linear-algebra

[pallid rampart]

what's the matrix

[wintry steppe]

but i think, the usual definition is how many entries in the vector

[pallid rampart]

$\bR^n=\brc{(x_1,x_2,\dots,x_n):x_1,x_2,\dots,x_n\in\bR}$

[stoic python]

Whoever:

[torn hornet]

defining n as the number of dimention is probably cyclic

[pallid rampart]

well you can call it specifically for R^n first, then show that R^n is a vector space and the name is consistent with the dimension of a vector spce

but yeah it's probably circular

[wintry steppe]

we live in the matrix

[eternal finch]

@wintry steppe Anyways, as a very specific example of a set of vectors spanning a space, suppose you want to see if {(1, 2, 3), (4, 5, 0), (6, 0, 0)} span R^3. Then you need to see if each vector in R^3 can be written as a linear combination of (1, 2, 3), (4, 5, 0), and (6, 0, 0).

To do that, you would consider an arbitrary vector (a, b, c) and see if x * (1, 2, 3) + y * (4, 5, 0) + z * (6, 0, 0) = (a, b, c) has a solution.

Indeed, you'd see that it does have a solution, so the set we considered spans R^3.

[torn hornet]

also for 1 thing of clarity, the notation $(x_1\dots x_n)$ is the same thing as $\begin{bmatrix} x_1 \ \dots \ x_n \end{bmatrix}$ which might be the "column" you are referring too

[stoic python]

JohnDoeSmith:

[pallid rampart]

It is enough to show that x * (1, 2, 3) + y * (4, 5, 0) + z * (6, 0, 0) = (0, 0, 0) only has 1 solution, namely x=y=z=0

[eternal finch]

It is.

Kind of a minor point, it's usually easier to stick as close to a definition as possible when somebody is confused.

[pallid rampart]

I agree

[eternal finch]

For example, as you said earlier, a basis always has a length equal to the dimension of the space.

I'd definitely use linear independence + that fact over span + some other fact normally.

[torn hornet]

(this is the defination of dimension btw)

[pallid rampart]

defination

[wintry steppe]

dimColA = rankA right?

dimColA = NulA = n

i mean +

[pallid rampart]

what's n

[wintry steppe]

number of columns

[pallid rampart]

oh yeah

rankA + nullA = n

[wintry steppe]

and for an invertible matrix it gets messy

i mean it gets changed

dimnulA = 0

[pallid rampart]

Yeah

for an invertible matrix the null space is just {0}

But TUT

[torn hornet]

it might be useful to note this is an iff condition

[pallid rampart]

It's weird that you know rank and nullity before knowing what a basis is

[wintry steppe]

i know what a basis is

[eternal finch]

I think I learned what rank and nullity were before bases.

In my first course in linear algebra.

[torn hornet]

thats bad imo

[pallid rampart]

Ughhh

[eternal finch]

It really is.

[pallid rampart]

How can you talk about dimension without basis

[wintry steppe]

but i didn't understand the sentence they said about span of H

[pallid rampart]

Da fuk

[torn hornet]

defining dimentions as "number of columns" or something i imagine

[pallid rampart]

Ok

[torn hornet]

which is kinda bad

[pallid rampart]

Do you now understand

Oh god

That's bad

[wintry steppe]

usually a matrix is m * n

right?

so it can get confusing

if they use the same n

[pallid rampart]

Yes, but you shouldn't be too fixed on notation stuff

[wintry steppe]

they should use something like "x"

[pallid rampart]

but also don't be like john

kek

[torn hornet]

in math you will see tons of different results use the same variables

you have to get comfortable with it

[pallid rampart]

yeah

[torn hornet]

its a kind of abstractization

but anyhow i think we should solve your problem first lol

[pallid rampart]

Do you know what the span of a set of vector means?

TUT

[wintry steppe]

but yeah if a matrix is m by n right? when they say that the columns of A span R^m, they mean the number of rows, because if u have a 3 by 2 matrix for example, and they say that the columns of A span R^m, they also mean that there's a pivot position in every row, and there are 3 rows, each individual "vector" in the columns of A has 3 entries

@pallid rampart the set of all linear combinations

[pallid rampart]

Ok so a set spans H means that the span of the set of vectors is H.

Also I have no idea what you are trying to say

[wintry steppe]

when they say the columns of A span R^m, m is equal to the the number of rows

of the matrix

[pallid rampart]

Yeah

That is right

[wintry steppe]

so as u can see, it can get confusing

[pallid rampart]

You mean it can get confusing whether it spans R^n or R^m?

[wintry steppe]

when it says that it spans R^(something)

and they use n or m

because sometimes, it does mean the number of rows

[pallid rampart]

Yes

In the specific situation, it should make sense

When you're talking about matrices, and the letter n is already used, then they will not say the columns of the matrix spans R^n

[wintry steppe]

so when does it not mean the number of rows or columns?

[pallid rampart]

m and n are the most common letters to denote the number of rows of columns

It of course does not mean that m and n are always the number of rows and columns

[wintry steppe]

do u have a general rule?

[pallid rampart]

For?

[eternal finch]

Dude, my numerical analysis textbook says "n-by-m" matrix. Threw me off hecka.

[wintry steppe]

for when they talk about Span, and they use R^n or R^m, when, in general, do they not also mean the number of columns/rows

[pallid rampart]

Then n and m are just some arbitrary positive number

[torn hornet]

it does not matter if its n or m

look at the context

[elfin ingot]

whats the question

[clear vessel]

let T be the linear transformation from R^2 to itself so that:
if v is a vector on the line y = 2x, then T(v) = 2v
if v is a vector on the line y = -2x, then T(v) = 0
let A be the standard matrix for T, find the eigenvalues for A, and give a basis for each eigenspace
someone said:

a vector along y = 2x is an eigenvector so (1,2) is an eigenvector with eigenvalue 2
also a vector along y = -2x is in the nullspace as T(v) = 0, so the vector (1, -2) has an eigenvalue of 0
the basis for the eigenspace can just be the set of the eigenvectors
so the eigenvalues are 2 and 0, but how do i give the basis for each eigenspace

[elfin ingot]

whats eigenvalues

[dusky epoch]

@clear vessel you already have the eigenspace bases

{(1,2)} for the eigenspace for λ=2, and {(1,-2)} for the eigenspace for λ=0

[clear vessel]

oh is dat it

[dusky epoch]

yes that is it

[cyan siren]

I know about eigenvalues, but have no idea what an eigenvector is

Wait is it just a vector of eigenvalues?

[eternal finch]

An eigenvector v of a linear transformation T corresponding to an eigenvalue k is a nonzero vector such that Tv = kv.

[elfin ingot]

what does Tv mean

T(v)?

[eternal finch]

Yeah.

T(v).

[wintry steppe]

for example the nulspace is a subspace of R^n, and the column space is a subspace of R^m

it does mean the number of columns and rows

cuz when u find nulA, you will get vectors that have number of entries = # of columns of A, and when u find ColA,you will get vectors that have number of entries = # of rows of A, and the same applies to the basis

[elfin ingot]

isnt the nullspace all vectors t(v) = 0

how can i find that in a matrix?

im p new and learning hre

[wintry steppe]

nullspace for an invertible matrix = 0

but generally, to find the nullspace

Ax = 0, solve for x

[elfin ingot]

whats x

a matrix ?

[wintry steppe]

nope

A vector

[elfin ingot]

yea

lol yea obv sry

yea got it

[wintry steppe]

Np

i have a linear algebra midterm tmr

the questions that have us solve stuff are easy, but if it gets too conceptual, i might have some problems

[wintry steppe]

Tough scene

[coral acorn]

anyone know how to do transition matrix stuff? I'm pretty stuck on a question :((

[coral acorn]

how are you supposed to prove an rref?

[latent marten]

hi is det(ABA^T)= det(A) * det(B) * det(A^T)

[wintry steppe]

determinants are multiplicative

in fact you can go further

[stoic python]

mart:

[wintry steppe]

@latent marten

[latent marten]

thx bro

[fossil quest]

Can someone tell me the subspace of the underlined term

K is a field and i have to find out the subspace for the inverse of it which i dont fully understand

Same for F2^2

Noone? :/

[clear vessel]

"suppose that A is a 5 x 3 matrix so that for all b in R^5, the equation Ax = b has at most one solution. write down the reduced row echelon form of A, and prove your answer."
i think the rref(A) is
1️⃣0️⃣0️⃣
0️⃣1️⃣0️⃣
0️⃣0️⃣1️⃣
0️⃣0️⃣0️⃣
0️⃣0️⃣0️⃣
but idk how to go about "proving the rref"???

[sour jetty]

I'm trying to find the limit of this series of n going to +infinity. The answer is 1/4, but I keep getting zero with all my attempts, because I feel like the 1/(n^4) in the front would make the whole thing equal to zero when n goes to +infinity. Anyone has a clue on how to properly find the answer of 1/4?

[dusky epoch]

how are you getting zero exactly

[sour jetty]

If n goes to +infinity, then I believe 1/(n^4) must go to 0

And then I'm just seeing 0 * (a sum of integers)

which would also be 0

[dusky epoch]

$\sum_{k=1}^n k^3$ approaches $+\infty$.

[stoic python]

Ann:

[sour jetty]

Sure, I can see that

(0 * +infinity) would still be 0, right?

Or at least not 1/4

[dusky epoch]

no, 0 * +∞ is indeterminate.

[sour jetty]

Oh I see

Thank you, that should help

[clear vessel]

ahhhhhhh ||https://ptb.discordapp.com/channels/268882317391429632/540211747613704221/709315831909056553||

[eternal finch]

You never said what you were aiming to prove.

[clear vessel]

proving the rref of A, but like what does that even mean

[eternal finch]

Yeah, dunno. Like, prove that the RREF must be that, I guess?

[latent marten]

can someone help me with 4b

[clear vessel]

ye but idk how to prove

[latent marten]

thats what im stuck on

[latent marten]

I proved it

My head hurts

[sick dragon]

Exam tomorrow boys, gonna be annoying and ask a ton of questions today 🙂

[shrewd mortar]

So let P be

c+b a
a c-b

since symmetric (we can take c+b c-b WLOG)

We know det P = 0, that is, c^2 - b^2 = a^2
c^2 = a^2 + b^2

and c+b - (c-b) = 2b, b > a

Anyways let c = s/2, b = k/2, so 2k/2 = 2b => k = 2b or that b is an integer
since c+b is an integer, c is an integer as well

hence we are done

@latent marten

(Showing c and b are integers is necessary because when we take c+b c-b to be arbitrary integers i, j we are really taking c to be the average of them and b half the difference)

[wintry steppe]

Midterm was ez

We had to find RREF for 4/5 problems, that was the only thing tiring

[dreamy depot]

Anyone care to check this subspace proof I think I did something wrong: http://mathb.in/42036

I think I need to define addition but i'm not sure

[quartz compass]

@real plaza yeah if it's diagonalizeable you have B = EDE^-1 and so A = PBP^-1 = (PE)D(PE)^-1 has the same diagonal matrix, which are the eigenvalues

[slow scroll]

well you can have similar matrices that are not diagonalizable. A matrix is diagonalizable if it is similar to a diagonal matrix. Also, while similar matrices do have the same eigenvalues, im fairly sure there exists matrices which are not similar, but just so happen to have the same eigenvalues
i.e. what you said here

Is it just that, two matrices are similar if they have the same eigenvalues...
is not true

[void sinew]

been stuck on this for a while can someone point me in the right direction?

[pallid rampart]

-A^2 - A = I

Then think about the condition for a matrix to be invertible

[void sinew]

thanks heaps i think i got it

[pallid rampart]

Show

@void sinew show your proof

[wintry steppe]

that looks like a quadratic function

[void sinew]

might be a bit hard to read but

[wintry steppe]

A^2 being x ^ 2, A being x, and I being a constant

0 being 0 ofc

lel

[quartz compass]

not quite

well what you wrote in your paper I think I see is right

oh I was thinking rent free was the same person

@void sinew what you have is right

[void sinew]

sweet thank you

[river jasper]

is there a method to find a second eigenvector from a eigenvalue by squaring the matrix?

[sick dragon]

forgot how to parameterize: could I rewrite it as ((1-2t), t, t) or (t, t, (1-2t)) ?

[wintry steppe]

do those solve the equation?

[light herald]

does the length of the vectors in an orthogonal matrix have to be 1?

[wintry steppe]

yes by the definition of orthogonal matrix

[light herald]

and the determinant will always be 1

[wintry steppe]

I think it could be -1

[light herald]

why isnt an orthogonal matrix called orthonormal matrix?

seems kind of misleading to just call it a orthogonal matrix if its also orthonormal

[subtle walrus]

probably historic reasons

matrices where the rows/columns are just orthogonal (not orthonormal) are not very useful

so its not very useful to think about it

and the term orthogonal is probably older

so it stuck

[clear vessel]

||https://ptb.discordapp.com/channels/268882317391429632/540211747613704221/709315831909056553||

[subtle walrus]

and orthogonal matrices are just the linear transformations that preserve the dot product, so it makes sense to call them orthogonal i guess

[sick dragon]

I mean why do they choose x here to be that and not y or z?

it doesnt matter does it

[wintry steppe]

this isn't the same problem, is it?

x y z are basically the same things here

so you can freely swap the names

but were they in your other problem?

you could for example choose x=t y=w and z=1-t-w here, yes

[sick dragon]

so if you have 2 variables, you decide one is t but if you have 3 you decide one is t and the other is w?

[wintry steppe]

it depends on the number of equations and number of unknowns

not just the number of variables (unknowns)

"could I rewrite it as ((1-2t), t, t) or (t, t, (1-2t))"
no because you are not meeting the x=z condition imposed on the system

[dreamy depot]

For the question at hand would taking $\big{(-x, -y, -x+-y, -x+-y, 2x , x,y \in \mathbb{F^{4}} \big}$ work as a good example

[stoic python]

Zophike1:

[elfin ingot]

isnt U (+) W = { u+w | u in U w in W }?

wouldnt ur rexample result in a 0 in the first entry always?

[dreamy depot]

Yeah pretty much

@elfin ingot so it works ?

[elfin ingot]

yea

its closed under addition

yea yea

prove it tho

[dreamy depot]

all right sweet just wanted to be sure I had the right idea

[elfin ingot]

i am just eyeballing it lol

[dreamy depot]

@elfin ingot does it have to be zero for all entrys have to be zero I think the above example I gave is wrong it -x+-y should be -x+y

@elfin ingot u there ?

[elfin ingot]

u need to have the zero vector thats forsure

[dreamy depot]

ahhh okay is that because of the direct sum condition ?

[sick dragon]

A linear system with more equations than unknowns is NOT always inconsistent.
Is this because you'll end up with a free variable?

[dreamy depot]

I only had that in mind when doing this problem

[wintry steppe]

um

[dreamy depot]

@elfin ingot ?

[elfin ingot]

thats just for any subspace really

u need the identity vector

[wintry steppe]

@sick dragon an inconsistent system is one with no solution

parameters aren't important there

[dreamy depot]

@elfin ingot by idenity vector you mean the identity matrix ?

So is my example incorrect ? @elfin ingot

[wintry steppe]

more unknowns than equations is where parameters come into play

[cold crypt]

for this question, what does part d even mean? help would be apprecated. thanks

[wintry steppe]

if you have more equations than unknowns, you might find that if a solution were to exist for a particular system, then something like 0=5 would have to be true

[dreamy depot]

@elfin ingot u there ?

[elfin ingot]

yea sorry

yea the vector I such that a+i = a for all a

(0,0,0,0) in this case

[dreamy depot]

Ahhh okay that makes sense I also considered that as well but my example is correct right ?

I just want to make sure before I get off and play some games

@elfin ingot ?

@elfin ingot u there mate ?

[elfin ingot]

whats ur example again

[dreamy depot]

hold on let me latex it

$\big{(-x, -y, -x+-y, -x+-y, 2x , x,y \in \mathbb{F}^{4} \big}$

^ This

[stoic python]

Zophike1:

[dreamy depot]

@elfin ingot ^

[elfin ingot]

yea

[dreamy depot]

thx bro

[elfin ingot]

np

[static bison]

this is a study guide question

[hallow cliff]

You still need help?

[static bison]

yes

@hallow cliff

[hallow cliff]

Do you know how to find W^perp?

If not make a matrix with u and v as row vectors and find it's null space

For the second part, this is an eigenvector/ value problem

I'm assuming you know what diagonalizing a matrix is, we are going to do this but in reverse

$ A = PDP^{-1}$

[stoic python]

magnusChadson:

[hallow cliff]

Have the first two columns of P be u and v and the second two columns be the vectors you found in part 1

D is going to be the diagonal matrix with -1 -1 1 1 on its main diagonal

Then find P^{-1}

Then multiply them all together

@static bison

[static bison]

my bad was working on anothe rproblem

is that all the steps @hallow cliff

I'll attempt it as soon as i finish doing a test vector space problem. I don't want to lose my train of thought

[hallow cliff]

Yep

[cold topaz]

to calculate tr(U^T V), I do it manually. Like I wirte U^T, then do matrix multiplication by V, then add the diagonal of the matrix, instead of the way the book did. I shouldnt have any problems in the future. RIght?

[hallow cliff]

What's tr???

[gray dust]

trace

[hallow cliff]

Oh lol, never seen that notation before

[cold topaz]

trace

[cursive narwhal]

No

You gotta do backflips first

Then you can get the trace

[hallow cliff]

Yeah you should be fine if you evaluated the matrix

Lol abhijeet

[static bison]

okay, now attempt that one

I yet to say it

Thanks a ton @hallow cliff

[hallow cliff]

Any time

[clear vessel]

T/F: "If v is an eigenvector of A and of B, then it’s also an eigenvector of AB"
is this an ok way of proving it true?:
Av = (λ_1)v , Bv = (λ_2)v, (AB)v = A(Bv) = A(λ_2)v = (λ_2)Av = (λ_2)(λ_1)v, so v is an eigenvector cuz it follows that ABv=λv form

[gray dust]

looks ok

[hallow cliff]

^

[quartz compass]

looks perfect, not only did you show it's an eigenvector but you also were able to give its eigenvalue

[hallow cliff]

Yep

You can try multiplying it by vectors in W and W^perp and see if it gives you the right values I guess

[clear vessel]

"suppose A is an n x n matrix with n > 1 and A^2 = In, the n x n identity matrix. what are the possible eigenvalues of A?"
so i was thinking, v = Iv = (A^2)v = A(Av) = Aλv = λAv = (λ^2)v, so λ^2 = 1, so λ is ±1.
the second part of the question says: "is the matrix A + 42In invertible?" (In is identity matrix) and idk how to go about answering without trying to find like an explicit example, or if its case by case?

[hallow cliff]

Is -42 an eigenvalue of A?

It's not so A + 42In is invertible

[slow scroll]

quirky alternative proof: (A+42I)(A - 42I) = A^2 - 42^2 I = (1 - 42^2)I so
(A - 42I)/(1 - 42^2) is the inverse to (A+42I)

[hallow cliff]

Lol that's cute

Do you know how to find the eigenvalues of a matrix?

[static bison]

ye

det(A-lambda*I)

[hallow cliff]

Yep

That's what 13 is asking you to do

[clear vessel]

wait if the eigenvalue is 0 is it not invertible

[hallow cliff]

Correct

[clear vessel]

osht

[static bison]

wait so for 13, we do det(A-lambdaI), we'll get some like ax^3 + bx^2 + cx +d.
for B = 1/c(A^2 + aA + bI), we're constructing a match based on the matrix A and the coefficients of det(A-lambda*I)'s polynomial? then just take the inverse of A and check if that equals B?

hmm

[slow scroll]

no, don't take the inverse of A.

[static bison]

inverse of B?

[hallow cliff]

Yeah but it should be easer to just multiply AB

[slow scroll]

Just check that BA =AB = I

[static bison]

ah okay

[hallow cliff]

^

[static bison]

okay. gimmie a minute to some calculations

so how about 14?

[agile gyro]

notice that A = BAB^{-1} and A diag

[static bison]

yo

i didn't spot that

thank you

wait

hmm

When I think of diagonalizing matrix M, i think Q^1 * MQ = D, so wouldn't it have to be something like AB = BA => B = A^-1 * BA @agile gyro

[agile gyro]

hmm

yeah sorry i forgot the definition of similar matrices

i thought that showed that A ~ B

[agile gyro]

try showing that every eigenvector for A is an eigenvector for B

[wintry steppe]

think you mean eigenvalue

similar matrices have the same eigenvalues but not the same eigenvectors, necessarily

noticing that is enough to do 14 here

[static bison]

so using B = A^-1 * BA, I made the conclusions that B is a diagonal of any combination and A is the Identity matrix because the eigenvectors of a diagonal matrix will always equal the identity matrix.
so B = A^-1 * BA => B = I * B * I, where b can be kI or have a diagonal of (a,b,c)

i won't know if I'm right until he posts the study guide answers in like two days

wait does "Suppose A has n distinct real eigenvalues" mean that there is no multiplicity among the the eigenvalues?

meaning that A can't be I because I's eigenvalues are all 1

[wintry steppe]

because A is nxn, yes

[static bison]

?

[wintry steppe]

because A is nxn, the statement "A has n distinct eigenvalues" means that each eigenvalue has multiplicity 1

[static bison]

can you elaborate

shit

[wintry steppe]

A and B are similar, so they have the same eigenvalues, and in particular B is an nxn matrix with n distinct eigenvalues

then you're done, really

can you finish from there?

[static bison]

i'm thinking

[agile gyro]

@wintry steppe i dont think that shows A is similar to B

[wintry steppe]

oh SHIT lol i read that wrong

let me think a bit more

[agile gyro]

because A similar to B means that exists P such that A = P^{-1}BP. But we only have that A = BAB^{-1} which is just another way of writing that the matrices commute

yeah thats what i was thinking at first

[wintry steppe]

oops my bad

[agile gyro]

but every eigenvector of A is an eigenvector of B

because BAv = Bkv = kBv

since v\neq 0 and B inv, Bv\neq 0

hm wait

lol

[wintry steppe]

yeah i totally read the question wrong, sorry if i confused you djcum2quick

let me think about it some more

[static bison]

yeah, I was confused way before your input

yeah me too

by the way

thank you everyone

XOxo

[agile gyro]

Bv is an eigenvector of A

[wintry steppe]

i remember some first year linalg exercises about simultaneous diagonalization of commuting matrices, which is whats going on here

best bet is probably to use commutativity to produce a basis of eigenvectors for B, which i believe is what wasd is trying to do

[agile gyro]

yeah thats what i was trying

i think A,B are simultaneously diagonalizable if and only if they commute

[wintry steppe]

i haven't done diagonalization seriously in over a year so unfortunately im not instantly seeing the solution

that is true

[agile gyro]

but i was too lazy to read about simultaneous diagonalizablility lol

if you look at the proof of that you will probably get an idea for this

[wintry steppe]

^^

good advice there

[static bison]

that's not even in my textbook

"simultaneous diagonalizability"

[wintry steppe]

haha then you might not want to listen too much to us

although you still might get something out of looking at the proof of what wasd mentioned

im gonna sit down in a min and try to work this one out

[static bison]

for a second I thought I was onto something but it turned out to so wrong

[wintry steppe]

if $v$ is an eigenvector of $A$ with eigenvalue $\lambda$ then commutativity of the matrices gives you that $Bv$ is an eigenvector of $A$ with eigenvalue $\lambda$, and since the eigenspace is one dimensional, there is a scalar $\lambda^\prime$ with $Bv = \lambda^\prime v$

wtf

[dusky epoch]

use single dollars

[stoic python]

TTerra:

[wintry steppe]

ty ann

anyways, that's probably a good starting point

yeah, i think that's the approach you were supposed to make

i apologize if i confused anyone with anything

and yes, this is simultaneous diagonalization, at least a special case of it

[static bison]

yeah, I'm not going to pretend like I understand that

I'm kinda exhausting from studying fro=om nonstop all day

i'm going to copy the important bits of this convo then return to this tomorrow

I want to understand this

thank you

XoxO

[clear vessel]

T/F (+ proof): "if w in R^2 is a linear combination of u and v, then u is a linear combination of v and w"
my initial thought was w = c1u + c2v, so u = (w - c2v)/c1, and thought i could prove it that way, but then i realized if c1 was 0, there was no way to write u as a linear combination of v and w. idk how i'd prove false; should i do an explicit counter example?

[elfin ingot]

yea

by trying to prove it u found like

the thing thats not making thiis true

now use this knnowledge to find ur counterexample

@clear vessel

[dusky epoch]

@clear vessel consider the case where w = v and {u,v} is linearly independent

[clear vessel]

ooooh yes good thanks you

[dusky epoch]

of course you would need to prove that a linearly independent set {u,v} exists

[clear vessel]

oh wait shid

idk if that exists

[dusky epoch]

hint: this is R^2 you're working in

but also i was kinda being facetious

[clear vessel]

bruh im slow

[dusky epoch]

this is R^2 and of course linearly independent sets of two vectors exist in R^2

{(1,0),(0,1)} is one example

[clear vessel]

but what if like u = 2v

[dusky epoch]

that is not linearly independent

i mean

if u = 2v then {u,v} will not be linearly independent

[clear vessel]

ye so how do i know that {u,v} is linearly independent

wait or am i just saying that "because we are in R^2, there exists a linearly independent set {u,v}"

[dusky epoch]

ye so how do i know that {u,v} is linearly independent
you don't

i am saying that there exists a counterexample to your statement

take any linearly independent set {u,v} in R^2 (and at least one such set exists) and set w = v

[clear vessel]

so like could i say, "Let there be a linearly independent set {u,v} in R^2 and a vector w in R^2. Let w be a linear combination of u and v. Then there exists a c1 and c2 such that w = c1u + c2v, and therefore u is a linear combination of v and w where u = (1/c1)w - (c1/c2)v. Let w = v. So, c1 = 0 and c2 = 1, and u would have to be the zero vector. This is inconsistent with statement that {u,v} is linearly independent, and thus u is not a linear combination of v and w" or am i just overly complicating something / need to reword for it to be a good proof

[dusky epoch]

overcomplicating

original statement: "if w in R^2 is a linear combination of u and v, then u is a linear combination of v and w"

[clear vessel]

but its not

the statement false

[dusky epoch]

no, i mean that this is what you were asked to determine the truth of

response: "false. consider any linearly independent set {u,v} in R^2 and let w = v (= 1u + 0v). then u will not be a linear combination of v and w, as any linear combination of v and w will be a scalar multiple of v, which u isn't."

[clear vessel]

oh frick yes, i think i've gotten myself in the mindset of a certain structure of proofs cuz i taught myself, yes thank you sm and sorry for my ignorance

[clear vessel]

T/F (+ proof): "For any natural number n > 1, Let V be the vector space of n x n matrices, with respect to the usual addition and scalar multiplication operations of matrices. Then there exists a basis of V consisting of 3n matrices." So like would the vector space of 2x2 matrices have 4 matrices, and if a vector space V has a basis of n vectors, then every basis of V must consist of exactly n vectors., so a 2x2 gotta have 4. so that means a 4x4 matrix vector space gotta have a basis of 16 matrices, so every basis of B gotta have 16, but here n=4 and 3n = 3(4) = 12 ≠ 16 so is that statement not true, and i prove by counterexample? my issue is how do i know that a vector space of of nxn matrices has a basis that needs n^2 matrices?

[hot vessel]

for matrices in R^(nxn), why is A + A^T always orthogonally diagonalizable

[slow scroll]

@clear vessel choose what you think should be a basis and show its a basis i guess?
@hot vessel any symmetric matrix is orthogonally diagonalizable

try taking its transpose

[dusky epoch]

$(A+A^T)^T = A^T + (A^T)^T = A^T + A$

[stoic python]

Ann:

[clear vessel]

@slow scroll it seems weird writing out 16 matrices to prove that they are the basis of the vector space of 4x4 matrices tho

[dusky epoch]

my issue is how do i know that a vector space of of nxn matrices has a basis that needs n^2 matrices?
the dimension of R^(n×n) is n^2

you can prove this by making an isomorphism between $\bR^{n \times n}$ and $\bR^{n^2}$ in one of a number of relatively obvious ways

[stoic python]

Ann:

[clear vessel]

i don't really know what an isomorphism is shid im dense

[dusky epoch]

rip

okay i mean

you can prove that $\dim(\bR^{n \times n}) = n^2$ i guess

[stoic python]

Ann:

[dusky epoch]

by asserting that there is a basis for R^n×n consisting of n^2 matrices

each with a 1 in one position and 0s elsewhere

[clear vessel]

do i have to like write them out or just say that rigorously

[dusky epoch]

the latter

i mean

there is a basis for R^n×n consisting of n^2 matrices
each with a 1 in one position and 0s elsewhere

this is rigorous enough imo

[clear vessel]

oh ye it is, maybe i'll say something to affect of "each with a 1 in one difference position" but ye ok yes thank you, i am quickly losing ability to think straight this late at night and apologize and thank you

[gaunt gulch]

Hey dumb question: if i have a 2D vector is it possible to represent any shear operation as a combination of rotations and stretches, and if so do i need more than 1 of each?

[quartz compass]

no it's not possible

it's possible to represent rotations with shears though

an easy way to see why, think about complex number multiplication, it allows you to rotate and scale and multiplication is commutative

but shearing is not commutative in general so it's not something you can access by just playing with complex numbers alone

[gritty frigate]

Guys

[sick dragon]

Is this all correct?

[wintry steppe]

x1 = x3 - 2, x2 = -2x3 + 3, x3 = x3

x3 (1, -2, 1) + (-2, 3, 0)

[gritty frigate]

Do you guys know about Markov chains ?

[sick dragon]

x3 is free, right

i can sub anything for x3 into either of those equations, so that checks out

[cold topaz]

is this an inner product?
a^2 u1 v1 + b^2 u2 b2 + c^2 u3 v3

[pallid rampart]

$(a^2,b^2,c^3)\cdot\begin{bmatrix}u_1v_1\u_2b_2\u_3v_3\end{bmatrix}=a^2u_1v_2+b^2u_2b_2+c^3u_3v_3$

[stoic python]

Whoever:

[cold topaz]

that's a yes!

right @pallid rampart ?

[pallid rampart]

Yeah but like

You can then argue every number is the inner product of some vectors

So the question is a bit weird

[dense spindle]

Do you guys have room for helping me out?

[pallid rampart]

You could've also just wrote

$(1,0,0)\cdot\begin{bmatrix}a^2u_1v_2+b^2u_2b_2+c^3u_3v_3\0\0\end{bmatrix}=a^2u_1v_2+b^2u_2b_2+c^3u_3v_3$

[stoic python]

Whoever:

[pallid rampart]

which

you know

is trivially true

[cold topaz]

the problem says u and v are in 3D. a,b,c are real numbers. is the following statemen an inner product in 3D? that's it.

[pallid rampart]

oh

probably not, unless a^2=b^2=c^3=1

[dense spindle]

@pallid rampart could you help me out?

[cold topaz]

I made a mistakeeeeeeeeeeeee
@pallid rampart

a and b and c are squared. neither are qubed.

sorry

so this is the statement.
a^2 u1 v1 + b^2 u2 b2 + c^2 u3 v3

[dense spindle]

I’m not really sure what b is asking

I solved everything except b

[mellow panther]

Write T(v) as a linear combination of the given basis vectors?

[dense spindle]

c1v1 + c2v2 + c3v3?

I''m not entirely sure

[pallid rampart]

well

still

i don't know if that expression can be written as the inner product of v and u

[cold topaz]

still depends on that values of a, b and c?

[pallid rampart]

yeah

[cold topaz]

so it forms an inner product if and only if their square has the same value?

or there is something else?

[mellow panther]

Yeah I think it's asking you to write T(v) like that

[pallid rampart]

yeah

i would say that's probably what it's asking you to say

[cold topaz]

but then there are cases where we have 2u1v1+3u2v2
How is that an inner product when we have 2 and 3?

and not the same value

[pallid rampart]

Probably not

[cold topaz]

what do u mean probably not? i have such examplles of inner product in my book.
lol

[pallid rampart]

damn

what is it?

it kinda doesn't make sense to me what the question is asking for

i'll probably get it if you show me this example

@cold topaz

[pallid rampart]

...

That's a completely different question

I didn't realize that

[cold topaz]

im just saying that we can get an inner product that have different values for a and b

then how come in my problem, there must be equal?

[pallid rampart]

Yeah I understood your problem incorrectly

Idk the answer

Lemme see

Well ok I see

$u^T\begin{bmatrix}a^2&0&0\0&b^2&0\0&0&c^2\end{bmatrix}v$

[stoic python]

Whoever:

[pallid rampart]

so the weighted matrix is just that

you should've said weighted inner product

[cold topaz]

i suck at explaining. but my main problem doesnt ask or talk about the weighted inner product. This is all i have as my problem:
the problem says u and v are in 3D. a,b,c are real numbers. is the following statemen an inner product in 3D?

[pallid rampart]

Ok fine

What are the expressions?

Can you show all of them?

[cold topaz]

u,v ∈ R^3 and a,b,c are real numbers.
<u,v> = a^2 u1 v1 + b^2 u2 b2 + c^2 u3 v3

@pallid rampart

[pallid rampart]

I would say yes

So it’s a weighted inner product because each ui is multiplied with vi

[cold topaz]

hmmmmmmmmmmm

[quartz compass]

doesn't say a,b,c are nonzero

so you will have things other than the 0 vector mapping to 0, so by technicality it may not be an inner product

[cold topaz]

but the the satement has a and b and c squared.

[quartz compass]

0^2=0

if a=b=c=0 it's not an inner product

0 is a real number

[cold topaz]

if I make it clear from the begining that a and b and c must NOT 0, and then proceed to prove the axioms, that'll do it?
@quartz compass

how about this?
Let a, b, c ≠ 0 then prove the axioms....

[quartz compass]

yeah try it

[solar osprey]

How do we show that two matrices are not similar

[cold topaz]

it has something to do with their transpose i guess.. @solar osprey

[solar osprey]

mmm

[cold topaz]

oh!

when u get the matrix's transpose it it is the same matrix, then they are similar.
But please double check that.

[eternal finch]

A matrix is similar to another matrix if they represent the same linear operator but in different bases.

People usually state this by saying that a matrix A is similar to a matrix B when A = P^-1 B P for some P.

You can show that two matrices are not similar by examining some property that similar matrices should share.

Such as determinant, eigenvalues, rank, trace.

You should look at these things because they are independent of bases.

[solar osprey]

tho

yea makes sense

maybe characteristic polynomial or smt

[eternal finch]

Sure, that's another thing you could check.

If you suspect two matrices aren't similar, then you only need to check one, tho. So, you should choose the easiest to compute.

[solar osprey]

What knowledge do u need to prove that real symmetric matrices are always diagonalizable

Im guessing it wouldnt if ur working in R

[cold topaz]

im working in R^3

[eternal finch]

Ha ha, I couldn't really explain it cleanly, so I had to go back to my book.

The result you are interested in proving is part of the spectral theorem. The proof I learned relied on choosing an eigenvector v of your operator T, which is possible because T can be represented by a real symmetric matrix; showing that T restricted to the orthogonal complement of the span of v can be represented by a real symmetric matrix. You can then get another eigenvector and repeat the argument.

In the pictures below, self-adjoint is the same as representable by a real symmetric matrix.

@solar osprey If you're interested in it.

[cold topaz]

@eternal finch i read the whole thing just to find out it is not directed at me. lol

[eternal finch]

let's say it doesnt matter. This is how i did a):
I got 2u and 3v first.
Then got 2u-3v=<x, y, z>.
Then <2u-3v, 2u-3v>.
So
(x^2 + 3y^2 + 2z^2)= result.
And at the end:
sqrt(result)
That's how u would do it if the bar didnt matter. right?

I mean, the steps are ok, but if you say 2u - 3v = (x, y ,z), then <2u - 3v, 2u - 3v> is not x^2 + 3y^2 + 2z^2.

[cold topaz]

how about <2u - 3v, 2u - 3v, 2u-3v>?

[eternal finch]

But why would you want that?

Also, doesn't make any sense.

Three arguments?

Oh, my bad.

Sorry, I didn't read all of the directions.

No, you're right.

Didn't see that a different inner product was being used.

[cold topaz]

so two arguments or three?

im cofused lol

[eternal finch]

What you did before is good.

Also, two.

[cold topaz]

all of this if the bar oesnt matter. right?

it is throwing me off. i hate it

[eternal finch]

Yeah. If the bar did matter, then you'd just compute what bar(2u - 3v) and then do the same thing.

[cold topaz]

what's bar(2u-3v)???

[eternal finch]

As 9029 said, the bar doesn't matter because you have real-valued vectors.

conjugate(2u - 3v)

[cold topaz]

so for onee final time, i know im being annoying...
SHould i ignore the bar in this case?

since it is in R^3

[eternal finch]

Yeah.

[cold topaz]

phew!

[eternal finch]

You should know how to do it if you had complex-valued vectors anyways, tho.

[cold topaz]

maybe i havent learned it yet

[eternal finch]

Perhaps.

[grizzled tapir]

anyone have a clue what trace_2 is? I couldn't find anything googling

[gaunt briar]

hey

what does this find

-b/2a

vertex right?

[quasi vale]

yes

[gaunt briar]

so if i have a parabola and use that formula on it will i get the xvalue of the vertex

oor the y

nvm

its te axis of symmetry

[clear vessel]

So i got a 2x2 matrix
[2 3
5 k], and it asks me for which values of k is this matrix invertible ( k = 7.5, detA = 0 ), but it also asks for which values of k will every entry of A^-1 be an integer? I can solve this feasibly using desmos and trial and error, but I was wondering if theres a better by-hand method i can use to solve this

[stoic python]

slimvesus:

slimvesus:

[clear vessel]

but if detA = ±2, A^-1 isn't an integer matrix, but if detA = ±3, then it is an integer matrix

and detA = ±4 doesn't work, but detA = ±5 works, but detA = ±6 doesn't work, when do i know this ends

[wintry steppe]

How can I find a vector parallel to (2, 1, -3) ? I know 2 vectors are parallel if their cross product is 0, but I'm kind of lost.

[odd kite]

@wintry steppe parallel vectors are all along the same line

what happens if you scale a vector by a constant?

[elfin ingot]

i dont understand the proof of L(V,W) is finite dimensional and has dimension mn

oh taken

[odd kite]

@elfin ingot well you left out the part about V having dimension m and W having dim n

[elfin ingot]

yeaa

sry

[wintry steppe]

what happens if you scale a vector by a constant?
@odd kite it gets either longer or shorter

[odd kite]

but it's still on the same line, right?

[wintry steppe]

I suppose but it's not parallel then ?

My goal is to find the symmetric and parametric equation of the vector (2, 1, -3)

[odd kite]

@elfin ingot suppose f: V-> W is a linear function and W is n dimensional. We can describe this by n linear functions, 1 for each component of the output. Each of these evaluates to a scalar, and takes m parameters as input. Okay so far?

[elfin ingot]

yea

@odd kite

[odd kite]

okay now each of these functions can be uniquely described by m constants

a linear function which returns a scalar must look like $c_1 x_1 + c_2x_2+\dots$

[stoic python]

Timon:

[odd kite]

does that make sense?

so we have m constants $c_i$ for each component of the output, and the output has n components

[stoic python]

Timon:

[odd kite]

so overall we need mn constants

to describe the linear operator uniquely

We write this as an n x m matrix

[wintry steppe]

I don't get why scaling a vector gives us a colinear vector

The website is showing an illustration where multiplying a vector by a constant creates another one above/below

But I thought multiplying by a constant just made it shorter/longer

[odd kite]

@wintry steppe remember vectors don't have a location, just direction and magnitude

the "base" point location isn't part of the vector, they just moved them side by side so you could see them

all parallel vectors are colinear and scale multiples of one another

[wintry steppe]

Okay, thanks

[distant granite]

is there a solution to this system of equations ?

[odd kite]

that's not a system of equations

A system would look like Wx = b

[distant granite]

yeah but you can transform it that way

https://www.wolframalpha.com/input/?i=systems+of+equations+calculator&assumption={"F"%2C+"SolveSystemOf4EquationsCalculator"%2C+"equation1"}+->"2*x+%2B+y+%2B+3*z+%3D+0"&assumption={"F"%2C+"SolveSystemOf4EquationsCalculator"%2C+"equation4"}+->"0%3D0"&assumption="FSelect"+->+{{"SolveSystemOf4EquationsCalculator"}}&assumption={"F"%2C+"SolveSystemOf4EquationsCalculator"%2C+"equation2"}+->"x%2Bz%3D0"&assumption={"F"%2C+"SolveSystemOf4EquationsCalculator"%2C+"equation3"}+->"x%2By%2B2*z%3D0"

considering b is the null vector

[odd kite]

@distant granite there's always a solution to Wx = 0

namely x= 0

there can be others depending on W

[distant granite]

i see thanks a lot

what is the term used to call the system of vectors i've sent ?

[odd kite]

you could call it a 4x3 matrix I guess

you could also call it a linear operator

[distant granite]

i'm sorry i have another question pls hehe

so if you look for the common linear subspace of two under linear spaces $R^4$ is it a linear subspace of $R^3$?

[stoic python]

Kasadraf:

[distant granite]

<@&286206848099549185>

[eternal finch]

Could you restate your question? To address a bit of your question, a subspace of R^4 can not be a subspace of R^3.

[distant granite]

that's what i was looking for. thanks.

[clear vessel]

So i got a 2x2 matrix
[2 3
5 k], and it asks me for which values of k is this matrix invertible ( k = 7.5, detA = 0 ), but it also asks for which values of k will every entry of A^-1 be an integer? I can solve this feasibly using desmos and trial and error, but I was wondering if theres a better by-hand method i can use to solve this
slimvesus said "If detA = ±1, clearly A^-1 is an integer matrix, and if A^-1 is an integer matrix, then A divides 2, 3, 5, and k. But then (detA)^2 must also divide detA, so detA = ±1"
so when detA = ±1, k = 7 or 8, which yields in an integer inverse matrix
but I got confused because, if detA = ±1/2, A^-1 isn't an integer matrix, but if detA = ±1/3, then it is an integer matrix, and detA = ±1/4 doesn't work, but detA = ±1/5 works, but detA = ±1/6 doesn't work, so when do i know this ends?

[gloomy arrow]

Does anyone have a linear algebra study guide/cheat sheet with all the things I would need to know for a final exam?

[stoic python]

slimvesus:

[odd kite]

@gloomy arrow different courses cover different things and may have a different focus, so maybe you should look at what your HW problems were or ask your prof. about what kinds of things will be on the exam.

[clear vessel]

@wintry steppe but like idk how to find those values without graphing 2/2x-15 3/2x-15, 5/2x-15 x/2x-15 and seeing where they are integers by trial and error

ye detA = 1/3 works cuz k would equal 23/3 which makes it invertible

but like detA = 1/2 doesn't work cuz k would equal 31/4 and multiplied by 1/detA wouldn't get rid of the fraction

[gloomy arrow]

@odd kite True. I have been spending all day reviewing hw and old tests

[distant granite]

are the two space vectors u1 = (a) and u2 = (-a) contained in the same linear subspace ?

[odd kite]

what do you think?

[distant granite]

me ?

[odd kite]

yes

[distant granite]

i say no

but i don t know what rules to apply

so that i decide

it's just intuition

timon pls help me out xD i m struggling here

[clear vessel]

it seems to only work when the determinant is 1/n where n has to be an odd number, so if 2k-15=±1/n, i got 15n/n ± 1/n, so k = (15n±1)/2n, where n is an odd number (or would it be better to say 2n-1 instead of n to guarantee an odd number), so would an ok answer instead of listing out all the possibilities for k (Cuz they infinite), could i say k = 7.5 ±1/2n, where n is an odd integer (or k = 7.5 ± 1/(4x-2) where x is any integer)? @wintry steppe

[odd kite]

@distant granite just to be clear, is 'a' a vector or a component

[distant granite]

@odd kite it's a real variable

@wintry steppe ?

[odd kite]

@distant granite is (-a) in span( (a) ) ?

[distant granite]

yes

[odd kite]

and span (a) is a vector space, in case you didn

't know

[distant granite]

ik

so the answer is yes to my first question ?

or no

[odd kite]

well, we established they are in the same vector space span( (a)) , so how many dimensions is that space?

[distant granite]

it's in R

so 1

[odd kite]

yeah, and that

that's what they mean by linear

[distant granite]

so they are in the same subspace ?

we know that they are linear

and -a is in the span of a

[odd kite]

we've established they are in the same 1d vector space. Is the word subspace confusing you?

[clear vessel]

@wintry steppe well we know that 1/det(A) has to equal an integer, cuz theres no GCF of 2, 3, 5, and k that we can divide them by, but the issue is proving that 1/det(A) is an odd integer

[distant granite]

yes

[odd kite]

it's pretty meaningless in this case, since they didn't specify subspace of what

every vector space is also a subspace of something

[distant granite]

this was not the question they proposed. Here it is

so basically we have u1 and u2 ,both vectors

[odd kite]

my Deutsche isn't great

[distant granite]

i want to find an example disproving 1

haha

tell me if u don't get smth

so i said if we have u1=a and u2=-a

then u1+u2 =0

which is included in U

but u1 and u2 are in separate linear subspaces

[stoic python]

slimvesus:

[clear vessel]

oh yeah, then write 2k - 15 = ±(2m - 1) if det A = 2k -15 is odd, then k = 15/2 ± 1/2 ± m which is an integer

wait no

[cold topaz]

if we can find the cos of an angle using the "classic" way, and if we use inner product for the same reason, we should get the same answer. right?

[clear vessel]

m has to be 1/m

1/detA has to be odd

[stoic python]

slimvesus:

[eternal finch]

@distant granite Your problem, roughly translated to English using a dictionary and machine translation, is this.

Let K be a field, V be a vector space over K, U, W, and W' be subspaces of V, and u_1, u_2 be in V. Which of the following statements are true for any choice of K, V, U, W, W', u_1, and u_2? Provide a brief justification or a counterexample.

(a) u_1 + u_2 not being in U implies u_1 + u_2 is not in U.

Btw, linear space and linear subspace are the same as vector space and subspace, respectively.

[clear vessel]

@wintry steppe 2k - 15 = 1/(2m+1) tho

we showed 1/det(A) = integer

[eternal finch]

Nah, state your question. Although, yah, it'd be preferable if you went in a question channel.

[ocean sequoia]

i thought you were allowed to ask on topic questions here?

[eternal finch]

I think the etiquette is that if the channel seems occupied, then you should go find an unoccupied channel.

Btw, I think you'll very much enjoy Axler's book. It's very clean.

And easy to follow.

[ocean sequoia]

yea ive been enjoying it alot so far

[clear vessel]

@wintry steppe im not trying to find whether k is an integer, but rather, is its inverse equivalent (k/det(A)) is an integer, i think im almost there

[ocean sequoia]

i figured my thing out

[clear vessel]

ye i think i gots it

thx @wintry steppe

[shrewd mortar]

@eternal finch are you an undergrad?

[gaunt gulch]

@quartz compass hmm interesting, what about scale, rotation, and another scale? I feel like i should be able to get any 2d point from a unit vector in some direction by rotating and stretching, right? If so, just need to stretch down to unit length vec first

[eternal finch]

@shrewd mortar Yes, I am, but in terms of mathematical maturity, I'm probably below a lot of high school and early undergrad students. My mind tends to not be very malleable.

[ocean sequoia]

isnt that literally what the SVD is? Scale Rotation Scale?

[shrewd mortar]

i see

[eternal finch]

🤣

😭

[clear vessel]

T/F (+ proof): "Let Pn(R) denote the vector space of polynomials with real coefficients, of degree at most n. There exists a linear transformation T : P8(R) -> P5(R) so that the kernel of T has dimension 2"
The kernel of T is the set of all u in V such that Tu=0 (zero vector in W) right? so the kernel of this T is the set of all polynomials in P8(R) such that Tu = 0 in P5(R)? idk what to do with this

[elfin ingot]

i think theres rank nulltiy in this idk why

id try looking at the possible values of rank(T) ( dimension of img of T )

nvm im stupid idk

[clear vessel]

[odd kite]

oh

it says T/F

and I just gave the answer

oops

@clear vessel yes use the rank nullity theorem

[clear vessel]

Rank(T) + Nullity(T) = dim(V)

so to solve for the nullity(T) which is the kernel of T

dim(V) - Rank(T) = Nullity(T)

[odd kite]

@clear vessel the question is asking, can the nullity be 2. So what is dim(V) here? What are the possibilities for Rank(T) ?

[clear vessel]

is dim V 9 and rank T 6

[odd kite]

no, rank(T) <= 6

[clear vessel]

how do i know that

wait lemme see my textbook

"The rank of A is the dimension of the column space of A." k that didn't help me

[odd kite]

the codomain has dimension 6. The image can be at most the dimension of the codomain but it can be less

[clear vessel]

ok hmmmmmmm

i know P8(R) has dimension 9 because you can have 9 diff coefficients in a polynomial

cuz the 9th comes from the zero vector

[odd kite]

you're right that PR5(R) has dimension 6. But not all of that is necessarily mapped to

does that make sense

there could be polynomials that are never reached regardless of what input you choose

[clear vessel]

ye eyeyeyeye ok i got that but am i right to say that the dimension of P8(R) is 9, cuz the question says its degree at most n, but you can represent a polynomial of degree 8 with 9 coefficients

[odd kite]

yes

[clear vessel]

T/F (+ proof): "Suppose {v1,...,vn} is an orthogonal basis of R^n, and let W = span(v1,...,vk). Then W^perp = span(vk+1, ..., vn)."
ik two vectors u and v in R^n are orthogonal if u*v=0
and ik a vector x is in W^perp if and only if x is orthogonal to every vector set that spans W (and that W^perp is a subspace of R^n but idk if that helps me prove it true/false

[odd kite]

what I mean is you have the right starting points for your proof

[dusky epoch]

let $x = \sum_{i=1}^n c_i v_i$ be an arbitrary vector in $\bR^n$. show that $x \in W^\perp$ iff $c_i = 0$ for $i=1,2,\dots,k$

[stoic python]

Ann:

[clear vessel]

uhh

[wintry steppe]

Can someone explain what does the following mean and how would it affect the table overtime will be required for the last 25% of production capacity available in Plant A to make product 1, and the last 50% of capacity in Plant B to make product 2???

[clear vessel]

@dusky epoch if {v1,...,vn} is an orthogonal basis of R^n, and if c1, c2, ... ck is 0, x dotted with v1,...,vk is 0, and therefore orthogonal to those vectors, so x is in W^perp because it is orthogonal to every vector that spans W (v1,...,vk) ?

[dusky epoch]

that proves the <= part

[clear vessel]

@dusky epoch wdym ≤ ?

[dusky epoch]

no

i mean <=, not ≤

you proved that if c_i = 0 for i = 1,2,...,k then x in W^perp

now you need to prove the other direction

if x in W^perp then c_i = 0 for i = 1, 2, ..., k

[clear vessel]

if x is in W^perp, it is orthogonal to every vector set that spans W, which means its dot prodcut with v1,...,vk is 0... @dusky epoch how is it not the same thing in reverse

[dusky epoch]

it kinda is but you still gotta say it to be rigorous

[clear vessel]

ok but idk how that gets me to W^perp = span(vk+1,...,vn)

like it makes sense why but idk how to write that rigorously

[dusky epoch]

$x \in W^{\perp}$ iff the coefficients of $v_1, v_2, ..., v_k$ in the decomposition of $x$ in your basis are zero

[stoic python]

Ann:

[clear vessel]

but ik that already isn't that what i just proved

[clear vessel]

[dusky epoch]

sorry, i'm in an online class rn

[clear vessel]

oki

[latent marten]

hi can someone help with that

[cyan siren]

Hi @latent marten, I haven't done eigenvectors yet, but I read an article a few days ago about these guys who discovered a new way to compute eigenvectors. It involves removing rows and coloums from the "parent" transformation matrix to find eigenvectors.

It was discovered in 2019, so your syllabus probably doesn't take into account the method

(I don't know much about it at all, but you may want to look into it further)

[torn silo]

@cyan siren bro 🤣

@latent marten you start by looking for the characteristic polynomial

[cyan siren]

what @torn silo

[torn silo]

you should learn the basic methods before searching the internet for new and improved methods

[cyan siren]

I didn't search the internet for new and improved methods, I stumbled across it accidentally : )

and from there looked into eigenvectors

[torn silo]

it's fun topic definitely

[solar osprey]

@dusky epoch Hello, first of all hope you are doing fine and sorry for tagging.

[dusky epoch]

augh. i'm in an online class rn

why are you pinging me in particular

[solar osprey]

You helped me last time

So I was wondering if u can help rn but i didnt know u had a class

sorry

[dusky epoch]

just post your question

[torn silo]

Well both are pretty simple

[solar osprey]

first one we cant know

second one its diagonalizable

is that it ?

[torn silo]

lets start with the first one

[solar osprey]

sure

[torn silo]

what would need to happen so you can diag it

[solar osprey]

each eigenspace has a dimension equal to the multiplicity

[torn silo]

yes but in this we only have dimensions

so we have a 5x5 matrix

which means we need 5 eigenvectors to get a diag matrix

[solar osprey]

yes

[torn silo]

so lambda 1 gives one and lambda 2 gives us 2

we have another eigenvalue

so how many dimensions would that eigenspace need

[solar osprey]

2

[torn silo]

yes

that would be the answer for that one

assuming lamda 3 has an eigenspace with dim 2 you can diag otherwise not

now the b

considering what we just talked about can we diag that Matrix

[solar osprey]

we have 4 eigenvalues

2 of them take up 4 dimensions

we have one dimension left and two eigen values

since dim of an eigen space is at least

is that impossible

[torn silo]

you have lambda 1 and lamba 2

each pop up twice

[solar osprey]

ahhh

each pop up twice

I though that we have 4 distinct eigens

[torn silo]

no no

that wouldn't make sense, but you're right the instructions are misleading

[solar osprey]

thats what happens when a french prof teaches in an american college haha

anws

[torn silo]

ya I would make sure to write it up cleanly

[solar osprey]

so we have 2 eigens that pop up twice

[torn silo]

to communicate that you understand the relationship between eigenspaces eigenvalues and diag matrices

exactly

so we've four eigenvectors

how many do we need?

[solar osprey]

we have four eigen vectors?

[torn silo]

$dim_{V_\lambda_1} = 2$ and $dim_{V_\lambda_2} = 2$

[stoic python]

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

[torn silo]

$dim_{V_\lambda_1}$ is the dimension of your Eigenspace to eigenvalue $\lambda_1$

[stoic python]

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

[solar osprey]

AHH

yes

dimension 2 so it has 2 eigenvectors

[torn silo]

exactly

[solar osprey]

another with dim 2 so that makes 4

[torn silo]

aye

and how many do we need?

[solar osprey]

1 more

[torn silo]

yep

and we don't have that one

so we can't diag the matrix

[solar osprey]

ohhh

I think ill review a little

multiplicities and dimension of eigenspace

I believe this is what confusing me

but I understood the exercise now

thank you

[torn silo]

no worries

that's a good idea

[solar osprey]

also

before I go

for first part

[torn silo]

what does $\lambda$ being an eigenvalue mean

[stoic python]

deekaan:

[solar osprey]

I tried to prove that

the only eigenvalues are 1 and -1

Assume C is an eigen val of A

then the exist a non zero vector v such that

Av = Cv

write A^2 V = A(Av) = C^2v

but A^2 = I

so C^2V = V

wait

so C has to be -1 or 1

[torn silo]

right

because you have c^2 = 1

[solar osprey]

but I am not 100% convinced because I took a detour to answerr the question

I am just wondering if there is a more direct proof

[torn silo]

what detour?

[solar osprey]

I had to find the eigen vals to show that if lambda is an eigen val then 1/lamdba is an eigen val

I didnt show it directly, I am wondering if its possible to derive the result from lambda

[torn silo]

I mean what you did there is a direct proof

[solar osprey]

Oh alright lol

Do you know any determinants I can compute as practice

Or like any 'famous' determinants

Something like Vandermonde's matrix

[torn silo]

http://www.sosmath.com/matrix/determ3/determ3.html

do you mean something like that?

[solar osprey]

Man ur my new life hero

ty

[torn silo]

no worries

[solar osprey]

bro

this is some crazy stuff

I like that

@torn silo Btw btw.

In the first problem I sent u up

You said that they didnt mention multiplicities

[torn silo]

Ya in the a) the algebraic multiplicities aren't clear, the dimension of the eigenspace gives you geometric multiplicities

I didn't express the idea correctly sorry

[solar osprey]

so for the first part

is it correct to say that

1. the multiplicities sum to 5

so this is why we cannot be sure if A is diagonalizable

because we could have multiplicity of Lambda 1 = 2

for instance

[torn silo]

I would make a clear distinction between algebraic and geometrical multiplicities

The issue with the a is you have three eigenvalues a b and c

[solar osprey]

im talking about algebraic, i havent heard of geometric

[torn silo]

geometric is what you're dealing with in this exercise

you're learning that the algebraic multiplicities have to be equal to the geometric multiplicities to diag a matrix

So if you have the eigenvalue 2 times you have the algebraic multiplicity 2 for eigenvalue

now you need to check how many dimensions the eigenspace has (geometric multiplicities)

[solar osprey]

yes

i c

[torn silo]

So in the a you have three eigenvalues

a b and c

for a you know the eigenspace is 1 dimensional

for b you know the eigenspace is 2 dimensional

for c you don't know what's going on (I would ask the teacher or whoever is in charge of that to clarify whats happening there)

now if c has an eigenspace of dimension 2 you're clear and can diag

but if c has only dimension 1, then you're missing a vector to create a diag matrix

because to diag a matrix of size n x n you need n eigenvectors

[solar osprey]

wait

if c has dimension 2 eigenspace

is that really enough?

shouldnt we have multiplcities (algebraic) = dimension of eigenspaces

can we guarantee that from what we have

yes we can

I just thought about it

[torn silo]

good

[solar osprey]

m1 >= dim(lambda 1) >=1

m2 >= 2

m3 >= 2 (if dim(lamda 3) = 2)

sum

u get >= 5

which has to be 5

[torn silo]

but like I said, I would ask the professor to clarify, because right now the question isn't clear and in math precision is kinda important

[solar osprey]

yes i am aware of that

ty

[torn silo]

no worries

[merry jewel]

anyne have a proof for this?

[wintry steppe]

I can verify if 2 lines intersect by checking if the angle between them is between 0 and pi/2 ?

Eh, I don't even have an angle in this case do I?

[dire thunder]

no, angle between lines exists always

but it does not show anything

what you've to do is to show that at some point equations describing them are equal

[severe magnet]

https://cdn.discordapp.com/attachments/501941258046930944/710142671792111656/218824407810834432.png

I wanted to argue that by looking at the linear system of equations Ax=b

which works fine for when Ax has one solution because then $| | Ax-b | | = 0$ for particularly one x

[stoic python]

Nabil:

[severe magnet]

but idk how to argue that if Ax=b has no solution (because that can also be a case)

would anyone know how I could argue that/make a better approach?

[slow scroll]

this looks like least squares. probably think projections

[odd kite]

I think a brute force approach should work. Write out | | Ax-b | | in terms of components and apply 1st and 2nd derivative test, and examine limits as components go to infinity

[slow scroll]

ew

[neat halo]

Is it true that the number of free variables you will get for a system of n independent equations with m unknowns is equal to m-n?

Is this true for polynomial systems as well?

[odd kite]

@neat halo no, only linear systems

[neat halo]

Why?

[odd kite]

@neat halo you can construct a counter-example. For example consider the system z = -x^2, z = x^2+ y^2. It has only one solution at (0, 0, 0). two equations, three unknowns, no free variable

[neat halo]

Idk where to ask this really... These are the equations I'm looking at
https://m.imgur.com/a/QFQf6G0

[odd kite]

what's the goal? find solutions?

[neat halo]

I want an expression for the x with the two dots