linear-algebra

[wintry steppe]

das why im asking ^^

if i wanted to use it then i wouldnt have asked for help with understanding the process

[lavish jewel]

you'll see it in class eventually

it involves a long talk about subspaces of a matrix and whatnot

right now all you need is a vector projection

i.e. the vector projection of a onto b is (a dot b) in the direction of b

[coarse sandal]

how do i go about approaching this problem

its not an exam or anyhting

this is hw

like how would i foil the (3u-v)^2

im not sure how to deal with it because of the normalizing around the 3u-v

[lavish jewel]

what are u and v? vectors in a fin dim space?

oh it says it there

the norm you have there is the same as taking u dot u

and also equivalent to u^T u

for some vector u in R^n

you can do the same for a difference of vectors

e.g. $\Vert 3u - v \Vert^2 = (3u - v)^\text{T}(3u - v)$

[stoic python]

Edd

[lavish jewel]

you can distribute the transpose and then foil your heart away

keeping in mind that all u^T v are the same as u dot v

[coarse sandal]

why is it (3u-v)^T?

ohh

[lavish jewel]

hmm?

ah, you can convince yourself on a piece of paper by writing it as a sum

write out what u dot v is

then do u^T v

and convince yourself they're the same thing

[coarse sandal]

Should i even be foiling?

cause i feel like i should be foiling so that i find a term in there that shows u dot v is 0

[lavish jewel]

how else would you multiply (3u^T - v^T) (3u - v)?

[coarse sandal]

not for that portion but for the proof over all

[lavish jewel]

i think it'll become clear after you do it

[coarse sandal]

okay

[lavish jewel]

how do you check if 2 vectors are orthogonal? just keep that in mind

[wintry steppe]

If i have two vectors instead of just one, would it still be A A^T?

[lavish jewel]

yes, but A is now a matrix, not a vector

[wintry steppe]

but its from an old assignment and it says specifically Matrix projection and not vector projection

[lavish jewel]

you can find what this matrix is by doing the vector projections and then grouping terms, same as before

i think you should really go back several lessons and review the definitions

[wintry steppe]

i think i should too

[lavish jewel]

you will get an orthogonal projection matrix, yes, and what it does is perform an orthogonal projection. an ortho projection is a decomposition of a vector into orthogonal components. this decomposition is done via vector projections onto an orthogonal basis

take the operations and associate them nicely, and you get the matrix that does them

[coarse sandal]

Thank you i figured it out

[lavish jewel]

notice i used a plural here. the projection matrix will do the work of several vector projections

[lavish jewel]

[wintry steppe]

Can I get help on this problem

[gray dust]

rewrite vectors as linear combos of e_1 & e_2 then use the linearity of T

ie if $x=ae_1+be_2$ for some scalars $a,b$ then $$T(x)=T(ae_1+be_2)=aT(e_1)+bT(e_2)$$

[stoic python]

RokabeJintaro

[wintry steppe]

I'm still having trouble understanding

Would I set e1 equal to y1 and e2 equal to y2?

[gray dust]

not sure what you mean

[crude falcon]

I think you can use the matrix associated to the linear map, you'd just need to multiply the vector to the matrix

[lavish jewel]

T(e1) = y1, this is what they told you

[gray dust]

unpack what T being linear means

[crude falcon]

someone correct me if im wrong

[lavish jewel]

yes jackieto, that's right

[wintry steppe]

So would I make an augmented matrix with e1 and y1?

[crude falcon]

you make the matrix with the images, in this case y1 and y2

[lavish jewel]

there is no need to write any augmented matrices

[wintry steppe]

So would I make a matrix like this?

[crude falcon]

yes

[wintry steppe]

And then what would I do?

[lavish jewel]

do you understand why jackieto told you to write this matrix?

[wintry steppe]

No

[lavish jewel]

this matrix represents the information you were given

i.e. that T(e1) = y1, and T(e2) = y2

multiplication of a matrix by a vector yields a linear combination of the matrix's columns

and the coefficients of that linear combination are the entries of the vector. multiplying a matrix M by e1 (the vector [1 0]) gives you the first column of the matrix M

so if M e1 = y1, that means y1 is the first column of M

similarly for y2

now follow rokabe's tip on decomposing the other vectors into linear combinations of e1 and e2, and use the linearity of T and M

[wintry steppe]

I'm having trouble understanding rokabe's tips

[rose umbra]

anyone idea?

[nocturne jewel]

Vandermonde back at it again

[native rampart]

Why is the Notation for row operations do weird

Instead of R_1->R_1-2R_2 it is R_1-2R_2->R_1

[nocturne jewel]

put this combination of rows into this row makes more sense?

[native rampart]

Just do C_2->C_2- C_1 ,C_3->C_3-C_1, C_4->C_4-C_1

Factorise the required factors and repeat

[rose umbra]

aim to triangular matrix ?

[native rampart]

Yes

A lower triangular one

[rose umbra]

i dont quite understand yet, how do you continue from here?

[native rampart]

Take (y-x) out of 2nd column

(z-x) out of 3 rd column and so on

Now you will have a bunch of 1s

[crude falcon]

to calculate the intersection of 2 subspaces and 2 basis, can I do that by slapping all the vectors in a unique basis and checking and deleting the linearly depending ones?

[native rampart]

Are you considering basis of the 2 subspaces?

[crude falcon]

yes

[rose umbra]

@native rampartthanks

[sand apex]

yeah im aware of that. just wanted to see if there was a "gold standard" for linear algebra, much like how baby rudin is considered a standard for undergrad analysis

[twilit minnow]

i dont get how they get the result after e^[-i(100pi/4)]

[dusky epoch]

e^(-i * 100pi/4) = e^(-25pi i)

-25 = 2*(-12) + (-1)

[rose umbra]

the basis of C2 above R is 4, and C3 above R is 5 ... right?

[dusky epoch]

wrong word

you meant dimension

and the dimension of C^3 over R is 6, not 5

[rose umbra]

oh right my bad

[pure thistle]

Does anyone know how to do this? Im not sure of my answer. I got
0 2 1
0 -1 2
3 1 1

[lavish jewel]

say you make matrices M_T and M_S, which have the corresponding basis vectors as columns

the change of basis mat should be M_S^-1 M_T

,w {{2,1,1},{0,2,1},{1,0,1}}^-1 {{6,4,5},{3,-1,5},{3,3,2}}

[stoic python]

Results provided by WolframAlpha

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

Requested by Edd#5182

[lavish jewel]

we can double check if this works.

what about a rank 3 3x4 matrix?

[pure thistle]

I have to compute the inverse of S and then multiply it by T?

[wintry steppe]

On a problem like this, how would I find if x is unique?

[nocturne jewel]

Ax=b has a unique solution iff A is invertible

that's one of the equivalences of FTIM

[wintry steppe]

Would it also be unique if there is a pivot in each column?

[nocturne jewel]

it'd be unique if RREF of A is I

[wintry steppe]

ohh okay thanks

FTIM

isn't that just the definition of an invertible matrix

[crude falcon]

I have a linear map f: R^3->R^4, in which space are the dimensions of the kernel and the image based?

For example, theres a theorem that says Dim(V) = Dim(Ker(f)) + Dim(Im(f)), is V R^3 in this case?

[wintry steppe]

yes

the precise theorem goes: if f: V -> W is a linear map, and V is finite dimensional, then dim V = dim ker f + dim im f

[teal grotto]

i need to find a basis for the space of 2 x 2 matrices with real entries that have eigen vector (1,-3). any thoughts on how to proceed? i have done a couple of things to start but nothing has really helped much

[torn hornet]

Probably first find the space of matrices that contain (1,-3) in their null space and then modify that by adding I to the basis of that space

[teal grotto]

ill try that. why do you think that will work?

[torn hornet]

Well I want you to figure that out

why is what i said correct

[quartz compass]

I did it differently, I put A=[a,b;c,d] then multiplied by the eigenvector and got two equations for the eigenvalue and eliminated it leaving me just with a relationship between the entries of A. From there I had the basis expansion since all 3 were linearly independent and we know it's not 4 dimensional because then we'd easily be able to make a matrix without that eigenvector

[torn hornet]

makes sense, in general for a vector in R^n the space of matrices that have it as eigenvalue is of dimension n(n-1)+1

[quartz compass]

ooh that's neat

[teal grotto]

this was my initial approach. i had the relationship between the entries and eliminated, but you lost me at basis expansion

[quartz compass]

solve for one of the variables and plug it into the matrix

then you can expand it as a sum of matrices with the entries as variable coefficients

@teal grotto

[teal grotto]

ok. thanks guys. ill try out both of these approaches

[woven haven]

I have a simple question: if I have two linear transformations whose null spaces are equal, am I right to state that they have the same nullity because their null spaces are equal?

[faint lintel]

I believe so yea

[woven haven]

Ok thank you @faint lintel

[teal grotto]

yes because if two sub spaces are equal then they have the same dimension

[woven haven]

@teal grotto Thank you very much that makes sense.

[hollow latch]

does a vector space have a certain set of scalars that can be used, or can all scalars in $\mathbb{R}^n$ always be used?

[stoic python]

light

[faint lintel]

depends on your basis vectors

if your basis is the vector of all 1s

then all you can do is create vectors where each entry is the same

@hollow latch

[hollow latch]

ah thanks

[hollow latch]

how does covector and matrix multiplication work?

[lavish jewel]

you mean covector times matrix?

[native rampart]

Isn't covector just a row vector?

In matrix terminology

[lavish jewel]

that's what i was gonna say

[blazing bear]

Hi, quick doubt:
I always saw the definition of a Linear System like y = L(x), where L is a linear operator, etc... Fine. We could have our input-output relationship as y = d(x) /dt, for example.
Today I saw on some lecture notes the definition of a Linear System as:
L_1(y) = L_2(x)

PS: L_1 and L_2 are linear operators, y is the o/p and x the i/p.

Question: why the Linear Operator L_1 on the output? I mean, wouldn't L_2(x) directly imply a linear system?

[dusky epoch]

do "o/p" and "i/p" stand for output and input respectively

also i'm assuming this is in the context of control theory?

the output isnt always directly affected by the input

if it is, you'll have L1 be the identity operator

but for example you could have a driven harmonic oscillator system

then you could describe it as $\ddot{y} + \omega^2 y = kx$, where $x$ is the driving force, $y$ is the position of the oscillator and $k$ and $\omega$ are constants required to make this equation make physical sense

[stoic python]

Ann

[dusky epoch]

in this case, y doesnt appear on its own but rather passed through the operator D^2 + ω^2 I

@blazing bear hope this helps shed some light on your doubt

[blazing bear]

Oooh, I got it.
The reason for me usually seeing y alone was because I was assuming "direct effect" from input on the output haha. But to be precise, there was actually an L1 there as an identity operator. Implicitly.

In your example, the solution y indeeds goes through those operators, whereas our forcing function doesn't (well, at least not through those ones)

Damn, how did I never thought about seeing it this way haha

Like, more general, you know?

[blazing bear]

It surely did. Thank you a lot.

[copper heath]

Let θ be the angle between the vector A = (-2, -6, 1), and B = (2, 6, -1)
What's the measure of theta?
I was never taught this, how do I approach it?

i can visualise it and reason through it but that's pretty taxing and takes a while

is there a way to compute it

i assume it is to do with the dot product..

[lavish jewel]

it has to do with the dot product indeed

there is a geometric definition for it that involves norms and a cosine

[copper heath]

you mean |A| * |B| * cos(theta)?

[lavish jewel]

indeed

[copper heath]

so i can just solve for theta using arccos, no?

[lavish jewel]

yep

[copper heath]

alright thanks
i hadn't considered that specific form of the dot product

ugh i hate needing to calculate 3D magnitues

[lavish jewel]

[copper heath]

what :(

[lavish jewel]

sqrt(x^2 + y^2 + z^2)?

[copper heath]

i know how to do it, the formula is just cumbersome and long when i have to also calculate other stuff with it

[lavish jewel]

or like a sane person would do it

,w norm{-2,-6,1}

[stoic python]

Results provided by WolframAlpha

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

Requested by Edd#5182

[lavish jewel]

do this in the bots channel tho

[copper heath]

if only i had access to wolframAlpha at exams :^)

[lavish jewel]

are you using a programmable calc or a simple one?

[copper heath]

oh yeah wait my calc can do norm

[lavish jewel]

yep, just in different ways depending on the type of calc

but most of the recent high school ones, even non programmable, have fancy matrix features

[barren terrace]

idk how to present it as system of equations

    0 = x2 
     a = x3
     3b -a = x4
       b = x5```

I've tried something like that and then tried to solve it with matrix

[barren terrace]

<@&286206848099549185>

[crude falcon]

let U,V vector spacecs, how can I compute U+V?

[wintry steppe]

if you have basis for both, u can join the basis together and u get a generator for U+V

if the basis are finite u can reduce it to a basis of U+V using the well known algorithm to do that

[pure thistle]

Does anyone know how to do (c)?

[subtle walrus]

try using (a) and (b)

[pure thistle]

Idk how to find it

Do you know?

Is it if dim(ker)+dim(range)=dim(T) then its one to one and onto?

[dusky epoch]

"dim(T)"?

[wintry steppe]

injective on finite dimensional implies surjective

[pure thistle]

I don’t understand?

[boreal crane]

Just if the dimensions of vector spaces are the same

[wintry steppe]

yea ik

thats what i meant by on

[boreal crane]

its injective if ker is trivial that means it contains just 0 and Surjective if dim(im(T))=rank(T)=dim(R^3)

[pure thistle]

Okay thanks 👍🏻

[gray dust]

@wintry steppe please don't post across servers. also you were already answered

[dusky epoch]

they're posting across servers?

[gray dust]

in hwh

[rose umbra]

How to solve question like this ?

[dusky epoch]

think about the definitions of everything involved

ker(T) consists of all vectors v such that Tv = 0

ker(ST) consists of all vectors v such that STv = 0

is it true that for every vector v in V, if Tv = 0, then STv = 0?

[rose umbra]

if i had to guess i would say no

i mean i dont see why it does

[dusky epoch]

why guess?

you should know what happens when you pass zero as input to a linear transformation.

where does any linear transformation send zero?

[rose umbra]

T(0)

0

i mean 0

[dusky epoch]

that's right.

now look at this again

is it true that for every vector v in V, if Tv = 0, then STv = 0?

notice that i am suppressing parentheses here as (ST)v is the same as S(Tv)

[rose umbra]

ohh

then the answer is yes

So the correct answer is the first one right?

[dusky epoch]

is this a "pick one" question?

you should be able to at least formulate all of these statements the way i just did

[rose umbra]

yea its a pick one

and yea i get it now

btw could you say anything about the third and forth ?

[dusky epoch]

you could say that neither #3 nor #4 are necessarily true

[rose umbra]

is this count as T: V->V ?

[gray dust]

@rose umbra the domain is the set of polynomials with deg=<2, the codomain is the set of real pairs

[rose umbra]

it should be then impossible to create one to one transformation ?

[gray dust]

why?

[rose umbra]

well there is not tranformation matrix ,right?

assuming T(x) = (0,1) how would you present (0,1) with the basis of R2[x]

[gray dust]

the domain is the set of polynomials with deg=<2, the codomain is the set of real pairs
these are finite dimensional vector spaces, so if T is linear then T has a matrix representation. but there's no need to think in terms of matrices

[rose umbra]

@gray dust

or T(x^2) doesnt matter

x^2 is like (0,0,1)?

[nocturne jewel]

in the canonical basis of R_2[x], yes

{1,x,x^2}

[rose umbra]

so assuming i want to get the transfomation matrix ,since its T:R_2[x]->R2 i will have to say that {1,x,x^2} is like span{(1,0,0),(0,1,0),(0,0,1)}?

[gray dust]

there's no need to think in terms of matrices

we can use rank-nullity to show that nullity(T)!=0, thus T isn't injective

[rose umbra]

I have a question which assuming this transformation and I have to pick up which statement is true

[nocturne jewel]

if you want the matrix representation of T, then you need to know what T does to basis vectors

[rose umbra]

yea assuming

[rose umbra]

is this true?

[nocturne jewel]

so then your matrix representation of T is those output vectors as columns

in the order of the input basis

[rose umbra]

how did you get it?

in order to find the matrix representation , I need to "build" the output vectors using the standard basis right?

[nocturne jewel]

$T[(f)_{B}]=\begin{bmatrix}2&0&2\3&1&2\end{bmatrix}(f)_B$

[stoic python]

Mosh

[nocturne jewel]

where ()_B represents the co-ordinate vector of the function in R_2[x] wrt the basis B={1,x,x^2}

[rose umbra]

yea but lets say for example it would be B{3, -2x , 2x^2}

how would you then get the matrix represenation?

[rose umbra]

assuming T(3) , T(-2x) and T(2x^2) have the same output respectively as here

[nocturne jewel]

$T(3)=3T(1)=3[2,3]=[6,9]$

[stoic python]

Mosh

[rose umbra]

i mean like that T(3)=(2,3)

and T(-2x) = (0,1)

T(x^2) = (2,2)

[nocturne jewel]

then it'd be the same matrix, just the basis would be different

and thus the co-ordinate vectors of the functions

[rose umbra]

ty

[glacial mantle]

What is the best linear algebra book for complete beginners

[wintry steppe]

How would I find if a transformation is bijective? I know that I should check if the columns are linearly independent for one-to-one

[nocturne jewel]

T is bijective iff T is injective and surjective

[wintry steppe]

How do I find if its injective and surjective?

[nocturne jewel]

check if the ker(T)={0} and im(T)=R^2

both of which can be answered from whether the matrix representation is invertible or not

[wintry steppe]

What does "ker(T)={0} and im(T)=R^2" mean?

[nocturne jewel]

kernel is just the 0 vector, image is all of the output space

[stray granite]

anyone

[wintry steppe]

So would the problem that I posted be bijective?

[nocturne jewel]

I believe so

[jovial nest]

@frosty vapor help I can't compute JCF

I have a matrix $\begin{bmatrix}3&1\-1&1\end{bmatrix}$

[stoic python]

Seasons of Light and Darkness

[jovial nest]

It has one eigenvalue, 2

So we should have $A=P\begin{bmatrix}2&1\0&2\end{bmatrix}P^{-1}$

[stoic python]

Seasons of Light and Darkness

[jovial nest]

Then we have $P=\begin{bmatrix}p_1&p_2\end{bmatrix}$ and we know that $(A-2I)p_2=p_1$ and $(A-2I)p_1=0$ so $(A-2I)^2p_2=0$ so first we find $A-2I=\begin{bmatrix}1&1\-1&-1\end{bmatrix}$ so $(A-2I)^2=\begin{bmatrix}0&0\1&1\end{bmatrix}$

[stoic python]

Seasons of Light and Darkness

[jovial nest]

Now, this is where I begin to diverge from W|A

They pick $p_2=\begin{bmatrix}-1\0\end{bmatrix}$ whereas I picked $p_2=\begin{bmatrix}1\0\end{bmatrix}$

[stoic python]

Seasons of Light and Darkness

[jovial nest]

Why do they choose the negative one

For my choice of $p_2=\begin{bmatrix}1\0\end{bmatrix}$ if you work everything out it does not give the correct $P$ because you get $A\neq PJP^{-1}$

[stoic python]

Seasons of Light and Darkness

[jovial nest]

So like I understand that it is wrong

But I don't understand why they chose p2 the way they did

[frosty vapor]

hmm

firstly

squaring the 1 1 -1 -1 should give zero matrix

in which case you can choose any generalized eigenvector p2

[jovial nest]

Wait it does

[frosty vapor]

and you can indeed choose p2 to be 1 0

[jovial nest]

Oh yes it does

Ok

[frosty vapor]

in which case p1 will be 1 -1

[jovial nest]

Yes this is what I did

[frosty vapor]

hmm then

i see no issue

unless you inverted wrong this is fine

[jovial nest]

So you have $P=\begin{bmatrix}1&1\-1&0\end{bmatrix}$ and $P^{-1}=\begin{bmatrix}0&1\-1&1\end{bmatrix}$ right

[stoic python]

Seasons of Light and Darkness

[frosty vapor]

you didn't negate the diagonals

er

the top right and bottom left

[jovial nest]

Yes I did

You swap their locations and then take the negative

Right

Or do they remain in place

[frosty vapor]

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

[stoic python]

Metalmonomorphism beamerslides++

[frosty vapor]

this is the one i got

you don't swap the other diagonal

just the main one

[jovial nest]

Oh

That's awkward

[frosty vapor]

:3

it be like that sometimes

[soft burrow]

omfg that emote

[jovial nest]

[faint lintel]

JCF 💀

[gray dust]

love jcf

[wintry steppe]

jcf is good for you

[frosty vapor]

jcf builds character

its good for u

[wintry steppe]

wait is the kernel to the image (?) as the domain is to the codomain

[wintry steppe]

what

[nocturne jewel]

no

[wintry steppe]

umm sorry i erased all of the lin alg vocab

from my brain

whats the kernel again

things that are sent to 0

[wintry steppe]

How do u find A as a linear combination of u and v using this diagram

[gray dust]

the diagram shows how many steps in the direction of u (& that of v) we need to reach a

[wintry steppe]

nvm yeah hahah i figured it out

tysmmm

[boreal crane]

https://media.discordapp.net/attachments/827249747210534952/854680165954617344/tenor.gif

[wintry steppe]

Two finite-dimensional vector spaces over F are isomorphic iff they have the same dimension.

To prove the direction where we assume that they have the same dimension, can we just define a specific linear map T in L(V,W) as follows: Tv_1 = w_1, Tv_2 = w_2, ... for each v in V?

For each basis vector of V I meant.

[native rampart]

Yes

[boreal crane]

Yup then you can Continue it Linear

[boreal crane]

And for the other direction you can prove with dimension theorem if dimV<dimW that They cant be an isomorphism cuz linear maps are either not injective or surjective

[full grotto]

guys i have an exam in 2 days and i need to know how to determine the dimension and a base of the inverse image of a linear transformation relative to an eigenspace i found

[wintry steppe]

Why would this statement be true?

[native rampart]

It is false?

Take A=0

A clearly has no pivots

[old flame]

In theorem 9.9 from Done right of Axler, I don't understand the underlined part, could anyone help thanks. It is to prove the number of different block matrices corresponding to either dimension 1 or 2 in a block upper triangular matrix under a real vector space

[old flame]

<@&286206848099549185>

[twilit minnow]

im confused on how they got to this conclusion

anyone care to explain

[old flame]

@twilit minnow since 3 is an eigenvalue, by definition, $det(A-3I)=0$, as $Av=3v$

[stoic python]

Otoro

[native rampart]

How did axler define U_m

@old flame

[twilit minnow]

thank you

isnt the selected red true?

b is not always true

[wintry steppe]

well if A has an inverse, could you come up with an inverse for A^2? for A^n?

is this a test

[twilit minnow]

sample test

[native rampart]

Literally just take B=A

[wintry steppe]

[twilit minnow]

no just think about it logically

if all diagonal entries are non zeroes (thats what made it invertible)

that means its square etc etc

would also not be 0 entries right?

[wintry steppe]

that also works

well

[twilit minnow]

unless theres a flaw in my logic

[wintry steppe]

yes, it works

the reasoning i was going for is more general, but your argument works

[old flame]

@native rampart Let $U_j$ denote
the span of the basis vectors corresponding to $A_j$ . Thus $dim U_j = 1$
if $A_j$ is a 1-by-1 matrix and $dim U_j = 2$ if $A_j$ is a 2-by-2 matrix. Where $A_j$ are the block matrices on the diagonal of the matrix of T

[stoic python]

Otoro

[native rampart]

nvm,axler explains that part

You understood U_m is a subspace of U+ null p(T)^n?

[old flame]

yup I got that part

[native rampart]

So U is also a subspace of that

Which means all elements of U+U_m are also in U+null(p(T))^n

Because well U+ null P(T)^n is a vector space

And sum of two elements in it is also in the space

[old flame]

so if I got that right, since Um is a subspace of U+nullP(T)^n and obviously U is a subspace of U+nullP(T)^n, then U+U_m is a subspace of U+nullP(T)^n, because taking an element from each and summing it would be inside U+nullP(T)^n right ?

[native rampart]

Yes

And all elements of U+U_m are a sum of a element of U and element of U_m

[old flame]

ah okay thanks, and one more question. I don't understand how does $dim nullP(T)^n=dim null P(T|U)^n+d$ proves the part where the characteristic equation of $A_m$ equals p

[stoic python]

Otoro

[old flame]

its also in the same screenshot btw

[wintry steppe]

how would I determine how many solutions a 4x4 coefficient matrix has?

The coefficantt matrix only has numbers in the diagonals

[teal grotto]

so you have a 4 by 4 diagonal matrix correct?

[wintry steppe]

@teal grotto

[teal grotto]

ok. what are your thoughts on this?

[wintry steppe]

@teal grotto

[teal grotto]

yes. every column is a pivot column. so what can you say about its rank?

[wintry steppe]

What does rank mean?

its in rref form

[teal grotto]

the rank of a matrix is equivalent to how many pivot columns there are in rref. and this is pretty much in rref

[wintry steppe]

Then there are 4 pivots

[teal grotto]

right. so that means it has to have infinitely many solutions.

[wintry steppe]

How do we know that thoough?

[teal grotto]

this is the rank nullity theorem

[wintry steppe]

I haven't learned that theoram yet, is there another theorm earlier on that tell us this?

[teal grotto]

what is your definition of having infinitely many solutions

[wintry steppe]

That means that there is a free variable

where we could chose the free var to be anything we want it to be

[teal grotto]

ah. okay. was interpreting the question incorrectly

call your matrix A. since there are no free variables in the rref of A, then for each y in R^4, there is exactly one solution to Ax=y

[wintry steppe]

wait so it only has 1 solution?

Because there are no free variables

[teal grotto]

yes. if there were a free variable, then you could have infinitely many solutions, like you said

[north anvil]

How does this matrix rotate the vector x by an angle theta counterclockwise? I’m struggling to visualize it

[wintry steppe]

try to find a formula for a point rotated counterclockwise through an arbitrary angle and see what you get

you could also just try computing the angle between x and Tx

$$\theta = \arccos \frac{\langle x, Tx\rangle}{|x||Tx|}$$

[stoic python]

R2T2

[wintry steppe]

(see if this is true)

[teal grotto]

https://www.desmos.com/calculator/c8trg7v4sw

@north anvil this should give a decent visualization

[north anvil]

Would I do this by making a triangle and using Law of Cosines?

[wintry steppe]

.

but that probably works too idk

i don't remember the law of cosines

[twilit minnow]

im sure its C

because a is obviously false and im pretty sure 2I-A isnt always invertible

but can someone explain to me how its C

[nocturne jewel]

Why do you think it's C?

[twilit minnow]

because (eigenvalue * I - A)eigenvector = 0

no?

unless im missing something

[nocturne jewel]

No

[twilit minnow]

could you please explain then

[nocturne jewel]

2I-A isn't the 0 matrix guaranteed

[twilit minnow]

so what does it mean by eigenvector of A corresponding to 2 is zero

[nocturne jewel]

It means that the nullspace of 2I-A would just be 0

[twilit minnow]

which is correct? and why

[nocturne jewel]

Just try again

[twilit minnow]

an explanation would help me try again

im lost

[nocturne jewel]

They all have to do with the invertibility of 2I-A and the fact it can't be an invertible matrix

[twilit minnow]

so its not C?

[nocturne jewel]

No, C implies invertibility of 2I-A

[wintry steppe]

c is definitely missing a word

[nocturne jewel]

the vector is 0 is fine..?

Weird but fine I think

[wintry steppe]

"the only eigenvector" sounds a lot more correct

also, eigenvectors are usually defined to be non-zero, so i'm not sure what's going on there

maybe it meant the eigenspace?

yeah my money's on it being eigenspace

[frosty vapor]

zero eigenvector wtf

[wintry steppe]

any hints for this question?

[teal grotto]

get a basis for ker(A+I) and ker(A-I)

[wintry steppe]

We don't do that in this course

its not as advanced

[torn hornet]

Ker just means null space

[wintry steppe]

yeah we haven't learned that

[torn hornet]

Huh

[teal grotto]

um. there goes the easy way

[wintry steppe]

no basis, no vector space, no subspace, no kernel

[torn hornet]

But diagonalizable?

[teal grotto]

dude what

[wintry steppe]

yes

[torn hornet]

You need those concepts lol

Like do u know eigenvalues

[wintry steppe]

yes

[teal grotto]

bruh. you need to be able to talk about a basis if you're going to be able to talk about when a matrix is diagonal

[wintry steppe]

🙁

[teal grotto]

what is your definition of diagonalizable

[wintry steppe]

It is

wait lemme get the exact

[torn hornet]

honestly at this point reject the course and learn the ideas on ur own lol

[teal grotto]

bro you guys are talking about multiplicity of eigen values but not bases???

[torn hornet]

i mean yeah thats what it means to be a null space, and i guess m parameters refer to the size of the basis

[wintry steppe]

also this

[torn hornet]

this is very wierd way to talk about this

[wintry steppe]

😦

It's just the way the course is

I have no control

there is a course that covers that stuff but its the advanced linear algebra one

[torn hornet]

do you get penalized for using concepts like basis or w/e

[wintry steppe]

i would

[teal grotto]

so, take any non-zero vector in the null space of A+I and take two linearly independent vectors in the null space of A-I

[wintry steppe]

we haven't done linear independence or null space either 💀

[torn hornet]

how

[teal grotto]

just quit

[torn hornet]

rank nullity?

[wintry steppe]

we've done rank

but very basic

[half ice]

Hah yeah stem courses ftw

[teal grotto]

this is not basic

[half ice]

Mine was similar. Basis was talked about but definitely given very little screen time

[wintry steppe]

I don't think we cover that stuff till later in the course

or if at all

[torn hornet]

so how do they expect yall to do this

[half ice]

Everything else was discussed in terms of algorithms on matrices

[teal grotto]

do you have any input on how to solve this one?

[native rampart]

Matrix algebra moment

[half ice]

You should go and learn and a basis in your own time, as that will give you a much better idea about vector spaces in general

[torn hornet]

honestly just quit this class and join the better one

[wintry steppe]

I will learn that content in my own time but i can't use any of it

[teal grotto]

wait, im curious now. what else have you guys covered

[wintry steppe]

matrices

gaussian elimination

determinants

adjugates

systems of equations

diagonalization

invertibility

just off the top of my head

those concepts

[teal grotto]

but you haven't covered linear independence?

[wintry steppe]

No

[teal grotto]

that has to be the most backwards linear algebra class ive ever heard of

[wintry steppe]

I guess its not done right

but yeah i got really stuck on this question

[native rampart]

You haven't heard of my LA class

Take his class and remove Gaussian elimination

And systems of equations

[dire thunder]

you haven't heard of my linear algebra class

take only gaussian matrices and determinants

[half ice]

You haven't heard of my linear algebra class. It was just y = mx + b

[dire thunder]

and system of equations

@native rampart btw are you space marine today

[teal grotto]

shit guys im actually laughing rn

im sorry about everybody's LA classes tho lmao. this has become one of my favorite subject areas

[native rampart]

tbh,I don't expect my professors to know anything

[half ice]

Yeah LA isn't taught very well to stems. You need some proof to get a full idea

[native rampart]

That would be too dangerous

[half ice]

Which is a shame because it is also likely one of the most important classes they'll take

[teal grotto]

right? its used almost everywhere

[half ice]

Anyone got an idea for this one? I'm actually drawing a blank

[wintry steppe]

I wanted to say something about eigenvalues

[dire thunder]

what is the question again

[wintry steppe]

how rank(A-I)=rank(-(I-A))=rank(I-A)=1

but idek

if u figure it out i just want some hints i want to be able to figure it out too 🙏

[dire thunder]

you may do it directly

well i guess

you can just write out 3x3 matrix

[teal grotto]

(A+I)x=0 will have exactly one parameter i think. (the x=(x1,x2,x3) part)

[dire thunder]

find REF of A+I and A-I

[wintry steppe]

Hmm

that would have to mean that the one parameter must be multiplicity 3

to have 3 eigenvalues

This is a T/F question

[teal grotto]

if you dont have to show work, just write true

[wintry steppe]

Its true?

I thought it was false

[teal grotto]

its true. sum of the dimensions of the eigen spaces of A is 3, hence diagonalizable

[native rampart]

null(A-cI) is dimension of Eigen space with eigenvalue c

[wintry steppe]

Rip 😦

[teal grotto]

what i meant was, any solution to (A+I)x=0 will have exactly 1 free variable (or parameter) because -1 is an eigen value of A with multiplicity one. (trying to bend to your first definition of diagonalizable)

[wintry steppe]

Ohh okay

i think i get it

lemme try some stuff

[teal grotto]

ok. try to show that any solution to (A-I)x=0 will have exactly two free variables, and that 1 is an eigen value of A with multiplicity 2. and gl

[wintry steppe]

bet

[crude falcon]

hey guys, if I want to check for linear independence using gaussian elimination, does it matter if I put the vectors in the matrix as columnns or rows? or only works if the vectors are as columns?

[lavish jewel]

doesn't matter

the rank of a mat is the same as the rank of its transpose

you can look for this on wikipedia if you want a proof: "Proofs that column rank = row rank"

you could express the elementary matrix in terms of the canonical basis vectors, and study what happens when you multiply the matrix A by this

so, study what happens when you multiply A by canonical basis vectors from the left and from the right

(separately)

[onyx palm]

Let's say I have a bunch of Eigenvectors, can I just put them together to form a Matrix? My Prof. told us that when modeling systems you can get all kinds of matrices that are modelling the same underlying system. And by calculating the eigenvalues/vectors these matrices are connected. Now what do I get when i just put these eigenvectors back together into a matrix? Is it something special? Or is it just another matrix with the same eigenvalues, eigenvectors ?

[dire thunder]

wdym put them to form a matrix?

well under some conditions you can get representation of linear map wrt to eigenvector basis

[dawn fractal]

u get a change of basis matrix

whenever they form a basis

[thin pawn]

How is it that we deleted the element a21?

I get what was done, but I didn't get why it is equivalent to the original matrix.

In case you don't get what I mean, here's the methodology to solve it:

[thin pawn]

Why is this matrix:

[thin pawn]

Equivalent to the first one here?

[onyx palm]

I dont know, we can get these special numbers and vetctors from the "map", shouldnt we be able to get a special matrix from these?

[lavish jewel]

they have done a step of "elimination"

they took row 1, multiplied it by -2, and added it to row 2

with the goal of getting rid of that leading 2 in the second row

[thin pawn]

I do get that, but I do not get what is the logic behind this being something valid to do.

What rule backs this property up?

Ah, the step is called Matrix Elimination.

Alright, Imma head out.

Thanks!

[thin pawn]

Very convenient name, just flew over my head.

[lavish jewel]

it is valid in the setting of solving a system of equations, as you end up multiplying both sides of an equation by the same scalar, or adding something to both sides of the equation

in a more abstract setting, it preserves some properties of the matrix that are important. i'm guessing you will learn that later

https://www.math.ucdavis.edu/~linear/old/notes3.pdf

this might help you out (or maybe not, i just googled random stuff)

[drowsy flower]

hi would this mapping be considered linear? Its from 2x2 matrix to 2x2 matrix such that L(A) = A + conjugate(A)

I think it is but I want to double check

[twilit anvil]

yes, but more is true: linear operators form a vector space

[wintry steppe]

Hey

Anyone know matrices?

[dawn fractal]

what do u need help with?

ask the question u want to ask

[wintry steppe]

So i had this question
Prove (AB)' = B'A'

Can u please prove this by any two matrices where order of A is 3×1 and order of B is 1×3

[dawn fractal]

what does A' notate?

the inverse?

[wintry steppe]

Transpose of A = A'

A=[1 2 3]
A' = [1
2
3]

[lavish jewel]

if you're restricted to those dimensions for A and B, you can just do it by hand and show they give the same result

[wintry steppe]

I tried

Got stuck

[lavish jewel]

where?

[wintry steppe]

So according to the order of a and b the order of product AB must be 3x3?

[lavish jewel]

yes

[wintry steppe]

And let's say A=[1
-4
3]
And B= [-1 2 1]

What will he AB?

[lavish jewel]

$\begin{bmatrix}
-1 & 2 & 1 \\
4 & -8 & -4 \\
-3 & 6 & 3
\end{bmatrix}$

oof

bruh

[stoic python]

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

[wintry steppe]

Ah

So

Oh okay i understand

Thanks mate!

[ancient pulsar]

can someone help me ?Find the spectral radius of the matrix Tj of Jacobi method for the following lineer system:

x₁-3x₃=2,

2x₁-2x₂+2x₃=1,

4x₁-3x₃=2.

[dusky epoch]

what are you tasked with? constructing an example of a matrix like that?

you mean a system represented by your matrix?

okay, so then why don't you write down the RREF?

write down the rref and put asterisks or question marks or whatever in places where you don't know what the number is

$\left[\begin{array}{ccc|c} 1 & * & 0 & 0 \ 0 & 0 & 1 & * \end{array} \right]$

[stoic python]

Ann

[dusky epoch]

this is what you will get

no? why

so what if there's no pivot in the second column?

try not to get bogged down by pivots too much

this is simple

try going back to the system-of-equations view

you have three unknowns so let's call them x, y and z

and let's give those asterisks names, say c1 and c2

$\begin{cases} x + c_1y = 0 \ z = c_2 \end{cases}$

[stoic python]

Ann

[dusky epoch]

no there would not be one solution

you have less equations than unknowns, you never have a unique solution in that scenario

why do you keep pinging me at every single message

@wintry steppe isnt

@wintry steppe this

@wintry steppe annoying

@wintry steppe as

@wintry steppe shit

well annoying it was regardless of intent

anyway i cant tell you if "an under-determined system never has a unique solution" is treated as a Theorem To Be Named by any source

[onyx palm]

quick question about eigenvectors: https://cdn.discordapp.com/attachments/423244559682764800/855106503534313502/unknown.png

from -1 0 3|0 I got x_2 no problem
but im unsure how to find x_1
-x + 0y +3z = 0
so i set z = t
which means that y = 0
and x = 3 which gives me t * (3,0,1)
but i dont know how to find the other vector
do i just say y=s and the the other two lines are independant of y, therefore its just s(0,1,0) ?

[dusky epoch]

yes

[onyx palm]

could you say that any time i have 2 lin independant rows and a zero in the remaining row that i will have a base vector as eigenvector?

[dusky epoch]

you can choose any two values of x_3 you want for part c

if one of them is the number 2, then so be it

[wintry steppe]

Ohh okay thank you

[onyx palm]

I have another question: They are asking to find a matrix D that diagonalizes Matrix A

This is from the solution

they just went ahead and made a matrix out of the 3 eigenvectors

how or why does that work?

[lavish jewel]

it is based on the idea that if you have a matrix A and an eigenvector v of A, then Av = lambda v

if the matrix A is of size n x n and has n eigenvectors, you can do the same for each eigenvector

i.e. A v_i = lambda_i v_i

you can then put all the v_i as columns of another nxn matrix, call it P

then AP = PD, where D is a diagonal matrix whose diagonal elements are the lambda_i corresponding to the v_i that are the oclumns of P

then if all the eigenvectors are linearly independent and you have n of them, you can invert P

and then A = PDP^-1

oops

wait, i did something dumb there

there we go

[onyx palm]

ok imma have to think about this for a second 🙂

D is basically the matrix which has the eigenvalues on the diagonal?

[lavish jewel]

yes

[onyx palm]

aaaaaah

now it makes sense

[opaque topaz]

Hi. Is there a term for when you take a complex m x n matrix A, and replace each of the entries with the complex number's matrix representation, and get a 2m x 2n matrix?

[dusky epoch]

willing to bet there's no established term for that

[onyx palm]

I have a followup question. how do i change A = PDP^-1 to isolate D?

[teal grotto]

P is invertible

[onyx palm]

because when i have AP = PD i would need the inverse of D

how do i get to the P on the right?

[teal grotto]

you want to isolate D right

[onyx palm]

yeah

[teal grotto]

apply P^-1 on both sides

[onyx palm]

A = PDP^-1 ?

[teal grotto]

no haha the other side

P^-1 AP=D

[onyx palm]

but dont i have to respect some rules because its matrix multiplication?

like i can only grab things from the back

because AD is not DA

[teal grotto]

right matrix multiplication is not commutative. but if you know that X=Y, then ZX=ZY. thats all you're doing. you know that AP=PD, so P^-1 AP=P^-1 PD=D

[onyx palm]

oooh so i can put things in front?

and also in the back?

just not in between right?

[teal grotto]

yes because in general matrix multiplication is not commutative

[onyx palm]

AB = CD is this allows XAB = XCD
or this ABX = CDX ?

[teal grotto]

yes

[onyx palm]

nice thank you

[teal grotto]

np

[onyx palm]

but from the order it goes most right to most left right?

[wintry steppe]

society if matrices commuted

[onyx palm]

ABC = A(BC) ?

[teal grotto]

one could only imagine

[wintry steppe]

matrix multiplication is associative, the order matters not

[teal grotto]

no matrix multiplication is associative. you can do the multiplication in any order

[wintry steppe]

sniped xd

[teal grotto]

ouch

[wintry steppe]

[onyx palm]

so i can put the inverse either in the back or the front which gets me an I matrix which cancels out?

[wintry steppe]

im not sure what you mean but associativity means A(BC) = (AB)C

[onyx palm]

in this example

AP = PD

i can multiply both sides with inveser of P in the back

which makes them cancel

and just leaves A

But i could also multiply both sides with inverse of A so i would have P = APD ?

[wintry steppe]

A might not be invertible.

[onyx palm]

lets say its regular

[wintry steppe]

i don't know what that means

[onyx palm]

its invertible

[wintry steppe]

well, it's still wrong, you forgot an inverse sign on A

[teal grotto]

wait, you have AP=PD so how did you come up with P=APD? do you mean P= A^-1 PD

[onyx palm]

yes, i forgot

would that be viable?

[teal grotto]

yes

[onyx palm]

hah nice. Thank you all. Finally makes sense 🙂

[wintry steppe]

Given a matrix in echelon form, how do I know if the matrix is constient or not

[dire thunder]

@wintry steppe there is no row with zero coefficients and nonzero constant term

[wintry steppe]

ok thanks

[wintry steppe]

I have a question

if the rank of some matrix B is 1 does that mean it has 2 eigenvalues?

The matrix is 3x3

jeez

i can't type

since there are two parameters

must there be 2 eigenvalues?

or 1 eigenvalue multiplicity 2

I was thinking

[spare crystal]

i don't think 2 eigenvalues is possible

[wintry steppe]

yeah

it must be 1 eigenvalue multiplicity 2 for the two parameters

[spare crystal]

dunno if you've studied linear transformations, but if you think of B as a linear transformation, the image of the transformation is 1-dimensional, which means it has precisely one eigenvalue, i.e. whatever the vectors that lie on that image are multiplied by

[wintry steppe]

multiplicity 2 though right?

it has to have 2 eigenvectors no?

[spare crystal]

hmm not geometric multiplicity 2

it has infinite eigenvectors, but they'd all lie on the same line

[wintry steppe]

2 basic eigenvectors*

[spare crystal]

wdym by basic eigenvector

[wintry steppe]

nvm

[spare crystal]

i think it'd only have one basic eigenvector

take this matrix
$$
\begin{pmatrix}
1 & 0 & 0 \
0 & 0 & 0 \
0 & 0 & 0
\end{pmatrix}
$$

[stoic python]

Frank

[spare crystal]

which has rank-1

[wintry steppe]

yeah

[spare crystal]

all eigenvectors must be of the form (x, 0, 0)

[wintry steppe]

Mmmm

so it only has one

[spare crystal]

yep

[wintry steppe]

what if the rank is 2?

would it have (x,y,0) eigenvectors?

[spare crystal]

it might not even have eigenvectors then if the matrix has real entries

i mean ig for a rank 1 matrix

you'd always have 0 as an eigenvalue with multiplicity 2

[wintry steppe]

im so confused by this

Is it because

Yeah im not sure

I know it has two free variables

but how does the eigenvalue 1 have a multiplicity of 2??

$(A-I)x=0\implies Ax=1x$

[stoic python]

Reaper

[wintry steppe]

so clearly 1 is an eigenvalue

[teal grotto]

how are you defining multiplicity in your class

[wintry steppe]

multiplicity means that any eigenvalue has 'm' basic eigenvectors

1 eigenvalue -> 2 basic eigenvectors

multiplicity is 2

usually shows in the characteristic polynomial when theres a square

(x+1)^2

[hollow finch]

well theres algebraic and geometric multiplicity

algebraic is the multiplicity in the characteristic polynomial

[teal grotto]

ok. ill try to get back to you in later im working on something else rn

[wintry steppe]

ok

[hollow finch]

geometric is how many linearly independent eigenvectors the eigenvalue actually has

[wintry steppe]

I really don't understand how the concepts connect

[hollow finch]

like if you have a 3x3 with a characteristic polynomial of (t-k)^3 and two linearly independent eigenvector with eigenvalue k, then you have an eigenvalue of k with algebraic multiplicity 3 and geometric multiplicity 2.

[teal grotto]

his class doesnt do linear independence. thats why i asked

[wintry steppe]

^😔

[hollow finch]

you were right in your original question. rank 1 3x3 means 0 is an eigenvalue with algebraic multiplicity 2 (because nullity is 2), and you also have a nonzero eigenvalue (because rank is 1). so you have only two different eigenvalues but you will get 3 eigenvectors in this case

[rose umbra]

what role is used here?

[hollow finch]

how can you do linear algebra without linear independence

thats ridiculous

[teal grotto]

thats what we said yesterday

[hollow finch]

[wintry steppe]

basic determinant rules det(AB)=det(A)det(B)

[hollow finch]

that is one rule thats being used

[wintry steppe]

it is?

[hollow finch]

reaper is right

the other is that $I=P^{-1}I
P$

[teal grotto]

disregard me

[stoic python]

nix (@ me for the love of euler)

[teal grotto]

i missed that

[wintry steppe]

How do you know that 0 is an eigenvalue?

[hollow finch]

youre good

[rose umbra]

what is the A and B here?

i cant see it

[teal grotto]

$$\begin{align*}P^{-1}AP-\lambda I&=P^-1AP-\lambda P^{-1}IP\&=P^{-1}(AP-\lambda IP)\&=P^{-1}(A-\lambda I)P\end{align*}$$

[stoic python]

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

[hollow finch]

vectors in the null space are eigenvectors (with eigenvalue zero). if a 3x3 has rank 1, then the rank nullity theorem tells us that the nullity is 3-1=2. so the dimension of the null space is 2, so you have two linearly independent vectors in the null space, and therefore two linearly independent eigenvectors with eigenvalue zero.
and since your teacher apparently didnt tell you, you can basically think of linear independence as not being the same or redundant. so like (1,1) and (2,2) are not linearly independent. theres a lot more to it but thats a bare bones understanding.

$Av=0$ is the same as $Av=0v$

[stoic python]

nix (@ me for the love of euler)

[wintry steppe]

😳

[spare crystal]

@wintry steppe how does your class define rank

[rose umbra]

@teal grottothx

[wintry steppe]

rank is simply the number of leading ones in a matrix after gaussian elimination

[wintry steppe]

I think this is beyond the scope of my course 😦

[spare crystal]

ah ok, so im guessing you didnt cover rank-nullity theorem then

[hollow finch]

better to think of it as the dimension of the column space but sure

[hollow finch]

alright

[wintry steppe]

we haven't covered vectors, null space, dimension, lin independence

[hollow finch]

is this like a high school basics class or a college level into to linear algebra

[wintry steppe]

its actually a second year linear algebra course

[hollow finch]

...

[wintry steppe]

college level

[spare crystal]

ah rip this is a lot easier with vectors lol

[hollow finch]

second year as in into to linear part 2 or youre taking it in your second year?

[wintry steppe]

yeah i think this question would be easier with those concepts

[wintry steppe]

this is the first of two classes

[hollow finch]

got it

thats baffling tbph

the textbook isnt axler's linear done wrong is it?

[spare crystal]

gaussian elimination is multiplication by an invertible matrix right

[wintry steppe]

hey hey

[spare crystal]

axlers is theoretical

[wintry steppe]

axler slander 😠

[hollow finch]

they are equivalent

[wintry steppe]

we're using this

https://lyryx.com/linear-algebra-applications/

[hollow finch]

left multiplication specifically

[spare crystal]

yeah so you can use that to find the 0-eigenvectors

since rank1 means you can use gaussian elimination to get a matrix with 1 in the corner

[hollow finch]

anyone who says determinants are useless deserves constant slander /s

[wintry steppe]

U a friedberg kinda person?

[spare crystal]

so we can write
$$
BE =
\begin{pmatrix}
1 & 0 & 0 \
0 & 0 & 0 \
0 & 0 & 0
\end{pmatrix}
$$

[stoic python]

Frank

[spare crystal]

where E is an invertible matrix composed of elementary row operations

[hollow finch]

not necessarily

[teal grotto]

oh god theres two of these classes

[spare crystal]

where'd i mess up 😦

[hollow finch]

the top row could be
(1 * *)
(0 1 *)
or (0 0 1)
where * denotes an unknown entry that can be anything

[wintry steppe]

the second one might be more rigorous i dont know

[spare crystal]

doesn't that mean it's rank 3

[hollow finch]

nope the second two rows would still be zero

[spare crystal]

oh i see

[hollow finch]

$\begin{bmatrix}0&0&2\0&0&6\0&0&1\end{bmatrix}$

[stoic python]

nix (@ me for the love of euler)

[hollow finch]

for example would become the last case

[spare crystal]

right

oh god ew

[wintry steppe]

the question im mainly trying to solve is If $A$ is a $3\times{3}$ matrix where $rank(A+I)=2$ and $rank(A-I)=1$ then $A$ is diagonalizable

[hollow finch]

single $ to have it be inline

[stoic python]

Reaper

[wintry steppe]

got it

[hollow finch]

the question im mainly trying to solve is If $A$ is a $3\times3$ matrix where $\rank(A+I)=2$ and $\rank(A-I)=1$ then $A$ is diagonalizable

[stoic python]

nix (@ me for the love of euler)

[wintry steppe]

lmaoo

[hollow finch]

thats a good question

[wintry steppe]

yeah its kinda tough

its the only one that stumped me

[spare crystal]

wait how do you learn eigenvectors without vectors

[hollow finch]

so when youre finding eigenvectors with eigenvalue -1 lets say, how do you do it?

[teal grotto]

dude what is a basic eigen vector

[wintry steppe]

well I know what vectors are but we haven't done much with them

[spare crystal]

ah ok

[wintry steppe]

lemme get the definition for u

[hollow finch]

basic i assume would mean something like, forms a basis maybe

[teal grotto]

im gonna cry. actually

[spare crystal]

ohh this is a cute question if im seeing it correctly

[teal grotto]

they dont do bases in his class

[hollow finch]

asda

wow

i

yeah wow

[wintry steppe]

well, I would plug in for $\lambda$ then solve the system $(\lambda{I}-A)x=0$ to get the corresponding eigenvectors

[stoic python]

Reaper

[hollow finch]

totally

so lets say lambda I-A is rank 2

how many eigenvectors will you get

[wintry steppe]

Oh

1?

[hollow finch]

yeah. why?

[wintry steppe]

theres only one parameter

[hollow finch]

righto

so we only get one eigenvector from A+I

because its rank 2

now A-I is rank 1. so how many eigenvectors do we get with eigenvalue 1?

[wintry steppe]

we would get 2?

[hollow finch]

yeah!

[wintry steppe]

wait buttt

thats eigenvectors

not eigenvalues

shouldn't that be looked at first?

[spare crystal]

3 basic eigenvectors => diagonalizable

[hollow finch]

look back at your notes or text if you need to for this: what is the requirement for a matrix to be diagonalizable?

[wintry steppe]

any $n\times{n}$ matrix $A$ is diagonalizable if there are $n$ eigenvectors

[stoic python]

Reaper

[wintry steppe]

thats one of the theorems

[hollow finch]

exactly

[wintry steppe]

but wait

how does 2 parameters lead to 2 eigenvectors

[hollow finch]

well when you have two parameters you get two solution vectors

theres more technical language for that that your class has probably skipped?

[wintry steppe]

yeah maybe

🙁

[hollow finch]

but you get two linearly independent vectors in the null space of A-lambda I which means two linearly independent eigenvectors with eigenvalue lambda of A

[wintry steppe]

For A to be diagonalizable then shouldn't $(\lambda{I}-A)x=0$ have 3 eigenvectors? How does that relate to $(\lambda{I}+A)x=0$ having 1 eigenvector?

[stoic python]

Reaper

[hollow finch]

3 eigenvectors total

[wintry steppe]

oh jeez

[hollow finch]

-1 has one eigenvector, 1 has two. thats 3 total

[wintry steppe]

OHHHHHH

[hollow finch]

🙂

[wintry steppe]

I get it

wow

[hollow finch]

awesome!

[spare crystal]

what is the significance of quotient spaces V / U? im going through Axler's rn and he talks about quotient spaces and proves that V / null T is isomorphic to im T which is kinda cool, but in what way are these quotient spaces actually useful?

[teal grotto]

@spare crystal one example: you get to talk about affine subsets that are parallel to U

to expand, take $V=\mathbb{R}^2$ and $U=\textnormal{span}((1,1))$. Then $V/U={v+U:v\in V}$. so if we set $v=(1,0)$, then $v+U\in V/U$ is the set $v+U={v+u:u\in U}$ and geometrically, it is the line $y=1\cdot x$ shifted to the right by one unit.

[stoic python]

coycoy

[signal badge]

meme answer: it's a surprise tool that will help us later
explanation: I feel like you might not see the significance immediately until you come across theorems whose proofs require quotient spaces, and then you might want to come up with your own answer by extracting it from the proof. For example, there's a theorem which says that every square matrix is similar to an upper triangular matrix (over the complex numbers), and here quotient spaces are used as a means of reduction -- reducing a vector space V into two smaller vector spaces V/U and U, since dim(V/U) = dim(V) - dim(U).

[spare crystal]

ooh ok thanks for the insight! yeah it seems like an interesting concept, i'll just have to work with it a bit more and see it in action

[wintry steppe]

lol there are quite a few things in linear algebra that are extremely important, but when you see them you're like "wtf is this"

quotient spaces, dual spaces come to mind

[teal grotto]

i haven't really seen much use for dual spaces other than the adjoint of a linear operator, (which is just the transpose i think?), and nothing else from the quotient spaces other than the example i provided. do you mind giving some more examples or expanding on dual/quotient spaces?

[wintry steppe]

adjoint and transpose are slightly different

transpose is defined for any linear map between two vector spaces, adjoint is defined for linear operators on a vector space (so maps V -> V) and depends on a choice of inner product on V

they agree when you use the inner product to identify V with V* i believe

that said

an extremely important instance of dual spaces is in the building blocks for differential forms, which are fundamental in differential geometry

(and more generally tensors)

the first isomorphism theorem is probably the most common (and important) instance of quotient spaces

quotient spaces can be thought of as a way of "reducing" structure. i like to think of them as a way to divide out stuff you don't want

i'll give an example of that in a moment

oh, quotient spaces come up in the definition of homology/cohomology. that's incredibly important in topology

here's an example that illustrates the idea of "dividing out stuff you don't want"

[stoic python]

R2T2 ✓

[wintry steppe]

this is the linear-algebraic version of symplectic reduction

also, functional analysis

in infinite dimensions dual spaces become a lot more interesting, since then, they're never isomorphic to the original spaces

https://kconrad.math.uconn.edu/blurbs/linmultialg/dualspaceinfinite.pdf

(so they're of some intrinsic interest)

this concludes my wall of text

[teal grotto]

@wintry steppe woah thanks. im gonna have to digest the omega map a bit, and review what the hell an adjoint is, but again, thanks so much

[wintry steppe]

i invite you to prove that it's well-defined

[nocturne jewel]

The proof is left as an exercise to the setter of the question

[wintry steppe]

symplectic reduction :catglad:

[teal grotto]

id rather not lol. that looks kind of just... meh

[wintry steppe]

what does that mean

[teal grotto]

idk. it just looks like not fun trying to prove that

[wintry steppe]

it's easy lol

[teal grotto]

oh. wait. i actually havent thought much about it. i looked at the function and then looked at your menacing invitation/emoji, and i was like, this is probably a nightmare

[wintry steppe]

i wouldn't give an extremely difficult exercise without a warning

[wintry steppe]

the statement of part a feels a little off to me

[teal grotto]

why?

[wintry steppe]

oh boy how do i describe this

also the statement of part b is nonsense

Whats wrong with my answers for part a and b?

i didn't say anything was wrong with your answers.

i said the problem is poorly phrased

so i;m trying to figure out what it actually means before reading your answers

What are you having trouble interpreting?

the eigenvalues of $A$, as displayed, would be scalars $\mu$ such that $A - \mu I$ is not invertible. NOT scalars $\lambda$ such that $A$ (which is a matrix, as constructed, which depends on $\lambda$) is not invertble. presumably they meant "let $A$ be the matrix $$\begin{pmatrix} 1 & 4 \ -2 & 1 \end{pmatrix},$$ find the scalars $\lambda$ such that $A - \lambda I$ is not invertible." (these are what eigenvalues actually are.) even worse, the given matrix isn't even of this form, so i'm not sure what happened

[stoic python]

R2T2 ✓