectors which in this case solves v_A

[wintry steppe]

uhh yeah i guess?

my intention was to multiply v_B with B^(-1) since thats the change of basis matrix A->B

that would give me v_A, hopefully lol

[gleaming topaz]

Yeah in this case that works because A is the matrix consisting of the standard basis vectors in R3 so the change of basis matrix A->B just equals B

Otherwise you multiply v_B with (A->B)^(-1) to get v_A

[wintry steppe]

yuss

[meager tinsel]

hey umm so
there is this article
https://en.wikipedia.org/wiki/Matrix_chain_multiplication
which explains how to make the most efficient matrix multiplication in terms of associative order

there are 2 algorithms in there
that is
a naive one (with dynamic programming an stuff)
and a more advanced one that ties this to polygon triangulation

now

what I was thinking of is that surely

for tensor contraction this sort of order optimization makes sense as well

for matrixes the dynamic solutions is basically
there is a linear graph (range)
and we solve it for all subranges (subgraphs)

for tensors it won't be linear
wont even be a tree necessarily
will be just some graph
however, solving it for all subgraphs still works

What I wonder is

is there some analogy for tensors with efficient algo
maybe symplexification (is that a word) of some n-dimensional ...bodies (they would be very non convex)

I am very not sure but would be glad to hear someone's thoughts on this

[empty copper]

Once I followed a seminar, which offered two different research paths; one was to investigate algorithms for calculating high dimensional tensor products, and the other was about high dimensional quasi-monte carlo integration

Unfortunately for you, I followed the latter one

[meager tinsel]

(((

well, maybe it is time to catch up

[empty copper]

QMC is pretty interesting

Low discrepancy sequences, nets n stuff

[meager tinsel]

yeah I mean catching up on the former not abandoning your thing)

tbh I implemented a naive solution as an upgrade to a library

and the the guy (author) asked me "what about tensordot"
and I was like well sure ehhh....

lemme dig this
and Now I found this and ofc nlogn is better then n^3

also, I like how it is smarter

[jagged saffron]

does hermitian imply diagonalizable with real eigenvalues?

i think it does

[feral mountain]

@jagged saffron yeah look up spectral theorem

[jagged saffron]

yea i was making sure

I don't really understand where to use schur decomposition here

[cloud glen]

what is that symbol

beside (A)

what is it called

[gray dust]

sigma $\sigma$

[stoic python]

RokettoJanpu:

[cloud glen]

oh i thought sigma was the big E for summation

strange

thank u

[gray dust]

lowercase sigma $\sigma$, capital $\Sigma$

[stoic python]

RokettoJanpu:

[cloud glen]

oh, lol

tyty

[gray dust]

np

[cloud glen]

i dont understand the ~

how come they just dont use the equal sign (=)

[half ice]

You could

Some texts also use R

[cloud glen]

oh okay, so i guess its just the flavour of notation

they chose

[half ice]

Note that ~ it's not trying to say that both elements are the same, but is trying to say both elements are related

[cloud glen]

ohh okay ill take that into account

[half ice]

I can see sticking away from = for the sake of clarity

[cloud glen]

wasnt aware

i thought equivalence meant equal

thats why i was confused

[sonic osprey]

Yeah I mean, it'd be weird to say that $\begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix} = \begin{pmatrix} 2 & 0 \ 0 & 2 \end{pmatrix}$

[half ice]

~ is something that acts LIKE = but doesn't have to be =

[stoic python]

Zopherus:

[sonic osprey]

uh

choose an example where the matrices are actually similar

[half ice]

Rofl

[cloud glen]

so its saying they're similar because of eigenvalues i guess

since the topic is eigenvalues

[half ice]

You can set up an equivalence relation where similar matricies are related

[sonic osprey]

It doesn't really say anything about why matrices are similar

Just what properties does similarity have

that every matrix is similar to itself

is the first property

[half ice]

Note that this gives you the power to treat any two similar matrices as "the same matrix" but still be able to say that they aren't really the same matrix

I'm totally being clear here

[cloud glen]

huh interesting, okay

so is this saying that A ~ B because of those properties

[half ice]

There we go. That's a great example. A is related to B, which is to say that A and B aren't necessarily the same, but are related by some rule

[cloud glen]

wicked

[sonic osprey]

Also no

That's not what they're saying

[half ice]

In this case, it's matrix similarity

[sonic osprey]

They're saying that IF A ~ B, THEN these properties

Not the other way around

You can have matrices with the same of all 5 of those things, but aren't similar

$\begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix}, \begin{pmatrix} 1 & 1 \ 0 & 1 \end{pmatrix}$

[stoic python]

Zopherus:

[cloud glen]

so whats the base case to prove similarity if these properties don't define it

[sonic osprey]

have the same determinant, trace, rank, characteristic polynomial and eigenvalues

[half ice]

Did you really find an example where all 5 are the same that fast

[sonic osprey]

But they're not similar

Eigenvalues the same imply the rest so

you only need to fufill that single condition

Proving similarity is hard

Well, kind of

Your class will get to how we normally show two matrices are similar, i.e., diagonalization

[half ice]

Two matrices A and B are similar if
A = PBP'
For some invertible P

Or is it P'BP

[cloud glen]

hmm okay

[sonic osprey]

they're the same, since you can just take P = P^{-1} if necessary

[half ice]

Oh yeah good point

[half ice]

By the axioms of an equivalence relation,
If A~B and B~C, then A~C

Similarity makes "cliques" so to speak

We actually say that it "partitions" the set of all matrices

I'll leave this alone, I don't know how to explain it that great and you likely don't need it atm

[cloud glen]

yea haha appreciate the help

[half ice]

Np. Anything else you wanted? This is one of the best parts of lin alg

[cloud glen]

I think I'm okay for now. Reading about diagnonlization so if any questions pop up ill be sure to ask

[heavy glacier]

Eh so why does an element of SL(2, Z) Preserve Z^2 (I worked out the matrix multiplication but still not really sure why)? And don't we need further restriction than just being in SL(2, Z) to be a rotation map of R^2?

[half ice]

Let's say a matrix has determinant 2. Then, it's inverse has determinant 1/2. But, this implies the inverse does not have integer entries

If the determinant is not 1, then the inverse is not made of integers

[naive flicker]

yo

https://cdn.discordapp.com/attachments/571736375054041118/651247362844524564/unknown.png

[half ice]

Hmm. I'm making a logical leap here

[naive flicker]

can someone help

[wintry steppe]

@naive flicker I don't think this is #linear-algebra

[heavy glacier]

ask this in #prealg-and-algebra

ok I got it now Kaynex, ty

[half ice]

Did you? I still don't

[heavy glacier]

I was trying to find that A in SL(2, Z) maps lattice elements to themsleves (which I don't think is true)

it just transforms the whole lattice to itself

if T is the matrix multiplication

T takes M_2(Z) to itself

since what you get is (2x1 matrices of) linear combos of integers in the image which are just integers

and I cant say it preserves determinants (since the lattice elements dont have determinants defined) but were in SL group so its not sending elements way out in terms of metric either

[uneven bloom]

Conversely, if det(A)=1 and A has integer entries, then A^(-1) also has integer entries (by adjugates, for example).

[north sierra]

so when we're finding the characteristic equation and eigen values of a matrix A

we are finding the det(A-lamba*I)

so the determinant = the eigen values?

[gray dust]

A's charpoly=det(A-lambda*I). A's eigvals are values of lambda satisfying charpoly=0

[north sierra]

okay

thank you

rokettoJanpu

[gray dust]

np

[cloud glen]

$-\lambda^{3}-2*\lambda^{2}+5*\lambda+6$ = 0

how do i find the eigenvalues of this

[dusky epoch]

this is not a matrix so it does not have eigenvalues

[cloud glen]

this is the resulting determinant

from my calculations

[stoic python]

Миша:

[dusky epoch]

did you mean to say "how do i solve this equation"

[cloud glen]

Yes

[dusky epoch]

ok what have you tried

[cloud glen]

factorization, but that leads to no avail because of the constant +6

and i don't know quadaratic formula for degree of 3

[dusky epoch]

"quadratic formula for degree of 3" makes no sense

"quadratic" means degree 2

[cloud glen]

cubic formula

[dusky epoch]

there is a cubic formula but it's long and unwieldy and definitely not something you're ever expected to use

but since all your coefficients here are integers...

the rational root theorem might be of use.

[cloud glen]

oh okay ill look into that, thank you

[wintry steppe]

nothin' wrong with a little bit of quartic formula

[cloud glen]

just learned rational roots test

seems like a lot of work, especically for an exam

i hope i get a simpler polynomial

[half ice]

It's amazing that you can say such a direct thing about a polynomial's roots without any work

x² - 2 has no rational roots via rational root theorem. Therefore √2 is not rational

[cloud glen]

sorry not following

[half ice]

What's the possible rational roots for x² - 2?

[cloud glen]

sqrt(2)

oh rational

uh not sure, is it some imaginary number?

[half ice]

Rational root theorem says that the "possible rational roots" are of the form
±2, ±1

If the polynomial has any rational roots, they're in that list

[dusky epoch]

people don't understand if-then statements, kx

[half ice]

😢

Whoa emojis are big now

[cloud glen]

i think im missing something

whenever i try to find a polynomial for this problem

it is a challenging one with a large degree, making finding the roots a challenge

is there a special technique for this type of problem?

I tried the approach of simplifying it with a similar matrix

but it still doesn't help much

[winter moat]

@cloud glen it is given that one eigenvalue is 3. It should help

-3*

[cloud glen]

oh, i was wondering why thats there

thank u

[winter moat]

It tells you that $(\lambda +3)$ is a factor

[stoic python]

oscillatingEquilibrium:

[cloud glen]

can i deduce the rest of the factors from that fact alone?

[winter moat]

Yes, if you can factorise a quadratic polynomial

[native lodge]

by inspection of the matrix alone, no

but once you have the characteristic polynomial, then yes

[cloud glen]

oh i see what ur getting at, thank you

ill try it now

[winter moat]

Good luck

[cloud glen]

tyvm

[native lodge]

when it says if possible though

makes you think if they are going to try and do something tricky

[winter moat]

I wonder if second eigenvalue is -3 as well

[native lodge]

could very well be the case

[winter moat]

If that's true, then third eigenvalue can be found from trace of the matrix

And can be verified

But it's better to do it the long way

[cloud glen]

i decided to try this one since its easier

but i got the polynormial x^3+x^2-x-1 = 0

now im stuck

how do i find the rest of the eigenvalues?

if i use the rational root theorem

i get {-1,1}

but then how can i be sure there isnt a multiplicity of one of these eigenvalues?

[winter moat]

You didn't get third?

[cloud glen]

from the polynomial $ x^{3}+x^{2}-x-1 = 0$

[stoic python]

Миша:

[cloud glen]

i could only deduce -1 and 1

[winter moat]

It is $(x+1)^2(x-1)$

[stoic python]

oscillatingEquilibrium:

[cloud glen]

oh,

how were u able to factor it

[winter moat]

You know x-1 is a factor

So try to take it common out of groups of two terms

That's how you normally factorise it

Or it is easier to take (x+1) as a factor

$x^2(x+1)-(x+1)=(x+1)(x^2-1)$

[stoic python]

oscillatingEquilibrium:

[cloud glen]

oh i thought u could only do that for quadratic

[winter moat]

You can do it for any polynomial

Or any functions

Distributive law holds

[cloud glen]

ah okay sweet

ty again

[winter moat]

You're welcome!

[jagged saffron]

https://cdn.discordapp.com/attachments/442470054978650122/651408933013159936/unknown.png

Can anyone guide me on the right track here?

I tried using spectral theorem on B,C so they are diagonalizable with real eigenvalues, but the issue is that they aren't diagonalizable on the same basis unless B,C commute and there are no repeated eigenvalues
So I can't use it to find eigenvalues of A

I don't see how schur's decomposition theorem will help here unless I can split the decomposition into real and imaginary parts?

Also tried using gershgorin circle theorem but I don't think that works

[mint sentinel]

anyone know why we use the formula v divided by the norm of v squared where v is an eigenvector to find a matrix P orthogonal to a matrix A?

[vast torrent]

We want P to have determinant +1 or -1

[mint sentinel]

ahh

i see

[vast torrent]

Otherwise it doesn't have Q-¹=Q^t

[mint sentinel]

because if the determinant is 0, it is not orthogonal

uh could you expand on that

[vast torrent]

Take determinant of both sides

Assuming Q-¹=Q^t

1/det Q = det Q

[mint sentinel]

i see

[vast torrent]

Only works for +-1 :)

[mint sentinel]

so it's almost like a notation trick

or something

or a trick

rather

[vast torrent]

It's a definition that we impose as part of wanting Q-¹=Q^t

[mint sentinel]

if you know what i mean lo

[vast torrent]

We want that as part of the definition

[mint sentinel]

yeah, that sounds better

[vast torrent]

Consequence of the definition is that columns and rows are unit length

Make sense?

[mint sentinel]

yes

thanks

!

[vast torrent]

[mint sentinel]

i see someone calculate the eigenvalues and vectors by det(A-alpha I) and others by det(alpha I-A), are both right to use?

[dusky epoch]

these differ by a factor of (-1)^n.

[vast torrent]

the advantage of the first is that the determinant is the char polynomial at alpha=0. Advantage of the second is the char polynomial has leading coefficient 1 on alpha^n

Choose your warrior

[mint sentinel]

: o

[vast torrent]

Same roots of the polynomial is the point

[mint sentinel]

because what happens when i calculate with each formula is is that the basis matrices have their vectors mirrored

[vast torrent]

Eigenvectors don't have a God-given order to them

You pick an order

[mint sentinel]

ait

[vast torrent]

As long as you match the right eigenvalue to the right eigenvector, i mean

[mint sentinel]

yeah

[quartz compass]

there are times when you'd wanna order your eigenvectors in order from smallest eigenvalue to largest

[mint sentinel]

thx

when

[quartz compass]

you can worry about it when it comes up, it's not too serious

nothing for what you're doing I think, stuff to do with approximations

[mint sentinel]

is it relevant to finding an orthogonal basis

oh alright

[vast torrent]

When doing orthogonal bases, i order them so the det is 1 rather than -1

Just my personal preference

[mint sentinel]

👍

[vast torrent]

When is a vector space isomorphic to its dual? Is that only for finite dimensional spaces?

Wait no it can't only be finite cases because L2 spaces are

[jagged saffron]

you can explicitly construct a isomorphism in the finite dimensional case

However I know this does not hold for all infinite dimensional spaces

I think dual space is not isomorphic ever in the infinite dimensional case

[vast torrent]

isn't

$L_2^* = L_2$

?

square integrable functuons R^n to R?

[stoic python]

gfauxpas:

[jagged saffron]

I don't know any analysis

https://en.wikipedia.org/wiki/Dual_space#The_infinite-dimensional_case

but its true

dual space of infinite dimensional V.S is never isomorphic to the vs

[vast torrent]

oh I was thinking of the contunuous dual space

[jagged saffron]

oh no idea then

[opal plaza]

Hello again :)
Trying to conceptualize a problem: Describe the eigenspace of a transformation in R2 which reflects points over the Y axis.

I imagined that the transformation would be two vectors (-1, 0) (0,1) which means the eigenvectors would be (1,0) (0,1). And this means the space would be the individual span of these two eigenvectors? Which are kinda like the pivot the points rotate around as they're tansformed over the y axis.

feel like I'm on the cusp of understanding this

[half ice]

So an eigenspace can be best thought of as a set of vectors that "don't change direction" after a transformation

Let T be the act of flipping R² over the y-axis. This is a linear transformation

Obviously, anything on the x-axis isn't affected by T. So, this is an eigenspace. Plus, it has eigenvalue 1 since you don't have any scaling due to T

With me so far on this?

[opal plaza]

so the x axis is an eigenspace

[half ice]

Yus. We can describe it instead with a basis, one such basis is (1,0)

That's likely what you mean

[opal plaza]

Okay perfect

[half ice]

The other one is the y-axis. It has eigenvalue -1, since vectors get flipped

You can describe it instead with the basis (0,1)

[opal plaza]

and that's a vector going up the y-axis?

[half ice]

Any vector that lies entirely on the y-axis yeah

[opal plaza]

okay, perfect. And those two together describe the eigenspace because they don't change in direction based on the transformation.

[half ice]

There are two eigenspaces, each one is their own

[opal plaza]

I see

[half ice]

And those are the only two, yeah

I think you get it! You likely don't need to understand it in the depth I'm trying to push onto you lol

For many classes it's important that you can compute those basis vectors

[opal plaza]

So if it was in R3 the space would be a vector through the plane just with three coordinates, and the transformation would kind of rotate around that span?

no thank you for the help this is really interesting. I know how to compute them I'd just like to conceptualize it a bit better

[half ice]

Like, if we are flipping over the z-axis?

Then anything on the plane is in an eigenspace. This case is different, since the eigenspace is two dimensional and needs two basis vectors to describe it

As well, anything on the z-axis is another eigenspace. This one is one-dimensional

[opal plaza]

mm I suppose so. I'm trying to visualize some random matrix like (4, -2, -2) (0, 1, 0) (1, 0, 1). Like I solved the eigenbasis but I'm trying to picture what's going on beyond that

[half ice]

It's difficult to picture for completely random T lol

[opal plaza]

haha true

[half ice]

Here's an interesting example. Let's say you have R² and T is a 90 degree rotation. What's the eigenspace?

[opal plaza]

should be the same, no? (0,1) and (1,0)

[half ice]

I mean a 90 degree rotation about the origin, sorry. So (1,0) goes to (0,1)

[north sierra]

i dont know this stuff geometrically so i have no idea how to answer this

[gray dust]

did you get the visual intuition of eigenvals and eigenvecs?

[north sierra]

yeah so it's just a line right?

and the eigen value scales the vector?

but like not really

i dont get it confidently

[gray dust]

there may exist a set of vectors in a vector space which only get scaled by some constant via a linear transform on that space

[north sierra]

ah i see

[gray dust]

this is all compactly said by $A\xi=\lambda\xi$ where $\lambda$ is an eigenval and $\xi$ is one of $A$'s eigenvecs associated with $\lambda$

[stoic python]

RokettoJanpu:

[north sierra]

i see

[gray dust]

these exercises are to see if you can identify eigenvecs and eigenvals from how the matrix transforms these vector spaces

[north sierra]

yeah

brb

hmm

so

like I wanna say an eigen value would be 0?

@gray dust

[gray dust]

for what

[north sierra]

for 31

[gray dust]

why

[north sierra]

so im thinking cause it goes through the origin lol

[gray dust]

an eigen value would be 0?
so im thinking cause it goes through the origin lol
how do these relate

[north sierra]

hmm not sure

im just imagining a line through the origin

[gray dust]

sketch R^2 and some line thru the origin

[gray dust]

draw some vectors. transform them exactly as told by q31

[north sierra]

ok

not sure how to tbh

[gray dust]

do you understand what transformation q31 entails

[north sierra]

not really

[gray dust]

which part

[north sierra]

the part where it says reflects points across some line through the origin lol

[gray dust]

remember reflecting stuff across lines in geometry class?

[north sierra]

kinda like reflecting x^2 over the x-axis?

i never had a geometry class

[gray dust]

plot the point (1,0) and reflect it across the vertical axis

[north sierra]

ok

isn't it the same thing?

[gray dust]

no

[north sierra]

like tht?

[gray dust]

note that the points are equidistant from the y axis. the line connecting them is perpendicular to the y axis

[cloud glen]

if an eigenvalue has multiplicity of two, does it mean that there will be two possible eigenvectors for that eigenvalue

[north sierra]

is my pic good?

[gray dust]

yes

[north sierra]

yes to who?

[gray dust]

your R^2 sketch

[cloud glen]

sorry didnt mean to interupt

ill let u finish

[north sierra]

nah u good mike

sup btw lol

[cloud glen]

nm gettin ready for finals hah u

[gray dust]

what i said can be applied to reflecting anything about any line

[north sierra]

samee

[half ice]

Imagine folding a piece of paper along the line. Where does (1,0) end up?

[north sierra]

(-1,0)

what i dont get is that i thought eigen vector, values were supposed to only scale the vector, why is the question saying it reflects?

[half ice]

Maybe get a piece of paper and actually try it out

[gray dust]

the reflection is what the transform does to ALL vectors in R^2

[regal crater]

or just play with mirrors

[half ice]

Basically, connect the point with a line that's normal to your line. Then, flip that line

[north sierra]

still dont get this uestion lol

so the yellow line i made is the line they're talking about that passes through the origin?

[gray dust]

yeah, though the answer to the q should apply to any line you draw through the origin

[north sierra]

true

[half ice]

With the picture you drew, (1,0) does not reflect to (-1,0)

[north sierra]

so now i can draw any vector and try to reflect it over that line?

oh

[gray dust]

that was just an example for him to practice reflection. i told him to do it about the vertical axis

[half ice]

Imagine folding a paper over that line

But yeah, do the x-axis or y-axis, it's easier

[north sierra]

true

so (1,0) does reflect to (-1,0) right? for the y axis

[gray dust]

aye

[north sierra]

aright

tbh

i still dont get the Q and how its related to eigen stuff

eigen value

when i reflect a point over the yellow line

what does that tell me?

i guess this would be the reflection ?

[gray dust]

yes

[north sierra]

so what does this tell me?

[gray dust]

draw any vector. note the line through the origin which contains that vector. then transform the vector. note the new line which contains the transformed vector

reflect some more vectors across the yellow line, and you should see a set of vectors which "stay" in their line

[north sierra]

like that?

so since the reflections are the same length they're not being scaled and thus the eigen value is 1?

[half ice]

Note that all vectors start at the origin

[north sierra]

oh

[half ice]

Now, you have the right idea. If a vector is perpendicular to the line, then the vector still lies along the same line it was on before. Its direction hasn't been changed, so it is an eigenvector

[north sierra]

is that better?

[half ice]

Perfect. That's bae

[north sierra]

LOL

[half ice]

What's the app?

[north sierra]

Notability on the I Pad

[half ice]

There's my own awful thing

[north sierra]

LOL

[half ice]

Yellow is the axis of reflection

Red is an eigenvector

[north sierra]

kaynex, the good at paint

[half ice]

That eigenvector has an eigenvalue -1, since its direction is getting flipped

[north sierra]

oh

red line is the transformed vector?

[half ice]

I was thinking the vector before transformation, but it doesn't really matter

[north sierra]

oh okay

[half ice]

Any vector that has the same direction as the red is an eigen

Including if it were in the opposite direction

There's other eigenvectors. We have ignored the obvious case

[north sierra]

okay. But the opposite direction vectors would be - eigen values right

[jagged saffron]

does anyone know a proof for generalized schur's inequality, specifically the real part of the eigenvalue

http://mathworld.wolfram.com/SchursInequalities.html

any proof

I am desperate

[jagged saffron]

ok well i got it

[cloud glen]

my textbook shows for calculating eigenvectors to do $\lambda*I - A = 0$

but online it says to do $A - \lambda*I$

[stoic python]

Миша:

[cloud glen]

which one is correct

[stoic python]

Миша:

[cloud glen]

nvm results the same thing

[dusky epoch]

you want the dets of these matrices, not the matrices themselves, to be 0.

[wintry steppe]

can someone help me do this question?

idk how to do either part

[vast torrent]

If you know the answer to one part, do you know how to get the other part?

[wintry steppe]

no

[vast torrent]

Okay

What does it mean for three 3-tuples to span R³

[wintry steppe]

a random vector in R3 can be expressed as a linear combination of v1 v2 v3

[vast torrent]

Almost

A unique linear combination

[wintry steppe]

so the random vector has unique elements?

[vast torrent]

No, consider

b1=(1,0),b2=(0,1),b3=(1,1)

Then (2,2)=2b1+2b2=2b3

[wintry steppe]

instead of v1 v2 v3?

[vast torrent]

Just an example of a non unique lin comb

[wintry steppe]

why isn't it unique

[vast torrent]

Because i gave you 2 ways

2b1+2b2

And 2b3

For (2,2)

[wintry steppe]

ok

[vast torrent]

So

Do you know about row reduction

[wintry steppe]

using gauss method?

[vast torrent]

And solving 3 equations with 3 unknowns

[wintry steppe]

yeah

[vast torrent]

Okay

I claim

That showing every vector in R3 is a unique lin comb of v1 v2 v3

To show that that's true or false

You only need to consider how many solutions there are to

c1v1+c2v2+c3v3=0

c1,c2,c3 are scalars

I claim that if there's a unique solution to that equation , then and only then do those vectors span R³

Can you solve that?

[wintry steppe]

yeah i'll try

[vast torrent]

Good luck

Come back with more questions :)

I may not be around but someone should be

[wintry steppe]

ok, thank you

[vast torrent]

@wintry steppe if there is one and only one solution

Do you see what the solution is?

[wintry steppe]

0?

[vast torrent]

Yes, c1=c2=c3 =0

I found a proof as to why you only need to analyze that case

[wintry steppe]

can you send the link?

[vast torrent]

https://proofwiki.org/wiki/Expression_of_Vector_as_Linear_Combination_from_Basis_is_Unique

[native lodge]

An alternate look at the definition of linear independence, is that if your v’s are column vectors, then what you have is a representation of a matrix with v’s as its columns being multiplied by a column vector of c’s on the right.

So Ax=0

[wintry steppe]

thanks

[native lodge]

And to be linearly independent, the only set of c’s that makes Ax=0 true is the zero vector itself

[vast torrent]

If you don't understand sum notation I'll explain the proof

[native lodge]

If there are nonzero vectors in this nullspace, the vectors are not linearly independent

[half ice]

If you're just learning LA for the first time and want an easier read, you may be best going through an applied book first

Also, feel free to ask any questions about it here

[wind yacht]

If it helps, I am using hoffman's linear algebra

[half ice]

It sounds like you are in an applied course and likely need an applied book

What book comes with the course?

[wind yacht]

although I have also been recommended Linear Algebra Done Right (Undergraduate Texts in mathematics) & Linear Algebra by Hefferon

[lone quail]

Why is it that if the cubic invariant is 0, then the polynomial of second order, which represents the conic, can be factored with two first order polynomials?

[wintry steppe]

How would I even approach this? Prove or disprove: Every square matrix is similar to an upper triangular matrix.

By similar I mean similar over reals

[vast torrent]

Can you guess whether it's true or false?

[wintry steppe]

Well the only thing that I can use to prove similarity is if there exists a P such that A=P^-1 BP. Just from that I want to say false because I don’t see how you would prove said statement with that.

[vast torrent]

I don't know the answer off hand but my suspicion is no because the existence of JNF requires the scalars to be complex

[wintry steppe]

JNF?

[vast torrent]

So my intuition says to construct a simple 2x2 matrix with no jordan normal form

[wintry steppe]

You mean a 2x2 whose rank doesn’t equal 2?

[vast torrent]

Did you learn about eigenvalues

[wintry steppe]

Yes, but not JNF

[vast torrent]

Thats when you do a change of basis of the (generalized) eigenvectors

Said another way

You do a change of basis so that you get a triangular matrix with eigenvalues on the diagonal

That exists as long as eigenvalues exist

So i would make a 2x2 matrix that's not triangular that has no eigenvalues

That's my hunch

[wintry steppe]

So I want to find a matrix with no real eigenvalues?

[vast torrent]

Im guessing

[wintry steppe]

Ok thanks

[vast torrent]

That's not triangular

Idk if it will.work

But try a 2x2

[wintry steppe]

Wait, so then how would I show that said matrix is not similar to any triangular matrix over reals?

[vast torrent]

Oh uh hmmm im thinking

[wintry steppe]

Do I write general form of an upper triangular matrix and show that there is no P to make the 2x2 similar to a general upper triangular?

[vast torrent]

Im thinking about row echelon form

When is that not triangular

[wintry steppe]

When there are redundant rows?

Wait nvm that

[vast torrent]

I cant think of one

[wintry steppe]

Why does the row echelon not being triangular matter?

For P’s invertibility?

[vast torrent]

Because every matrix is similar to its rref

[wintry steppe]

Isn’t the only matrix similar to the identity matrix the identity matrix?

[vast torrent]

No but that's also not what i said

Every matrix is similar to.its rref so if every rref is triangular there's your answer

[scarlet ermine]

i've been working on this problem and i can't really figure it out, my understanding for part a is i just multiply a^k*v(0) k number of times to get the series, but for part b it seems to need something about eigenvalues and eigenvectors and its over my head

[vast torrent]

Yes, that's the answer mikhail

[wintry steppe]

Ok thanks

[vast torrent]

What is SAGE

[scarlet ermine]

matlab knockoff

my professor is a hipster

[vast torrent]

Fair enough

Ideally A is diagonalizable. Have you diagonalized a matrix?

[scarlet ermine]

is that where i take a and subtract i*x from it?

[vast torrent]

Nope

It's when you do a change of basis into a diagonal matrix

Not always possible, but often is

But D^k is easy if D is diagonal because it's just the diagonal entries^k

[scarlet ermine]

oh right

i've not been able to figure that out unfortunately

[vast torrent]

If you dont know how to use eigenstuff, then look up how to have SAGE do it

Diagonalization of a matrix

If such a change of basis exists, it will.give you A=PDP-¹ with D diagonal

[scarlet ermine]

right i've figured that out

[vast torrent]

And (PDP-¹)^n does something nice

[scarlet ermine]

i never was able to figure out (or my professor) how to put a matrix to the k power

[vast torrent]

It's difficult in general without diagonalizing it

Or finding the closest thing you can to diagonal

Like it might need some 1s on the off-diagonal

And might need complex numbers even if A has real entries

[scarlet ermine]

i vaguely remember calculating p and d from a's eigenvalues and eigenvectors but i never fully understood it

[vast torrent]

Well the assignment looks like it wants the computer to do the work

[scarlet ermine]

sage is basically matlab but worse

and nobody knows how to use it including the professor

[vast torrent]

I don't use either

I use R but i have on my bucket list to switch to Python when i learn it

Wolfram alpha can diagonalize too

[sonic osprey]

That's like

Sage and Matlab do very,very different things

Sage can do a lot of things matlab can't do

[vast torrent]

At least for small matrices

I don't think wa would like it if you put in a 1000x1000

[wintry steppe]

@vast torrent hi again, for part A of this question i got the matrix:

[1 3 4 0]
[0 1 1 0]
[0 0 0 0]

row echelon form

could you help me with parts b and c?

[scarlet ermine]

@sonic osprey wow a sage stan

i always thought sage was only used by my professor and a small group of hipsters in 2012

based on how impossible it is to find online resources about it

[vast torrent]

@wintry steppe this goes back to what Amph said about the solution space to Ax=0

Do you remember?

[wintry steppe]

i remember but didnt understand

[vast torrent]

Lin ind means the only solution to c1a+c2b+c3c=0 is c1=c2=c3=0

So you want to solve a system of 3 eqns in 3 unknowns

How do you write such a system.as a mtx eqn?

[wintry steppe]

for part A?

[scarlet ermine]

@vast torrent wolfram gave me

which is pretty neat

[sonic osprey]

@scarlet ermine I feel like you're not looking in the right place, there's plenty of documentation

https://doc.sagemath.org/

[scarlet ermine]

@sonic osprey well theres that huge textbook and some stuff like that

[sonic osprey]

Yeah I'm a bit biased, I've contributed a bit to sage

[scarlet ermine]

but once it gets to troubleshooting or youtube videos

[toxic pendant]

I'm having trouble understanding the notation in this problem so some clarification would be appreciated

[vast torrent]

Defunct, lucky, the mtx is diagonalizable without even needing complex numbers

[scarlet ermine]

you're kinda boned bc nobody makes videos, theres 0 active forums, nobody at my uni uses it, none of my friends or tutors or other profs use it

[toxic pendant]

So S1 and S2 are intersected

[sonic osprey]

There are active forums

[toxic pendant]

but what does the weird 0 symbol mean

[sonic osprey]

https://groups.google.com/forum/#!forum/sage-support

[vast torrent]

And do you see what happens when you take SJS-¹ to the nth power?

[sonic osprey]

not super active, but active enough

[toxic pendant]

Is it supposed to represent an empty matrix?

[vast torrent]

@toxic pendant

[sonic osprey]

I'd estimate that around half of math professors have used sage before

[vast torrent]

$\varnothing = { , , , }$

[stoic python]

gfauxpas:

[regal crater]

tbh thats pretty wasteful notation

[vast torrent]

Sorry you feel that way

Well

[toxic pendant]

Okay so my guess was right? The condition is basically saying that there are no intersections?

[vast torrent]

With vector spaces

[scarlet ermine]

@vast torrent not yet but im gonna do the math and see what happens

[wintry steppe]

1c1 + 3c2 + 4c3 = 0
0 + 1c2 + 1c3 = 0
0 = 0
? @vast torrent

[vast torrent]

Sometimes "empty intersection " means only {0}

[toxic pendant]

In that case if one has (1,1) and the other has (2,2) would there be an intersection?

[wintry steppe]

based on the row echelon matrix

[vast torrent]

And actually that looks more like a phi

[toxic pendant]

Since (2,2) is just scaled up

Here's a better picture

[vast torrent]

Aaaaaaa too many conversations

[toxic pendant]

ctrl c ctr v'ing the symbol into google brings up the page for phi if that helps

[wintry steppe]

sorry @vast torrent

[scarlet ermine]

well despite my suffering

at least today is the fancy holiday dinner

[vast torrent]

@wintry steppe anyway, you can solve this by making a mtx and doing row reduction

@toxic pendant im guessing its just supposed to be empty set

[toxic pendant]

Sounds good

[scarlet ermine]

@vast torrent i don't see any pattern, maybe im dumb tho

when i multiply pdp^-1 i get A, as expected

but when i square that i just get a^2

which now i say it sounds like exactly what would happen

[vast torrent]

You can prove inductively that only the middle matrix needs to be calculated D^n

The change of basis mtces you can keep as they are

[scarlet ermine]

so i calculate d^n?

that seems to just be -2^n, 1^n, and 5^n

[vast torrent]

Well that's pleasant, isn't it

You still have the change of bases to multiply out

[scarlet ermine]

what do you mean?

[vast torrent]

Youre finding (SD^nS-¹)v, not (D^n)v

[scarlet ermine]

ah let me try that

i don't see it

[vast torrent]

What was the original problem?