linear-algebra

[hard drum]

nah dwww

[west saddle]

in linear algebra

can you do r2-r3->newr2

I dont understand how wolfam did this

[nocturne jewel]

Yes, you can do row operations

[west saddle]

well what im saying is

we so far have only been taught Ra-Rb->Rb

Not Ra-Rb->Ra

and to then go off of that, if I followed the way of Row 3 - Row 2 -> Row 2 it would be -1-1 which would give me -2

And that is not the correct answer, it has to be +2

It still should have been a legal move

[nocturne jewel]

You're subtracting from the a row, so the result replaces the a row

[west saddle]

and in this case the a row is row 3?

[nocturne jewel]

no, you're subtracting row 3 from row 2

[west saddle]

so Ra(0,1,1,0,1) is row 2 and Rb(0,0,1,0,-1) is row 3 and that gives a new Ra

so Ra-Rb-> Ra

The number being replaced is is Row 2

[ocean sequoia]

why would it matter?

[west saddle]

Our notes from the class only mention Ri-Rj->Rj

not Ri-Rj->Ri

[ocean sequoia]

I get that

[west saddle]

This problem is the first way seeing this

[ocean sequoia]

but why cant you do the other way

[west saddle]

it gives -2 instead of positive 2

This problem has a unique solution

So my understanding is -2 is an incorrect answer

[ocean sequoia]

You're subtracting from the a row, so the result replaces the a row

as Mosh said

[west saddle]

I'm understanding I am subtracting from row 3

Wolfram replaces row 2

[silver heath]

Could someone explain this explanation of hte solution better thanks!

[lucid glacier]

The identity only has the eigenvalue 1

[silver heath]

?

[lone quail]

In a diagonal matrix, the eigenvalues are the values on the diagonal

[lucid glacier]

Generally if you have a polynomial p(x), and a matrix A, if c is an eigenvalue of A, then p(c) is an eigenvalue of p(A)

So the eigenvalues of A^4 are lambda^4

Where lambda are the eigenvalues of A

Were A^4 the identity its only eigenvalue would be 1

But it isn't, so it can't be the identity

[silver heath]

right

[lucid glacier]

that's kind of a roundabout way of disproving the claim tbh

[silver heath]

oh ok well i just used the complex diagonalizaiton and raised it to the 4th power xD

[lucid glacier]

That's the same thing pretty much

[lucid glacier]

This is a stronger claim though, since it works even when A isn't diagonalisable

[silver heath]

k

[lone quail]

Btw guys I need some help in finding a mistake in my working. I need to calculate XL given in the formula at the start of this page, and I'm given matrix A. The end result on the solutions is -(1/3)(e^(-3t)-1) on the top right corner of the final matrix, rest is correct. Where have I gone wrong?

[silver heath]

Sorry i cant... exactly read your hand writing

[lone quail]

Is it bad? I can try better

[silver heath]

Let me rephrase

i can but im not exactly sure what you are doing

[lone quail]

Basically I'm diagonalizing A because of the property of exponents of matrices

It makes things easier

[silver heath]

also what is x(a)? or...?

[lone quail]

But for some reason my answer is slightly different compared to solution

X(0) is just a multiplication value, I do nothing with jt

You can just ignore it

[lucid glacier]

Seems like T^-1 is off by a minus sign?

[lone quail]

It should be negative, ny handwriting was just confusing there

If you see under theres negative signs

[lucid glacier]

Ic

[lone quail]

If you are wondering about that property basically it works like this

Where T are the matrices that diagonalize A and Ad is the diagonalised matrix

So basically this saves me the pain of using Taylor expansion with matrices

[lone quail]

Any ideas?

[lucid glacier]

Can't seem to find a mistake

[silver heath]

perhaps it might be because you did P^-1 D P but idk

[lucid glacier]

Huh

[lone quail]

Nah the formula is just the reverse

[silver heath]

ye xD

thats the only difference

[lone quail]

Maybe solution is wrong idk

[lucid glacier]

Actually wait I do think it should be TDT^-1 in this case

[lone quail]

Nah because they are under

They are not on A

[silver heath]

@lone quail what did you do on the last step

[lone quail]

Just multiplied by -1/3

[silver heath]

Oh i see nvm i thought the squeezed in equals sign was an F xD

[lone quail]

Xd

But yeah I mean from the solution every other element in the matrix is correct

[silver heath]

@lone quail it seems you did your matrix multiplication wrong

[lucid glacier]

I just checked on matrixcalc and it seems doing Te^DT^-1 gives the solution in the answers

[lone quail]

Could be

Where?

[lone quail]

Hmm thats same as mine

End result is exactly the same xd

[lucid glacier]

The other way doesn't make sense

[silver heath]

wait

[lucid glacier]

Since T is the COB matrix from the diagonalising basis to the standard basis

[silver heath]

actually

i think you did your diagonalizaiton wrong

because in this case

[lucid glacier]

If you're sandwiching the diagonal matrix

[silver heath]

PDP^-1 != P^- D P

[lucid glacier]

It should be to the left

[lone quail]

Where did I do my mistake?

[lucid glacier]

This

It should be Te^Ad T^-1

[silver heath]

You did the wrong factoriazation

[lucid glacier]

Not the other way around

[silver heath]

^

[lone quail]

Nah im 100% sure thats the correct formula

[silver heath]

Uhh

[lucid glacier]

It's not

[silver heath]

no

[lucid glacier]

Multiply them

[lucid glacier]

See if you get back A

[lone quail]

From textbook

[silver heath]

It says TAT^-1

[lone quail]

Look under

Second line

[lucid glacier]

It depends if you take T to be the COB from the diagonalising basis to the standard one or the other way around

...

[lone quail]

Its that way if you diagonalise A

When exponents come jn its reversed

[lucid glacier]

Fedelisk

Take the diagonal matrix Ad

Multiply T^-1 Ad T

And see if you get back A

[silver heath]

^

[lone quail]

But im not multiplying by Ad here

Thats the issue

Thats the difference

I know the theory

[lucid glacier]

It's not reversed when you do matrix exponents

[lone quail]

It is here is the proof

[lucid glacier]

They're originally taking TAT^-1 = Ad

[lone quail]

You can look at the pr9of

Theres the proof below

[lucid glacier]

In your case it's T^-1 A T = Ad

[lone quail]

Noooo

[lucid glacier]

I'm talking about the very top line

You have the order reversed in the very top line already

So it carries over

[lone quail]

Guys to diagonalose a matrix is TAT-1

[lucid glacier]

It's saying that given
A= T^-1 Ad T
Then the below formula holds.
But in your notation you have
A= T Ad T^-1

[lone quail]

No because at the top Ad is the result

[silver heath]

@lone quail

[lone quail]

????

[lucid glacier]

This part

[lone quail]

Yes

Thats true

[lucid glacier]

In your notation

[silver heath]

Yes that part not given that A = TAT^-1

[lucid glacier]

A = T Ad T^-1

So the signs are accordingly

[lone quail]

Where is this?

[silver heath]

This textbook/lecture nots are trash

[lucid glacier]

The way you did the diagonalisation

[lone quail]

Nope

[lucid glacier]

Yes

[lone quail]

That's not what I did

[lucid glacier]

Ok the multiply T^-1 Ad T and see if you get A. The way you did the diagonalisation you get T Ad T^-1 = A

So the signs are reversed compared to the lecture notes

And that's where the confusion is coming from

[silver heath]

Fedelisk you did the diagonalization for PDP^-1
and used P^-1DP
So you didn't do the factorizaiton correctly

[lucid glacier]

If you don't believe me look up any lecture notes concerning matrix exponentiation

[lone quail]

Yes but thats another formula, the diagonalised matrix in the middle

[lucid glacier]

You are not listening

[lone quail]

THOSE T ARE NOT THE DIAGONALISATION MATRIX OF THE MIDDLE MATRIX I USED

they are the T of a different matrix

[lucid glacier]

Look at this

They are starting the process with the assumption I circled

[lone quail]

Yes

[lucid glacier]

In your case, your factorisation is yhe other way around

[silver heath]

@lone quail if you are asking for help stop being such a bullheaded frog

[lucid glacier]

So the proof will have the signs reversed

That is, the thing you call T, they would call T^-1

[lone quail]

Guys just look at start and end of proof

[lucid glacier]

That is the only difference

[lone quail]

Not really

Why?

[lucid glacier]

I've explained like 5 times by now. It's just the way you did the factorisation

[lone quail]

Im not getting what you mean still

[lucid glacier]

You know ehat a change of basis matrix i

Is?

[silver heath]

essentially you factorized A = PDP^-1 then tried to asume that PDP^-1 was equal to P^-1 D P which it is not

[lone quail]

Where have I assumed that?

That's what im not getting

[lucid glacier]

?

[lone quail]

Roughly why?

[lucid glacier]

It just makes it slightly easier to explain

[silver heath]

@lone quail try computing the factorization you got.... like multiply it out

and see what you get

[lone quail]

Which factorization?

[lucid glacier]

ok so, when you take a matrix and set its vectors to the vectors of some basis B, then that matrix is the change of basis matrix from the basis B into the standard basis E right

so in your case, T is the change of basis matrix from the diagonalising basis B to E

[lone quail]

Yes but what im doing here is T-1 x e^D x T

[lucid glacier]

and T^-1 is the change of basis from E to B

[lucid glacier]

I know

i'm getting to this

[silver heath]

except for hte middle matrix remove the e^blah

[lucid glacier]

I understand where the confusion is coming from

and I'm trying to resolve that

Let me make it clear i'm talking about your specific example

not the textbook

So the matrix A_d is the same as the matrix A represented in the diagonalising basis B

So to get from A_d to A you'd have
A = T * A_d * T^-1

Since you need to go from E to B, then from B to E

[lone quail]

No should be T-1 at start

[lucid glacier]

this is your starting point in your exercise

why

[lone quail]

Because Ad is already diagonalised

[lucid glacier]

Ad is the matrix A in the basis B

[lone quail]

So you need to undo

[lucid glacier]

T^-1 takes us from the standard basis E into the basis B

then T takes us back to E

[lone quail]

Yes but those Ts are not diagonalisation matrices anynire

Just change of base matrices

[lucid glacier]

they are

that

[lone quail]

Nope

[lucid glacier]

's the same thing

diagonalising matrices are change of basis matrices into a diagonalising basis

[lone quail]

Yes, we are not going into a diagonalisation basis here

Its the opposite

[lucid glacier]

no you are

[lone quail]

Ad is diagonalised

[lucid glacier]

T takes you FROM B to E

[lone quail]

A is non diagonalised

[lucid glacier]

so T^-1 takes you from E (your starting point) To B

and then T takes you back

so you end up going from E to E

which is how A is represented

[lone quail]

You agree that Ad = TAT-1?

[lucid glacier]

Absolutely not

it's the wrong way around

if you still don't believe me

try multiplying it out

[lone quail]

Ah I see now

Sec think

Kk makes sense thanks

[lucid glacier]

So what i'm getting at is that in your exercise A = T A_d T^-1, whereas in the textbook we have (Using different notation for clarity) B = P^-1 B_d P. So in the end they do get that e^B = P^-1 e^B_d P, but in your example it'll be e^A = T e^A_d T^-1 because of the same proof

great

glad we broke through in the end

[lone quail]

Yeah

This textbook is eh

Lmao

Anyways thanks for the patiencw

[lucid glacier]

np

[hard drum]

consider the simplest invertible matrices

[lone quail]

I think so but wouldn't know how to prove it. If you think about it tho, the determinant can sort of be imagined as an area in space, of a and b are different then the determinant of a^2-b^2 will not be 0

[hard drum]

(it is not true)

[lone quail]

So they can be not invertible?

[hard drum]

i didn't just wanna give it away lol

but consider 1x1 matrices, for example

[lucid glacier]

There are some cases where it turns out not intervitble

[lone quail]

Ah wait your right because because they can have the same determinant

[silver heath]

oh haha!

[lucid glacier]

fair enough, my counterexample was just taking A and B to be the identity

[hard drum]

that was my instinctive thought too aha

[silver heath]

a=b

[torn hornet]

just take A=B

yeah kek

[lucid glacier]

or yea A=B

[hard drum]

but i wasn't sure how to give that as a hint

so ye

[lucid glacier]

What if they're different tho

👀

[lone quail]

I mean if A and B got the same determinant

The result of the operation is prob 0

Maybe

[hard drum]

for 1x1 yes

not necessarily in general e.g. diagonal matrices with det 1 but different entries to one another

[lucid glacier]

spoiler it's still not true

just take A = -B

What if A^2 \neq B^2 tho 👀

probably still not true just slightly more annoying to come up with a counterexample to

[hard drum]

det(a^2 - b^2) = det(a+b) det(a-b)

so we need a+b or a-b to have det 0 and so that should provide us with examples ig? i'll think

[torn hornet]

just take a= b+c with c being non invertible

[hard drum]

yeah, try something like

[torn hornet]

gives infinite examples

[hard drum]

oh was gonna do the same gr so like ke.g.

diag(1,1) and diag(2,1)

nice

[torn hornet]

yeah

[lone quail]

But this is true as well

[lone quail]

Can you give me an example?

[lucid glacier]

just take say diag(2,0.5) and diag(0.5,2)

[hard drum]

diag(1,1,1) and diag(1,1/2,2)

or that

lol

[torn hornet]

does that actually work, shiN

[hard drum]

sorry but what's the link to this problem?

[lone quail]

Just two random matrices I squared up

[hard drum]

oh fair, sorry yeah that does work as a counter example

[lucid glacier]

I was giving a counterexample to what fedelisk asked. It doesn't work for the original question

[lone quail]

Btw those squared keep determinant 1

[hard drum]

yes they do

[lone quail]

Yeah

[lucid glacier]

yea determinant is multiplicative

[torn hornet]

i see

[hard drum]

if a matrix has determinant 1 then so does its square

[lone quail]

Yeah

[hard drum]

i was just worried as i thought you thought det(a^2 - b^2) = det(a^2) - det(b^2)

[lone quail]

What I was saying in fact if the determinant is different from the start then a squared minus b squared is invertible

[hard drum]

that's not true, as our counterexamples showed

[lone quail]

It didn't?

[hard drum]

e.g. a= diag(2,1) and b = diag(1,1) have different determinants

but a^2 - b^2 has determinant 0

[lone quail]

Ok so if I square a matrix I square the determinant but I can't subtract?

[hard drum]

yeah, det(a + b) is not det(a) + det(b) in general

[lone quail]

I see

[hard drum]

but det(ab) = det(a)det(b) as shin said which is cool

[lone quail]

Yeah

If you think of determinants as areas it does make sense

Theres a nice visualization of 3blue1brown

[opal mortar]

hi guys, anyone can solve 8b ?

[lucid glacier]

over what field?

[opal mortar]

don't know what formula to use. I can't solve it

[opal mortar]

matrix

[lucid glacier]

what field are the entries of the matrix in

real numbers? complex numbers?

[opal mortar]

whatever. The point is that when the matrix is ​​multiplied by 3 times it will produce an identity matrix 3x3

[lucid glacier]

it's not whatever

it actually matters a lot

[hard drum]

i'd assume R then

perhaps a better question is - have matrices always been real so far in the class? do you know what a field is? etc

[opal mortar]

hmm, maybe just real number

[hard drum]

sure

[opal mortar]

so, how to solve it ?

[manic wedge]

[-1 -1 0]
[ 1 0 0]
[-4 24 1]
seems to work (power 3 is indeed the identity matrix)

I don't see how to generate general solutions so I just searched random matrices until I found one with B^3 = I

I can see they have to satisfy B^2 + B + I = 0 for example (multiply by B - I to see that) but I wouldn't know anything useful

Maybe try to restrict B like a 2x2 matrix with B =
[1 0 0]
[0 a b]
[0 c d]
and search for 2x2 Matrices B' = [a b // c d] with B'^3 = I because they should get you B automatically with the condition

[ 0 -1]
[ 1 -1]
should do the trick as B' (again random matrices)

[opal mortar]

wow I've tried multiplying this one. and the result is true that B^3= I

[manic wedge]

sage: M = MatrixSpace(ZZ, 3)
sage: M
Full MatrixSpace of 3 by 3 dense matrices over Integer Ring
sage: M.identity_matrix()
[1 0 0]
[0 1 0]
[0 0 1]
sage: I = M.identity_matrix()
sage: while True:
....:     i = M.random_element()
....:     if i != I and i^3 == I:
....:         print(i)
....:         break
....:
[-1 -1  0]
[ 1  0  0]
[-4 24  1]
sage: a = Matrix([[-1, -1, 0], [1, 0, 0], [-4, 24, 1]])
sage: a^3
[1 0 0]
[0 1 0]
[0 0 1]
sage: M = MatrixSpace(ZZ, 2)
sage: I = M.identity_matrix()
sage: while True:
....:     i = M.random_element()
....:     if i != I and i^3 == I:
....:         print(i)
....:         break
....:
[ 0 -1]
[ 1 -1]
sage: a = Matrix([[1,0,0], [0,0,-1], [0,1,-1]])
sage: a^3
[1 0 0]
[0 1 0]
[0 0 1]

say thank you to SageMath I didn't do shit

[opal mortar]

right.. thank you so much SageMath and T0lgi

[manic wedge]

SageMath is a Program not a person

[opal mortar]

ok,,

is it website ? or app ?

[manic wedge]

They do have a website where I downloaded it on my Computer (it's about 5GB in total, though) and actually an App on iOS (and probably other) making use of server computation (so you need WiFi)

Try to minimize usage from external programs because trying to find solutions by hand is also exercise by itself

[vapid gust]

sorry I just need confirmation for an extremely quick question but

https://i.imgur.com/SJPvdE9.png this is incorrect, right?

(it wants me to get the Cartesian form a + bi)

im getting the answer 31/34 - (39/34)*i

[hard drum]

Yeah it seems p wrong

(for future reference, wrong channel sorry, a question channel might be better :p)

[vapid gust]

ah, okay

sorry about that

and thank you!

[hard drum]

no worriesss

[nocturne jewel]

Technically it's fine to ask about C in here, at least my first formal learning of C was in a LinAl course when we "rigourously" defined scalars

[runic agate]

Companion matrix of x^3 - 1?

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

[stoic python]

Mike Desgrottes

[limber sierra]

yep thats the natural linear algebra way to view this

alternatively you can look at it "algebraically" by looking for the matrix representation of a permutation in S_3 of order 3

which would give the same result

and i think is probably the best way to approach this

(if youre not familiar with the group terms, i just mean we're looking for a permutation ["swapping"] operation that "cycles back to the start" when applied 3 times)

(and then representing that with a matrix)

@opal mortar pinging in case this adds to your understanding, though your current answer is fine as well.

[rancid sonnet]

So the gradient of f transposed is a row vector right?

When I do it for a system of equations then, the n'th row of J should be the gradient of f_n?

But J is the Jacobian

And the Jacobian has the n'th column as the gradient of f_n

Should that be J transposed in C.11?

[lavish jewel]

definitions of these quantities are not standard and depend on the book

transpose stuff as needed

if f(x) is scalar-valued, then grad(f(x)) as written there is a column vector, and you transpose it to get a dot product

so that the first order taylor approx is also a scalar quantity

it seems like the f_i(x) are the entries of a vector, let's call it F(x) (seems that's what they did, too)

so now if you have a vector F(x) and you take its derivative w.r.t. x, the derivatives are along the rows

so J is a matrix where each row contains the derivative of the corresponding f_i(x) w.r.t. x

seems your book follows the convention of "all vectors are column vectors" and "differentiation w.r.t. vector quantities go in the next available 'way' "

[hard drum]

Oh rly fair lol, interesting

[sinful knoll]

can someone help me how to simplify this

[dusky epoch]

how much do you know about determinants?

[nocturne jewel]

When the prof wants to make sure you know the properties

[dusky epoch]

@sinful knoll ???

[sinful knoll]

I do know there are properities but none by heart :(\

[dusky epoch]

det(MN) = det(M) det(N)

sound familiar?

[sinful knoll]

oh yeah

[dusky epoch]

and also det(A^T) = det(A)

[wintry steppe]

i used wolfram alpha to calculate the SVD of this matrix

but, i dont understand how the last column of the U matrix is calculated?

[lavish jewel]

wdym by "how"

do you wanna know about the algorithm for computing SVDs? otherwise just get a 4th vector orthogonal to the previous 3

[wintry steppe]

yeah i dont really understand this svd stuff right now but, basically, you can compute an extra singular vector beyond whats in V and just use that?

[lavish jewel]

what

[wintry steppe]

lol

like

in V we have 3 singular vectors, right?

[lavish jewel]

U and V are, in general, not related at all to each other

matrices have left and right singular vectors

those depend on the original matrix's column space and left null space (left sing. vecs.) and row and null space (right sing. vecs.)

[wintry steppe]

ok well

i am clearly way out of my depth here and fundamentally misunderstand all this

but just what i was reading said, to calculate each column in U, we multiply each singular vector in V with the original matrix A

which does indeed work looking at the wolfram output, except for the fact that the final column in U doesnt seem to have a corresponding singular vector to calculate it from, which is why i was confused

but yeah clearly i am totally misunderstanding this on some base level so i will try to read into it

[lavish jewel]

well, since V is orthonormal, and a matrix M is decomposed into M = USV^H, when you multiply by the singular vectors in V (i.e. the right singular vectors), you get USV^HV = US

so the result is a matrix U times a rectangular matrix S with entries only along its main diag.

i think the best you can get this way is the set of vectors U_c that span the column space of the matrix

not the ones for the left null space

you can extend the vectors U_c into an orthonormal basis for R^m or C^m using gram schmidt tho

just generate random vectors of the correct size, check they're all linearly independent when put in a set with the vectors in U_c, and put the vectors U_c first when you do gram schmidt so that you modify the randomly generated vectors instead of the ones you already know are correct

[wintry steppe]

ok thank you, i will look into it

[lavish jewel]

what you were doing is related to the so-called "economy-sized SVD"

[solemn lotus]

will $S ={x(a_1, a_2) + y(a_3, a_4) \mid x,y,a_i \in \bR}$ always be a subspace of $\bR^2$ if $(a_1,a_2)$ and $(a_3, a_4)$ are non parallel vectors?

[stoic python]

CoolShot

[dusky epoch]

your notation needs fixing right now

you do not want to quantify over all possible values of a_i in your set

as-is, what you have is an overly elaborate description of R^2

if you meant $S = { x(a_1, a_2) + y(a_3, a_4) \mid x, y \in \bR }$, with $a_i$ arbitrary but fixed, this will \textbf{always} be a subspace of $\bR^2$ even if $(a_1, a_2)$ and $(a_3, a_4)$ happen to be parallel.

[stoic python]

Ann

[solemn lotus]

nvm youre right i confused myself
even if they were parallel it'd be closed under vector addition and scalar multiplication and have a zero vector
so itd be a subspace

[nocturne jewel]

yeah, S is just the span of those vectors

and span is always a subspace

[solemn lotus]

yup
ty

[hard drum]

You can view this as essentially taking (a1,a2) and (a3,a4) and forcing S to be a subspace

'oh, if u and v in the set then so must xu + yv for all x,y in R? well let's just chuck them all in'

[silver heath]

Numer d is not a vector space right?

because if u add x^3 with -x^3 = get somethign thats not in teh set?

[dusky epoch]

yes

i don't think so

[limber sierra]

the only sensible candidate fields are Z/pZ, but those end up failing distributivity

so no

(this isnt the most formal argument; for a more formal one, see https://math.stackexchange.com/questions/151850/prove-mathbbz-is-not-a-vector-space-over-a-field)

another simple argument: observe that Z would necessarily be a 1-dimensional vector space since Z is generated additively by (1)

but a 1-dimensional vector space would necessarily have "itself" as a base field (upon fixing a basis)

except Z isnt a field

[hard drum]

$A = \begin{pmatrix} 1 & 1 &\ 0 & 1 \end{pmatrix}$ is not, for example

[stoic python]

potato

[hard drum]

only eigenvalue is 1 and if Ax = x then we must have x being a multiple of (1,0) i.e. the eigenspace is 1 dimensional so not diagonalisable

There are nice types of matrices you can prove you can diagonalise (stuff like normal/unitary/real symmetric matrices) though

[wintry steppe]

it is a crime that no one has mentioned jordan canonical form yet

[ocean sequoia]

this is also called a defective matrix and something worth knowing about symmetric matrcies is that they have orthognoal eigenvectors

[lavish jewel]

canonical tteppa

[zealous junco]

Yea think so

So like if b=0 just show for any f,g in space then cf+g also integrate to 0, and if b not 0 u can find counterexample by taking day 2*f

Say*

[safe creek]

is P the price vector?

in the exercise they asked me to determine the price vector with a system

but I wonder, that vector p that I have there is the vector of prices?

[unique stump]

if you have no pivot columns that means all ur vars are free
so wouldnt that make it so that all b values would have solutions

[vivid atlas]

What is meant by saying "a set of functions"?

[nocturne jewel]

What it sounds like

a set, containing functions

[vivid atlas]

different functions

[wintry steppe]

[vivid atlas]

wiki "In mathematics, a function space is a set of functions between two fixed sets"

[wintry steppe]

i'd be careful using the term "space" here since that usually refers to a set of functions with some extra structure (such as that of a vector space)

[nocturne jewel]

yeah... it's what it sounds like

the set of functions b/w 2 sets

[vivid atlas]

so I'm reading about vector spaces "If S is a set, then F^S denotes the set of functions from S to F" but I just have a question in my mind, what function?

[nocturne jewel]

Any

[lucid glacier]

all of them

[nocturne jewel]

Like, let's say we have 2 vector spaces V and W, then we can say the set of functions $W^V$ would be all functions who map vectors from V to vectors in W

[blissful vault]

Is this correct so far? At this point, can I do an epsilon delta proof? I'm not sure if it is the same for functions with complex domain.

[stoic python]

Mosh

[lucid glacier]

what's the topology on V?

is it a normed space?

[wintry steppe]

pick any norm, it's finite dimensional so they all give the same topology

[lucid glacier]

yea I know, it

s just weird they didn't specify

wait actually you don't need a topology on V here nvm

I misread the problem

you can do smth like an epsilon delta proof but just using open balls in C instead of open intervals I think

[wintry steppe]

just handwave and say "function jumps at an eigenvalue so it's discontinuous there"

if it's a linear algebra class i doubt you have to give the rigorous analytical details

[lucid glacier]

maybe add that there are finitely many eigenvalues so there is a nbhd of lambda_0 in which it is the only eigenvalue

so clearly there is a discontinuity at that point

[blissful vault]

what is nbhd?

[lucid glacier]

neighborhood

[blissful vault]

Ah, thanks. So I can find the delta neighborhood in my discontinuity proof.

[blissful vault]

Just self-teaching rn

[lucid glacier]

yea, so it's constant at all points in that nbhd except lambda

where it jumps

actually if you're disproving continuity you don't actually need that

you want to show that there exists some $\varepsilon>0$, such that for all $\delta>0$, there is a point $x$ in the delta-nbhd of $\lambda$ such that $\abs{x-\lambda}<\delta$ but $\abs{f(x)-f(\lambda)}\geq \varepsilon$

[blissful vault]

ah true.

[stoic python]

ShiN

[lucid glacier]

where the first absolute value is complex

[blissful vault]

Does this check out at the end?

[lucid glacier]

seems good

[blissful vault]

thanks for the help

[lucid glacier]

np

[ionic laurel]

just started my intro to linear algebra class... what does this question mean?

do i leave in REF or RREF?

[nocturne jewel]

Continue row reducing until you can read off the solution set

[ionic laurel]

So i know this is an inconsistent solution set

[nocturne jewel]

for example, 7 you dont have to do anything else cause you can read off that there is no solution

[ionic laurel]

Yeah

If i swap its inconsistent

So then ig i dont have to do anything afterwards

[nocturne jewel]

Nope, however for 8 there is still work to be done

[ionic laurel]

Oh ok

[nocturne jewel]

cause you can easily tell the system will row reduce to $[I|b]$

[stoic python]

Mosh

[nocturne jewel]

so it will have a unique solution

[ionic laurel]

Yeah i didn’t learn that yet

I believe

[ionic laurel]

@nocturne jewel

the solution to 9 is unique?

how do i figure that out exactly

[ionic laurel]

If i put this into desmos I get that there is a common point of intersection

But my reduction work says that there are infinitely many solutions

what am i doing wrong?

[nocturne jewel]

sorry didnt see the ping: You further row reduce like I said

[ionic laurel]

does this not mean infinite many solutions

[nocturne jewel]

it doesnt

[ionic laurel]

because of the 0 row

what

[nocturne jewel]

you have x_1=-13/7, x_2=-5/7

then 0=0

[ionic laurel]

youtube.com/watch?v=18_oG-cD9Ck

[nocturne jewel]

There's no free variables in that augmented matrix

all the variables are pivots

[ionic laurel]

oh

this video said that a zero row meant that it was infinite many solutions

hmm

[nocturne jewel]

well you didnt link a video

[ionic laurel]

it isn't linking for some reason

i posted it

[nocturne jewel]

yeah you missed the https://

www.youtube.com/watch?v=18_oG-cD9Ck

[ionic laurel]

copy pasting from mac doesnt show up

https://youtube.com/watch?v=18_oG-cD9Ck

there we go

i mean even from the thumbnail

[nocturne jewel]

yeah, that's different

[ionic laurel]

oh ok

[nocturne jewel]

cause one of the columns in that example is free

[ionic laurel]

alright i got it

ill review free and pivot variables then

thank you mosh

[earnest ferry]

how can I prove u+v belongs to W for u,v belongs to W.

[dusky epoch]

ask yourself: is the sum of two singular matrices necessarily singular?

the answer should become obvious if you consider something like the identity matrix and how it can be written as a sum.

[earnest ferry]

thanks

[wintry steppe]

sup

[wintry steppe]

Say I have 2 matrices, A and B, is there a way to find every possible result of matrix multiplication of A and B (ie. ABB, ABBAB, AAABBA, etc.) assuming that there's a finite set of results?

[nocturne jewel]

Gonna say no, cause there are an uncountable number of binary strings

[coarse rain]

There's a countable amount of finite binary strings though. The only case I can think of a finite number of possible products happening tho is when the eigenvalues are 0 and 1.

[tranquil steeple]

or -1

[lucid glacier]

So, I want to calculate the determinant of the matrix
$$\begin{pmatrix}\vert & \ldots & 1 \ e_1 + e_n & \ldots & 1 \ \vert & \ldots & n-2\end{pmatrix}$$
For arbitrary $n$ ($e_i$ are the standard basis vectors). I suspect it always comes out to -1 but i'm not immediately seeing an inductive argument for it

[stoic python]

ShiN

[lucid glacier]

also eigenvalues seem to be a pain

any ideas?

the n-1 first vectors are $e_i+e_n$ and the last vector is all 1s except the last coordinate which is n-2

[stoic python]

ShiN

[lucid glacier]

Here's n=8 if that helps visualise it

I think maybe splitting it into the diagonal part and the part with only the row and column of 1s might make it easier? then you could use induction on that part

[lavish jewel]

does it help to rewrite the last column as the sum of all the previous ones minus e_n?

[lucid glacier]

🤔

Don't see exactly how that helps me

[lavish jewel]

me neither, i was just throwing the idea around

[lucid glacier]

Ah ok

[lavish jewel]

maybe rules of how the determinant of a matrix changes when you do column operations, for example

idk

[lucid glacier]

That might help

I hate determinants

[lavish jewel]

😌

[lucid glacier]

ok so actually using multilinearity it reduces to the last row just being -e_n

and now the matrix is lower triangular

[lavish jewel]

aha

[lucid glacier]

so the determinant is the diagonal

so it becomes -1

nice

[lavish jewel]

nice

[lucid glacier]

thanks

[tranquil steeple]

the eigenvalues are always 1 with multiplicity $n-2$ and the last 2 eigenvalues are $((n-1)\pm \sqrt{(n-1)^2+4})/2$

[stoic python]

Sven-Erik

[lucid glacier]

matrixcalc gave me a different eigendecomposition for n=6

not really important tho

[tranquil steeple]

julia> A
6×6 Matrix{Float64}:
 1.0  0.0  0.0  0.0  0.0  1.0
 0.0  1.0  0.0  0.0  0.0  1.0
 0.0  0.0  1.0  0.0  0.0  1.0
 0.0  0.0  0.0  1.0  0.0  1.0
 0.0  0.0  0.0  0.0  1.0  1.0
 1.0  1.0  1.0  1.0  1.0  4.0

julia> eigvals(A)
6-element Vector{Float64}:
 -0.19258240356725248
  1.0
  1.0
  1.0
  1.0
  5.192582403567252

julia> [((n-1)+sqrt((n-1)^2+4))/2, ((n-1)-sqrt((n-1)^2+4))/2]
2-element Vector{Float64}:
  5.192582403567252
 -0.19258240356725187

[lucid glacier]

oh wait the matrix I entered in matrixcalc was slightly different

by a column operation

not sure if that would affect the EVs

[tranquil steeple]

All EV can also be expressed explicitly

[plain wolf]

anyone know a visual learning instructor for intro to linear and differential equations. its a struggle, keep running into instructors that have a vibe of "dont ask questions, do it yourself". i think my only option is taking online and self teach.

[umbral yew]

you could check out the "Essence of linear Algebra" Playlist on youtube

has a lot of good visuals

[steep stratus]

Here to verify a question, it looks right to me but im not 100% sure ```
Given any nonzero vectors u and v, determine the scalar c so that the vector u + cv is perpendicular to v.

v * (u + cv) = 0
(v1,v2) * (u1 + c * v1, u2 + c * v2) = 0
v1(u1 + c * v1) + v2(u2 + c * v2) = 0
c(v1^2 + v2^2) + v1 * u1 + v2 * u2 = 0
c = -v1u1 - v2u2 / (v1^2 + v2^2)

If v and u were in 3 dimensions (question does not specify)
would
c = -v1u1 - v2u2 - v3*u3 / (v1^2 + v2^2 + v3^2)
be correct?

thx for taking the time to read this

[nocturne jewel]

$c=\frac{-(v\cdot u)}{\norm{v}^2}$

[wintry steppe]

That looks right

[stoic python]

Mosh

[nocturne jewel]

yes

[wintry steppe]

yes

you can simplify it however

you don't need to write them as (v1,v2)

$v\cdot (u+cv)=0 \implies v\cdot u + v\cdot cv=0 \implies v\cdot u = -||v||^2c \implies \frac{-(v\cdot u)}{||v||^2}$

[nocturne jewel]

v.(cv) in the 2nd line

[stoic python]

jswatj

[dusky epoch]

\| for the norm double-bars

it looks a little nicer that way

$||v||$ vs $|v|$

[stoic python]

Ann

$||v||$ vs $\|v\|$

[wintry steppe]

oh, i had no idea

[dusky epoch]

you can also do \Vert (capitalization matters)

[wintry steppe]

what do you use for vector arrows?

like over the vector

[nocturne jewel]

\vec{}

[wintry steppe]

nice

[hard drum]

\bf

jk

[wintry steppe]

vector arrows

[plain wolf]

any recommendations for instructors at a community college?

[steep stratus]

alright thx, i just read this, ill write it in the simpler form

[steep stratus]

need help on a particular problem ... let me type out how far ive gone so far

so for 5a i would just do like

x1 + x3 = b1
-x1 + 2*x2 = b2
2*x1 + 3*x3 = b3

right?

for 5b am i bit confused. by solve it do they just mean put x1 x2 and x3 in terms of b? if they meant that then why doesnt it say so? would x3 = -b3 + 2*b1 be correct if this is the case

and for 5c i was gonna guess 
 3   0   -1
3/2 1/2 -1/2
 2   0   -1

based off my answers from b.
is that right?
thanks

[ionic laurel]

can a matrix with infinitely many solutions be consistent?

how does that work

i am confused

[ionic laurel]

How do I approach this problem?

[nocturne jewel]

find the temperatures at 1,2,3 and 4

then solve the system

for example, $4T_1=10+20+T_2+T_4$

[stoic python]

Mosh

[nocturne jewel]

which comes from examining node 1

[ionic laurel]

okay

[nocturne jewel]

do that for all 4 nodes, and you'll have 4 equations in 4 variables

[ionic laurel]

each temperature does not need to have a coefficient though right

or does it not matter

[nocturne jewel]

Yes, some can be 0

the one the text uses as an example has +0T_3

[ionic laurel]

yeah

there is no T3

[nocturne jewel]

There is, it just doesnt affect T_1

[ionic laurel]

oh

gotcha

so we just write it as 0T_3?

Or we can ignore it

[nocturne jewel]

It will help when writing the augmented matrix, but for 43 it doesnt matter

$4T_1-T_2+0T_3-T_4=30$ makes it clear the 1st row of $[A|b]$ will be $[4,-1,0,-1|30]$

[stoic python]

Mosh

[ionic laurel]

oh ok

@nocturne jewel correct me if im wrong but

T_2 would be (20 + 40 + T_1 + T_3) / 4 correct?

[nocturne jewel]

no

4T_2 will be that

or that

[ionic laurel]

yeah sorry

forgot the average

i was just having difficulty understanding how to read the diagram

thank you though for your explanation

this isn't too hard

[wintry steppe]

for 5b you can just put it in an augmented matrix and row reduce it accordingly

actually for 5c just write it out like they want u dont need to invert it

then pull out b and get the coeffecients for A^-1

i think what you have is right

if you row reduced properly

[frosty moth]

Can anyone explain how points P and Q end up as a vector?

[lavish jewel]

one can use vectors to indicate positions, so-called position vectors

these vectors have tail at the prigin and head at the point

you see they took the coprdinates of p and q and directly put them into vectors. this is the same as doing e.g. p - (0,0)

coordinates and pos. vectors behave similarly to each other and the distinction is slight.

[frosty moth]

ok, i'd initially plotted them as follows. i've read that they don't have to start from 0 on a cartesian plane

[lavish jewel]

tje what doesnt start at 0?

[frosty moth]

the vector

hmm let me try to redraw the above using vectors instead of points

[lavish jewel]

vectors dont have a location

you can assign a location to them if it helps you, but this is extra info

in this sense, you need to express the vector in terms of both the head and the tail

once you subtravt the 2, you habe a single vector with no location. you could place it anywhere on your plane if youd like

though it is customary to again set ita tail at the origin

[frosty moth]

how would one draw them graphically?

I got this... but doesn't seem to make sense cos the - signs are no longer there

This seems to conform to the answers

Q - P = Q + (- P) = - (Q - P)

[lavish jewel]

youd need to look up how that plotter's vector() function works

cuz you see changing the order of P,Q changes the direction the vector points in

[lavish jewel]

is this an exam? it seems to have points assigned to it

[wintry steppe]

bout to ace his exam 🙏

[wild fulcrum]

with discord help

[earnest ferry]

its a question from a sample question paper

[lavish jewel]

aight

well, for it to be a linear transformation, some properties need to apply

[earnest ferry]

what about the geometric discription

[lavish jewel]

two things are happening

what does R x do to x?

what does Rx do to x?

[earnest ferry]

rotate x by an angle theta

range will be the sum of the vector and the rotated vector, ryt?

[lavish jewel]

mhm. and well, that's a linear operation. now say the result is called y = Rx. what does x_0 + y do to y?

[earnest ferry]

?

y becomes the sum vector of x0 and y

[lavish jewel]

what does that mean geometrically

[earnest ferry]

no idea

it will lie on the plane as that of x0 and y?

[lavish jewel]

let's see it this way. Rx is some random vector. let's say we fix its length to 1. we still don't know its angle though, so for practical purposes, we can depict Rx as a circle with center at the origin.

what happens if we now add x_0 to Rx?

what happens if you add x_0 to every single point on a circle?

[earnest ferry]

radius of circle changes?

[lavish jewel]

try it yourself on a piece of paper. say the circle passes thru (0,1), (1,0), (-1,0), (0,-1), which a circle with radius one and center at the origin does

all of those points can be represented with position vectors

those are possible results of Rx

now add a vector x_0 = [2,2] to all of them

and draw the circle that passes through those points

what do you get

[earnest ferry]

displaces the origin?

[lavish jewel]

bingo

so you can describe this as first rotating the vector x and then displacing it

[earnest ferry]

got it

[lavish jewel]

now you can test the linearity yourself, but as a direct consequence of the displacement, 0 is no longer mapped to 0

this should already tell you if it's linear or not

alternatively, see whether T(v + w) = T(v) + T(w), which is a property that must hold if T is linear

[earnest ferry]

thanks for the help

[wild fulcrum]

or T(cv) compared with cT(v)

usually translation isn't linear

[lavish jewel]

that's the take-home message

[true umbra]

Can I ask a question here or is that not allowed?

[wild fulcrum]

[true umbra]

It's about vectors and the problem is:

I know what to do if vectors c + d is a single vector and it goes from the initial point of vector A to the terminal point of vector B, but I'm unsure how to approach this problem because that isn't the case

[wild fulcrum]

you know that for c+d

what about -c + d

notice how -c is opposite of c

[true umbra]

Yes, that'll make the same vector but just pointing the opposite direction

[wild fulcrum]

yeah so now you can draw the vector -c + d onto that graph

[true umbra]

Shouldn't -d be better than -c?

[wild fulcrum]

yeah sure

so now you have relation among 2a, b and c-d

[lavish jewel]

looking at the directions of the arrows

cant you straight up say c -2a = b + d

and simplify that

it's the same as ehat vee says, but without modifying any of the given vectors in the plot

[true umbra]

Ahh. I was looking at the question wrong, I don't know why I kept thinking it was a=c-d-b

I understand now, since a has a scalar value of 2, and the question is looking for a by itself, we need to divide by 2.

How can you draw that conclusion that you can straight up say c-2a=b+d, can you elaborate a little bit? You're right but I just want to understand how you were able to make that deduction right away

[wild fulcrum]

you draw a line from where 2a and b meet to where c and d meet

or you imagine that line

follow -2a + c is the same as follow b+d

i feel like we imagine these differently though, so this just how i see it

[lethal pivot]

alternatively, you could have also done: a + a + b + d = c by the process of vector addition (resultant vector is formed by connecting tail of first vector to the head of the last vector)

[wild fulcrum]

if you follow that

oh

nvm

i'm trippin

but yeah

[lethal pivot]

either way, u arrive at the same answer

[wild fulcrum]

many ways to imagine these

[true umbra]

I see

[lethal pivot]

either way, u arrive at the same answer

[true umbra]

Well, thank you all for the generous help

[lavish jewel]

as they said

vector addition sets vectors one after the other. the head of one is the tail of the other

subtraction instead joins the heads of two vectors, and the head of the resulting vector follows the classic vector = head - tail principle

[versed forge]

how do i get rid of the bracket inverses in this (matrix btw)

[wintry steppe]

just take the inverse of both sides

[versed forge]

so it becomes

NAinverse + C inverse = Binverse D

[wintry steppe]

how?

[versed forge]

by inversing both sides?

[wintry steppe]

both sides of the equation?

im not sure how A^{-1} or C^{-1} showed up

B^{-1}D on the right side is correct

when you take the inverse on the left it just cancels out the inverse already there

[versed forge]

so i

dont have to change anything

for the left side??

[wintry steppe]

you'd get N^{-1}A + C

[versed forge]

oh okok

i think i was confused

[lavish jewel]

you can also multiply both sides by the inverse of the LHS and see if that helps in any way

[wintry steppe]

multiply both sides by tteppa

[lavish jewel]

[wintry steppe]

[glass ridge]

guys, im confusing with definition of dependent variables. can someone help me if its right or not?

if there is y that dependent on linear 3 variables (in this case with noted as x0, x1, x2) and some constant, is it posible i write this as below?
y = x0 + x1 + x2 + C

[wild fulcrum]

what was the definition given to you?

[glass ridge]

well, i was thinking on implementation (the question is just made by myself) on some data. there are 3 variables that plotted with 1 output varible, and all of them is shown graphicly linear

and statement above is my conclusion.

[dusky epoch]

so you want to do a linear regression?

[glass ridge]

yes

[lavish jewel]

you're close, but forgot possible scalars for the x_i

[dusky epoch]

$y = \beta_0 x_0 + \beta_1 x_1 + \beta_2 x_2 + c$ i guess

[stoic python]

Ann

[wild fulcrum]

is what they asked even linear regression lol

i'm so confused

[zealous junco]

How do u show the (20)

[zealous junco]

oh nvm its just schur complement

[lavish jewel]

you can usually take all the vector equations and just stack all the vectors on top of each other

you could build it that way

dual problems are usually formulated that way

you get a new problem with different constraints, and then make a huge inefficient matrix that you would never actually code

but looks nice on paper

[zealous junco]

ok, though how does that relate to (20)?

hm actually lemme think abt it

[lavish jewel]

R is presumably a correlation matrix, yeah?

hermitian positive (semi) definite

you can factor it

yada yada

B^H B

[zealous junco]

yea

[sonic beacon]

what does the P_m (F) = span(1,z,..,z^m) notation mean here

and what is this saying:

[lavish jewel]

you know what the span of a set of vectors is?

[sonic beacon]

yes

it's the set of all linear combination of a set of vectors

[lavish jewel]

yep

the vectors are 1, z, z^2, ..., z^m

for example, you can construct any order 1 polynomial by considering linear combinations of 1 and z

[sonic beacon]

i am confused because i dont know what it means for 1, z, z^2, .., z^m to be vectors, their variables in a polynomial expression with degree m

[lavish jewel]

vectors in a vector space can be anything that satisfies the definition 😛

[lavish jewel]

look at this example

[sonic beacon]

can i see an example , yah

[lavish jewel]

you know what a linear combination is, yeah?

[sonic beacon]

yea

[lavish jewel]

so let's say we have the vectors 1 and z

[sonic beacon]

ok

[lavish jewel]

any polynomial of order 1 is of the form az + b

which is the same as az + b*1

[sonic beacon]

order here means degree, yeah?

[lavish jewel]

that is a linear combination of the vectors 1 and z

yes

[sonic beacon]

ah, i see

[nocturne jewel]

(where a and b are from F)

[lavish jewel]

yep

you could also build an isomorphism to R^{m+1} if it helps you

some sort of correspondence between these monomials and canonical basis vectors

e.g.

for quadratic polys

[sonic beacon]

so it's saying the set of polynomial functions with degree up to m that map F -> F can be expressed as span(1,z,..,z^m) ?

[lavish jewel]

associate 1 to [1,0,0]

associate z to [0,1,0], and so on

[nocturne jewel]

yes, cause polynomials are just linear combinations of monomials

[lavish jewel]

yep

[sonic beacon]

ok i see

!

thank you

[lavish jewel]

just open up your mind to what "vector" means

[sonic beacon]

ok that helps

[lavish jewel]

review the definition of vector spaces and subspaces and notice it's just a set of basic rules

[nocturne jewel]

vectors are just things that can have a notion of "adding" and "scaling"

[lavish jewel]

literally anything that satisfies those rules is a "vector" in a very abstract sense

[sonic beacon]

i am used to think of vectors strictly as F^n as oppose to any thing that satisfies the vector space axioms

[lavish jewel]

that tends to be the first example one sees, indeed

[sonic beacon]

but that makes a lot more sense

[sonic beacon]

can you help break down what this is saying as well

is it saying polynomial functions are uniquely determined by their coefficients? i dont understand

what it is saying

[nocturne jewel]

yes

[sonic beacon]

ok i see

[nocturne jewel]

Basically if I have $p(t)=\sum_{i=0}^{n-1}a_it^i$ then I can make an isomorphic map which maps $p(t)$ to $$[a_0,a_1,...,a_{n-1}]^T$$

[stoic python]

Mosh

[sonic beacon]

so if you have: a1 + a2 z^2 + ... + an z^m and b1 + b2 z^2 + ... + bn z^m, then in order for the two polynimials to be equal then a_i = b_is have the be equal?

[nocturne jewel]

yes

[sonic beacon]

awesome, thank you

[nocturne jewel]

if p(z)=q(z), then constants must match, linear terms must match, etc

[sonic beacon]

ok, that makes a lot more sense now thank you

that answers all my questions for now, thanks for all of the help

[zealous junco]

ok, so i think this is saying y, R are fixed, and saying that if t is varying, then min t = y^HR^-1 y must hold for the block to be PSD

maybe im not used to it but was hard to parse what the statment is saying...

[lavish jewel]

i would need to read all of the context to know. idk what the goal is. estimate y? estimate R? get t?

[zealous junco]

yea, i think the context agrees

theorem is this

they basically break Y, which is fixed matrix into columns y_t

[zealous junco]

now, in here, why can't we have y^H R^-1 y < min|x|^2

[lavish jewel]

substitute R = AA^H into y^H R^-1 y

and also Ax = y

[zealous junco]

oh right thx

A has full column rank

nvm

[lavish jewel]

that's ideal if true

if not, the previous images show tikhonov reg

[zealous junco]

o no im being omega dumb

forgot A^+ y = x

[lavish jewel]

you say that, but what you get is the minimum 2-norm solution for x

depending on how A looks, there might be no or infinitely many such x

[zealous junco]

true, this is only supposing there is some x?

[lavish jewel]

they used tikhonov so i guess they want a smooth one

[zealous junco]

hm thnaks ill look it up

[lavish jewel]

that R + \sigma I stuff

[zealous junco]

I think its ok to assume A full column rank here and theres a solution, so all is good

since the context is T(u) is toeplitz with decomposition A^HD^1/2 D^1/2 A

and applying there, y is in the span of A

[lucid pasture]

Heya all, super sorry if I’m in the wrong channel here; it wasn’t a specific homework problem or anything and I think it has to do with la so I decided to put it here but does anyone know what this is?

I searched up Vandermonde Determinant and just saw a bunch of matrices?

Not sure how it would be calculated in this context,,,

[manic wedge]

do you know the general form of a vandermonde matrix? And do you know what a determinant is?

[manic wedge]

The general form of it is
$$ M =
\begin{pmatrix}
1 & a_1 & a_1^2 & \dots & a_1^n \
1 & a_2 & a_2^2 & \dots & a_2^n \
\vdots & \vdots & \vdots & \vdots & \vdots \
1 & a_n & a_n^2 & \dots & a_n^n \
\end{pmatrix}
$$
The usual time complexity of calculating a Determinant of a Matrix is usually $O(n^3)$, basically a polynomial of degree 3 with respect to the number of rows/columns of the Matrix.
The Vandermonde Matrix is a special form of Matrix that comes up in other areas in math and Vandermonde found a trick to calculate the Determinant of this certain form of a Matrix (which is $O(n^2)$). It's not that complicated as we got this as a homework problem but I'll spoil the result here
$$
\det M = \prod_{1 \leq i < j \leq n} (a_j-a_i)
$$
So for example, the Matrix
$$ M =
\begin{pmatrix}
1 & 1 & 1 \
1 & 2 & 4 \
1 & 4 & 16 \
\end{pmatrix}
$$
(why is this in the vandermonde form?) has the Determinant
$$\det M = \prod_{1 \leq i < j \leq 3} (a_j-a_i) = (4-2)(4-1)(2-1) = 6$$
The sequence looks at increasing matrices. The matrices get solely defined by the second column (hope that's clear). The sequence asks $a_1, a_2, a_3, \dots$ to be $1, 2, 4, \dots$ so $a_i = 2^{i-1}$. Since we know the Determinant of these matrices, we can rewrite the series as
$$
\det M = \prod_{1 \leq i < j \leq n} (2^{j-1}-2^{i-1})
$$
So the sequence would be
$$1, (2-1), (4-2)(4-1)(2-1), (8-4)(8-2)(8-1)(4-2)(4-1)(2-1), \dots = 1, 1, 6, 1008$$
Hopefully you can see by looking at what factors gets added (or here multiplied) at each step that each term divides its successor. The second note that $2\cdot v(n+1)/v(n)$ divides $v(n+2)/v(n+1)$ is a bit tricker. If you want to know more, you can play around with these terms in the factorised form and try to get an intuition for why this should be true.
@lucid pasture

[stoic python]

T0lgi01

[lucid pasture]

Hi and sorry I just saw this ahaha, thank you so much for the thorough and helpful explanation!! that makes a lot more sense 😅 🥰

[worldly bear]

i dont know how to set up the initial system

is it like kx+y=0 for the first equation?

i dont think thats right bc then the last equation would be 2y-3z=0 but if y and z are 1 then that equation is never true

[nocturne jewel]

Can't you just... do the multiplication then get a polynomial in k = 0?

[worldly bear]

oh like just multiply the matrixs left to right

[nocturne jewel]

Yeah

[worldly bear]

idk what i was thinking

[nocturne jewel]

you'll get a 1x1 whose entry is a function of k

[worldly bear]

i think i was trying to take it as a coefficient matrix or something

[nocturne jewel]

so f(k)=0

There might be some neat trick cause ik $x^TAx$ is a common form of things, but I've never learned it

[stoic python]

Mosh

[nocturne jewel]

[worldly bear]

thats over my head as of now

im sure theres alot of neat tricks and stuff you can do with matrix and transpose and trace

[hard drum]

Yeah, here A is symmetric and diagonalising A makes this much simpler

But that's almost certainly over the top here

[ionic laurel]

Hello - i am trying to utilize the rref command in my MATLAB script but it is not working

Does anyone know why?

It is just printing out the exact same matrix

B = [1 -2 0 0 -9 -8 ; 3 -5 -4 1 -29 -27 ; -1 2 0 1 11 11 ; -4 6 82 -6 32 26];
B([1,4],:) = B([4,1],:);
B(4,:) = B(3,:) + B(4,:);
B(3,:) = -1 * B(4,:) + B(3,:);
B(2,:) = 3 * B(3,:) + B(2,:);
B(1,:) = -6 * B(2,:) + B(1,:);
B(1,:) = 1/2 * B(1,:);
B([3,4],:) = B([4,3],:);
B(1,:) = -2 * B(4,:) + B(1,:);
B([4,1],:) = B([1,4],:);
B([3,4],:) = B([4,3],:);
B([2,3],:) = B([3,2],:);
B(2,:) = 4 * B(3,:) + B(2,:);
B([2,3],:) = B([3,2],:); %REF
B(3,:) = 2 * B(4,:) + B(3,:);
B(2,:) = -1 * B(4,:) + B(2,:);
B(3,:) = 1/37 * B(3,:);
B(2,:) = 4 * B(3,:) + B(2,:);
B(1,:) = -2 * B(2,:) + B(1,:);
B(1,:) = -1 * B(1,:); %RREF
B = [1 -2 0 0 -9 -8 ; 3 -5 -4 1 -29 -27 ; -1 2 0 1 11 11 ; -4 6 82 -6 32 26];
rref(B);

i reset the matrix to the original one stated in the problem to test the command but i have no idea as to why it the rref is not working

[turbid comet]

im so confused how you do this

[solid smelt]

what's meant by consistent

[turbid comet]

check if it has >= 1 solution

[solid smelt]

any ideas so far

[nocturne jewel]

Row reduce until you get something lower triangular, then check for consistency

[turbid comet]

@nocturne jewel i tried to reduce but all i get is extra baggage whenever i try to do something

[solid smelt]

lots of zeros u could take the determinant

but try and visualize the column space

[turbid comet]

im so lost

[solid smelt]

u know about spans?

[turbid comet]

nope

just started this course

first lesson

[solid smelt]

but u can write it in matrix form

[turbid comet]

yes

[solid smelt]

and I guessing u don't know much about determinants

[wintry steppe]

wdym extra baggage

you should be able to just row reduce it and see if theres more than one solution

[earnest ferry]

please help

[limber sierra]

which half do you need help with? what have you tried?

[earnest ferry]

I've tried the first half

[limber sierra]

might be prudent to try a contrapositive argument

[earnest ferry]

but I'm not able to prove the first part

[limber sierra]

we know that |a+b| ≤ |a| + |b|, so suppose a, b are linearly independent and prove |a+b| < |a| + |b|

actually nah that works but its probably not the easiest approach

how about this

wtf

we want to show $\phi \alpha + \psi \beta = 0$ has nonzero solution scalars $\phi, \psi$; we can divide through to get $\alpha + (\psi/\phi)\beta = 0$ (or if $\phi = 0$ divide by $\psi$ analogously). take the norm of both sides and work from there

???

i just copy-pasted

whatever

[stoic python]

Namington

[earnest ferry]

I got norm of alpha + (norm of beta)psi/phi = 0

[lucid glacier]

You can just use the fact that 2 vectors are linearly independent iff one is a scalar multiple of the other. It's the same as what nami said but stated a but more succinctly

[wintry steppe]

I have v1=(a,b,c,d) and v2=(a',b',c',d'), 2 linearly independant vectors from R^4. How do i write Vect{v1,v2}?

[nocturne jewel]

what is vect(v1,v2) suppose to denote?

[wintry steppe]

The vector space defined by those 2 vectors

[nocturne jewel]

Just span{v1,v2} I suppose

$\text{span}{v_1,v_2}={v\in\mathbb{R}^4| v=xv_1+yv_2; x,y\in\mathbb{R}}$

[stoic python]

Mosh

[wintry steppe]

Yes i know that, but it isn't what they're asking actually. Im also kinda confused with their answer, i'll send a pic (its in french tho)

[nocturne jewel]

well you asked for the vector space of those vectors so...

[wintry steppe]

2

22.b

[wintry steppe]

Thats what they're asking as well, but i think we need to be more precise i guess

[wintry steppe]

The value of k for which the linear system 2x+ky+z=4
-kx+2y+2z=2
x+y=0

Does NOT have a single solution.

I found 6. Is this correct?

[lavish jewel]

you can check yourself, you can rref the resulting matrix

[wintry steppe]

what is rref?

[nocturne jewel]

reduced row echelon form

RREF the matrix is synonymous w/ using row operations to get it into RREF

[wintry steppe]

Isn't that a long procedure?

"Row operations" oh well I guess my question is answered already

[nocturne jewel]

I mean... it scales in complexity as the size gets bigger

but generally a pretty quick algorithm

[stoic stag]

hey guys, could someone help me prove why non symmetric real matrix can't be diagonalized by any orthogonal or unitary transformation? I just can't get a grip on how to begin at all. (I'm really sorry if this is too trivial and is basically dumb to ask)

[wintry steppe]

prove the contrapositive

[stoic stag]

so prove that only symmetric matrices can be diagonalised by the aforemrntioned transformations?

[shell kindle]

Can anyone tell me what "det" means?

[wintry steppe]

'det' means determinant

[shell kindle]

Ahhh thank you so much.

[keen carbon]

If I have a symmetric $(k \cross k )$ matrix A, I know it will have a set of orthogonal eigenvector with real eigenvalues. The eigenvectors may be normalized. Now if I construct a matrix $P$ where each colomn contains one of the eigenvectors, how can I prove that it's orthogonal? i.e $P^{T}P = \mathbb{I}$

[stoic python]

shadowplayer67

[keen carbon]

I tried proving $PP^{T}$ by examining what I get from multiplying the i-th row in P by the j-th colomn in $P^T$ but it was nonsensical because I ended up with $\sum_{n= 1}^{k} e_n^{(i)}e_n^{(j)}$ so adding the sum of the products between i-th coordinate and j-th coordinate of the n-th eigenvector. When I instead looked at $P^TP$ ,I got the following sum $\sum_{n= 1}^{k} e_i^{(n)}e_j^{(n)}$ which is just the dot product of the i-th and j-th eigenvector

[stoic python]

shadowplayer67

[keen carbon]

sorry I know it's a lot. I just wanted to show my working/results. I know from $PP^T=(P^TP)^T=I^T=I$ but that feels like cheating because I'm bypassing a proof I can't show. It feels so simple that I don't know what's going wrong

[stoic python]

shadowplayer67

[ocean sequoia]

If P is an orthogonal matrix then P^t = P^-1 I believe

And P * P^-1 is the identity

[keen carbon]

sure, but that's the thing.,I KNOW that P^t should P^-1 but I'm wondering why the equality doesn't hold when I decide to expand it in terms of row-coloumn multiplication

[ocean sequoia]

The best way to go about this honestly is to work it out with a 2 by 2 example

[keen carbon]

oh yeah, I'll do that

brb

[coarse rain]

Think in terms of the inner product, and apply P^TP to the basis vectors

For example, Pe1 is just the first vector in the orthonormal basis

And thus P^TPe1 is just e1, as multiplying any row other than 1 with Pe1 gives you 0 (due to the fact that the columns of P are an orthonormal basis)