linear-algebra

[pale shell]

When the rows and columns are swapped

[subtle walrus]

i mean yeah, then you have to work with that

[split heart]

Why the dimension of a Matrix 2x2 is 4?

because it has 4 elements?

[limber sierra]

you can form a basis by $\left{\begin{pmatrix}0&1\0&0\end{pmatrix},\begin{pmatrix}1&0\0&0\end{pmatrix},\begin{pmatrix}0&0\1&0\end{pmatrix},\begin{pmatrix}0&0\0&1\end{pmatrix}\right}$

[stoic python]

Namington:

[limber sierra]

hence dimension 4

[half ice]

You can make the set of 2×2 matrices into a vector space over R. If you do, it will have that basis ^^

[split heart]

Thank you

If the Matrix is 2x4, the linear Tranformation is R2--->R4?

T(X) = M. x

Or otherwise?

[subtle walrus]

well, can you multiply a 2x4 matrix with a 2 dimensional vector?

[cursive narwhal]

@split heart

[split heart]

No

If the Matrix is 2x4, the linear Tranformation is R2--->R4?
@split heart So this is right?

[subtle walrus]

if M is 2x4 and x is 2 dimensional, then M.x makes no sense, so certainly it can't be from R^2 into anything?

[split heart]

Alright, I get it

[fossil wagon]

does this theorem state that if two linear transformations generate the same space, then they are the same?

[split heart]

So R4 ---> R2

[subtle walrus]

yes

[split heart]

Thank you very much @subtle walrus

[subtle walrus]

it seems like this sentence is incomplete @fossil wagon

[west spade]

If the (Vn...) notation means that those vectors form a basis

[fossil wagon]

yes

[west spade]

Then that is true, but it doesn't have to do with what space they generate

[fossil wagon]

but I don't understand why

[west spade]

What would it take for the two transformations to be completely identical

[fossil wagon]

the matrix is identical?

I just don't understand what that theorem states to be honest

[west spade]

it says that if the two transformations do the exact same things to a set of basis vectors, then they do the same thing to every vector

a.k.a all you need to know about a transformation to completely classify it is what it does to a set of basis vectors

it's a very powerful theorem

[fossil wagon]

hmmmmm...

but what about the T1 = T2 part?

[west spade]

That means that the two transformations are equal

If you put the same vector into both of them, the same vector will pop out

[fossil wagon]

but the matrix is not necessarily the same?

a.k.a all you need to know about a transformation to completely classify it is what it does to a set of basis vectors
I don't get this

[west spade]

like, if you know what a transformation does to a set of basis vectors then using that information you can figure out what it does to every vector

[fossil wagon]

ah yeah I get it now

[limber sierra]

the matrix is the same

fyi

[fossil wagon]

thank you

okay so, I just don't understand an exercise that said okay if we have a W = {(x,y,z)/ 2x+-y+3z=0} then the basis vectors are (1,2,0) and (0, 3, 1), so we can do T(1,0) = (1,2,0) and T(0,1) =(0,3,1)

and in general T(x,y) = (x, 2x+3y, y)

why are the basis vectors of R^2 assigned those two vectors in that order?

[cursive narwhal]

What do you mean? The linear transformation tells you exactly how one vector in one vector space transforms into another vector in another vector space. It is a function. To be more explicit, the rule for your map is given by:

$T(x,y) = (x,2x+3y,y)$

So, $T(1,0) = (1,2 \cdot 1 + 3 \cdot 0,0) = (1,2,0)$ and the same thing follows for the other standard basis vector of 2-space.

[stoic python]

Abhijeet Vats:

[cursive narwhal]

@fossil wagon

[low plank]

Hey, can someone confirm that my solution is right pls

[fossil wagon]

what does this theorem state?

[sonic osprey]

what about the statement are you confused about?

[fossil wagon]

does this mean there's only one linear transformation that can generate w?

[sonic osprey]

what does generate w mean

[fossil wagon]

like with Tvi I can find any single element of W

[sonic osprey]

I mean sure?

The theorem doesn't state that T is unique though

There could be many different possible T's that satisfy this condition

[fossil wagon]

I'm not really understand what it means ;-;

that's the part I don't really understand why using a base of V I can find W

[sonic osprey]

You can't "find W"

[fossil wagon]

like with a linear transformation

[sonic osprey]

what does that even mean

[fossil wagon]

I mean the elements of w

[sonic osprey]

I mean

If I give you the set A = {1,2,3} and the set B = {4,5,6}

[fossil wagon]

sorry I'm nervous rn 😭

[sonic osprey]

can you find me a function f from A to B such that f(1) = 4?

[fossil wagon]

ehmm yeah? like (2) (x), x being an element of A(?

👀 .

[sonic osprey]

Uh

does that satisfy the requirement that f(1) = 4?

[fossil wagon]

oh wait

(4)(x)

I'm just confused ;-; I think

[sonic osprey]

Is that a function from {1,2,3} to {4,5,6}?

[fossil wagon]

oh no

[sonic osprey]

You don't need to specify functions with a formula

you can always specify functions by just saying

f(1) = 4, f(2) = 4, f(3) = 4

And that's a function from {1,2,3} to {4,5,6}

[fossil wagon]

hmmm

but I'm not getting the whole {4,5,6} set, right?

[sonic osprey]

You don't have to?

[fossil wagon]

hmm ok

[sonic osprey]

Maybe it'd help to review what a function is

Since linear transformations are just a special type of function

[fossil wagon]

I just have so many questions because we are covering this, and I tried learning the theorems, and solving some exercises but I don't understand this question on the book and I thought this theorem would make it more clear

"Let v1,v2,v3,...,vn a base of R^n, and let w1,w2,...,wn a base of P n-1. So there are two linear transformations S and T such that Tv1 = w1 and Sw1 = v1 for every i = 1,2,...,n

Since linear transformations are just a special type of function
yeah maybe

I just have so many questions because we are covering this, and I tried learning the theorems, and solving some exercises but I don't understand this question on the book and I thought this theorem would make it more clear

"Let v1,v2,v3,...,vn a base of R^n, and let w1,w2,...,wn a base of P n-1. So there are two linear transformations S and T such that Tv1 = w1 and Sw1 = v1 for every i = 1,2,...,n

And I thought this was true, but my books says otherwise

[sonic osprey]

What's P n-1

[fossil wagon]

A vector space made of polynomials like ax^(n-1) + bx^(n-2) + ... + k

[sonic osprey]

Yeah, there's only one such linear transformation

[fossil wagon]

why

[sonic osprey]

The important part here is the dimension of the vector spaces

Both have dimension n

I mean

Try it when n = 2 for example

Let (1,0) and (0,1) be your basis for R^2

and let 1, x be your basis for P n-1

[fossil wagon]

I thought that was true because T would be something having x's in it, and S would be full of scalars or whatever

[sonic osprey]

Can you find me a linear transformation T such that

T(1,0) = 1 and T(0,1) = x?

[fossil wagon]

0

oh wait

lol

hmmmm

(1 , 0)?

[sonic osprey]

what do you mean

[fossil wagon]

ehmmmm I thought T had to be a matrix

[sonic osprey]

a linear transformation is a function between vector spaces

It can be represented as a matrix multiplication yes

But you should think of it as a function

[fossil wagon]

hmmmm

I'm so confused 😭

but why is it false?

[sonic osprey]

idk how to explain to you that its false

unless you tell me why you think its true

like, i could give you a proof of the fact, but you probably wouldn't understand it

[fossil wagon]

I thought that was true because T would be something having x's in it, and S would be full of scalars or whatever

I don't know if this makes sense

[sonic osprey]

it doesn't

[fossil wagon]

ow ;-;

but how can I have a function from R^(n) to P n-1?

[sonic osprey]

T(0,1) = 1

T(1,0) = x

T(a,b) = ax + b

is a function from R^2 to P 1

You should also check that this is a linear function

[fossil wagon]

and how can I figure out what T is?

[sonic osprey]

That is T

[fossil wagon]

like how do I go from T(0,1) = 1
T(1,0) = x

to T(a,b) = ax + b?

[sonic osprey]

Forget the first parts

T(a,b) = ax + b

Is a function from R^2 to P 1

[fossil wagon]

hmmm I think I'm struggling understanding all of this because what a linear transformation is not that clear to me

when we represent T(v) = w, then this means T is gonna convert v from what ever space it is to the space of w?

[sonic osprey]

That's how functions work yes

[fossil wagon]

okay

and T is supposed to have only scalars, right?

[sonic osprey]

Again, this makes no sense

T is a function

T can be represented by a matrix yes, but ultimately, T is a function

[fossil wagon]

I mean if we say Tv = Ax (where A is a matrix)

[fading lily]

you can think of it as the left multiplication transformation

$$L_A(x) = Ax$$

[stoic python]

Circulation:

[fossil wagon]

okay

so can matrices contain variables? like (y^2,y,a)?

I just don't understand how a linear transformation can go from R^n to P n-1

using matrix multiplication

[fading lily]

did you learn about basis

if you didnt then dont worry about it just yet

[fossil wagon]

I did

[fading lily]

the matrix will give you coordinates in terms of the ordered basis you used

you multiply each row of the resulting matrix by the ordered basis

[fossil wagon]

okay

like

if I have T(0,1) = (0,1,0) and T(1,0) = (1,0,0) then

T(x,y) = xT(1,0) + yT(0,1)

is that what you mean?

[fading lily]

First of all that transformation is not linear unless you mean like T(z) where z = (x,y)

[fossil wagon]

oh (x,y) is supposed to be a column vector

like

x
y

[fading lily]

I'm talking about the matrix representation of a linear transformation

using matrix multiplication
@fossil wagon lets use this example

I just don't understand how a linear transformation can go from R^n to P n-1
@fossil wagon .

lets fix n to some number

how about $$T:R^3 \rightarrow P_{2}(R)$$

[stoic python]

Circulation:

[fossil wagon]

okay

[fading lily]

give me some linear transformation with those domain/codomain

also the notation that i will use might be confusing idk what textbook you are using

[fossil wagon]

T x = xp^2 + yp + z
y
z

[fading lily]

ok so $$T(r) = T((x,y,z)) = x p^2 + yp + z$$

[stoic python]

Circulation:

[fading lily]

where $$r \in R^3$$

[stoic python]

Circulation:

[fading lily]

good so far?

[fossil wagon]

yes

[fading lily]

ok now lets pick an ordered basis for $$R^3$$ and $$P_2(R)$$

[stoic python]

Circulation:

[fading lily]

do you know the standard basis for these

vector spaces

[fossil wagon]

for p2 is (x^2, x, 1) (I'm not sure about the 1) and for R^3 is ((1,0,0),(0,1,0),(0,0,1))

[fading lily]

yes that is correct we usually write it in the opposite order you wrote it

[fossil wagon]

oh ok

[fading lily]

lets call the ordered basis for $$R^3$$ this symbol $$\beta$$

[stoic python]

Circulation:

[fading lily]

and lets call the ordered basis for $$P_2(R)$$ this symbol $$\gamma$$

[stoic python]

Circulation:

[fading lily]

so $\beta = {(1,0,0),(0,1,0),(0,0,1)}$ $\gamma = {1,x,x^2}$

[stoic python]

Circulation:

[fossil wagon]

okay this makes sense to me so far

[fading lily]

ok we want to create a matrix A such that when we multiply A by a coordinate vector of $R^3$ we get a coordinate vector of $P_2(R)$

[stoic python]

Circulation:

[fossil wagon]

yes

[fading lily]

we also write this matrix as $[T]_\beta^\gamma$

[stoic python]

Circulation:

[fading lily]

in my textbook at least

im using friedberg

[fossil wagon]

it's alright

mine used A but okay I can stick to your notation

[fading lily]

the way we compute A is we first apply T to every one of our basis vectors for $R^3$

[stoic python]

Circulation:

[fading lily]

so what is $T(1,0,0)$ and $T(0,1,0)$ and $T(0,0,1)$

[stoic python]

Circulation:

[fossil wagon]

T(1,0,0) = x^2
T(0,1,0) = x
T(0,0,1) = 1

OH WAIT

[fading lily]

okay write those in terms of the coordinate vector of $P_2(R)$

[stoic python]

Circulation:

[fading lily]

then those will be the columns of A

then you can multiply A by coordinate vectors of R3

then you get ur answer in terms of coordinate vector in P2

got it?

[fossil wagon]

like

p^2 0 0 x
0 p 0 y
0 0 1 z

[fading lily]

uh no

x^2 = (0,0,1) in coordinate vector

cuz (0,0,1) * (1,x,x^2) = x^2

so ur first column will be 0,0,1

$$A = [T]_\beta^{\gamma} = \begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{bmatrix}$$

[stoic python]

Circulation:

[fossil wagon]

T x = xT(1,0,0) + yT(0,1,0) + zT(0,0,1)?
y
z

[fading lily]

uh idk maybe

[fossil wagon]

but like why is it the other way around?

why

x 0 0 p^2
0 y 0 p
0 0 z 1

?

hmmm it's the same thing(?

[fading lily]

wut

idk what you are trying to describe

[fossil wagon]

like in general

[fading lily]

mine is $Ax = b$ where x is the coordinate vector for R3 and y is the coordinate vector for P_2

[stoic python]

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

[half ice]

[a] [d e f]
[b] [g h i]
[c] [j k l]
Isn't a valid matrix multiplication

[fossil wagon]

but I wrote it the other way around

[half ice]

Yes reversed, it works

[hidden sable]

I need help diagonalizing this into A = QΛQT

A = QΛQ^T

Q is orthogonal

<@&286206848099549185>

[stable urchin]

What does orthogonal mean again

[limber sierra]

google exists

[stable urchin]

yikes okay

[limber sierra]

anyway @hidden sable are you sure thats supposed to be $Q\Lambda Q^T$ and not $Q\Lambda Q^{-1}$?

[stoic python]

Namington:

[hidden sable]

I thought that Q^t is equal to Q^-1?

[pallid rampart]

Q is orthogonal

[limber sierra]

oh fuck

Q is orthogonal

lmao im a dumbass

my bad my bad

[hidden sable]

lol

[pallid rampart]

google exists

[limber sierra]

in any case

the process is described here

https://en.wikipedia.org/wiki/Orthogonal_diagonalization

that specifically diagonalizes a quadratic form and youre diagonalizing a matrix but nonetheless

it just means you already know what A is

is anything in specific confusing you/do you need help with any step?

[hidden sable]

When I try to reduce the columns of X to length 1, then try finding the inverse, the inverse is not the transpose

That's what I am stuck on

[limber sierra]

sorry, what's X?

[hidden sable]

Set of eigenvectors

[limber sierra]

so the basis you constructed for the eigenspace wasnt orthogonal, it sounds like

[hidden sable]

The X in XΛX^-1

Does the order matter?

Yeah, basically

That's what I was stuck on

[limber sierra]

the order of the columns of eigenvectors doesn't matter, as long as each entry on the diagonal in Lambda corresponds to the column from X

that is to say, if (for example) 5 is an eigenvalue with eigenvector (1, 4) and you make (1, 4) the first column

then 5 has to be the top-left entry on the diagonal

if you make (1, 4) the second column, 5 would have to be the second entry on teh diagonal

etc

anyway

you're struggling with finding an orthogonal basis?

[hidden sable]

Yeah

[elder robin]

If I have a matrix 3x3 matrix A and a matrix E = [1], we can't multiply them, right?

Or does it act like multiplying by a scalar

[limber sierra]

what techniques have you learned? gram-schmidt?

[hidden sable]

Is that where you reduce the lengths of the columns to 1?

If so, yes

[limber sierra]

reducing the length to 1 is how you normalize an orthogonal set

(to make it orthonormal)

but first it has to be orthogonal in the first place

[hidden sable]

Oh

[limber sierra]

so there are a few techniques to orthogonalize a set

first verify that the vectors in your basis aren't orthogonal; then you can construct a new basis by

this technique

where proj is defined as such

so if you have two vectors in your basis, v_1 and v_2

you can create a new orthogonal basis by setting u_1 = v_1

and then $u_2 = v_2 - \frac{\langle v_2, v_1\rangle }{\langle v_1, v_1\rangle} v_1$ where $\langle a, b \rangle$ is the inner (dot) product of $a$ and $b$

[hidden sable]

?

[stoic python]

Namington:

[limber sierra]

sorry my brain is fried right now

[hidden sable]

lol its fine

so how do you find u2?

[limber sierra]

using the equation i gave

[hidden sable]

oh

[limber sierra]

$u_2 = v_2 - \frac{v_2 \cdot v_1}{v_1 \cdot v_1}v_1$

[stoic python]

Namington:

[limber sierra]

and then to normalize these, just divide them by their lengths

ie $\frac{u_1}{||u_1||}$ and $\frac{u_2}{||u_2||}$

[stoic python]

Namington:

[hidden sable]

I am assuming that in order to multiply 2 1x2 vectors, i will have to take the transpose?

since this is a 2x2 matrix

[limber sierra]

the $\cdot$ here is a dot product

[stoic python]

Namington:

[hidden sable]

can you take the dot product of 2 1x2 vectors?

[limber sierra]

yes... are you not aware of what a dot product is?

[hidden sable]

just making sure

[limber sierra]

you might be familiar with the term "inner product"

or even "scalar product" but this is a weird name

[hidden sable]

ok

ill try this

t

y

[hollow finch]

What do we know about the eigenvalues of a matrix A such that A^3=0

I'm sure at least one eigenvalue is 0 since there's no way it's invertible, but apparently it's incorrect that they are all necessarily 0

[pallid rampart]

Assume $\lambda$ is an eigenvalue for $A$, then that means that there exists a nonzero vector $v$ such that $Av=\lambda v$, then $0=A^3v=\lambda^3 v$. Since $v$ is not zero, it follows that $\lambda^3$ must be zero, so $\lambda=0$

[stoic python]

Whoever:

[pallid rampart]

@hollow finch

[hollow finch]

That's what the person put on the test but they lost points. Apparently they can also be complex? I mean I think the real part of the eigenvalues must all be zero, but idk.

I agree that at least all real eigenvectors must have eigenvalue zero I think...

Thinking in terms of the simplest matrix A can be similar to, wouldn't it have to be upper/lower triangular with zeros on all diagonal entries?

Oi I feel dumb rn

[hidden sable]

for my question, i still cannot reach the answer

nvm

wait, the inverse is still not the transpose

even though v1*v2 is 0

[hollow finch]

if youre referring to orthogonal matrices, then the columns must be normalized @hidden sable

[hidden sable]

the inverse of the normalized matrix is not the transpose

idk

[hollow finch]

whats not the identity when you multiply them

if youre getting things off the diagonal your columns are not orthogonal, and if the diagonal is not 1s then your columns are not normalized

[hidden sable]

the diagonal has to be 1?

[hollow finch]

$(A^TA){ii}=(AA^T){ii}=1$

[stoic python]

nix:

[hollow finch]

because its the identity

[hidden sable]

I understand that

for this matrix, i just cant seem to find the orthogonal matrix

I have to reduce it?

?

[hollow finch]

Find the matrix which orthogonally diagonalizes it?

[hidden sable]

QΛQ^T

[hollow finch]

What are the eigenvalues/eigenvectors

[hidden sable]

Eigenvalue = +$\sqrt2 -\sqrt 2$

[stoic python]

bladexyz17:

[hollow finch]

ew

[hidden sable]

sqrt 2 and negative square root 2

[hollow finch]

but go on

[hidden sable]

lol

eigenvalues = [-(sqrt2) + 1, 1], [ (sqrt2) + 1, 1]

$[-(\sqrt2) + 1, 1], [ (\sqrt2) + 1, 1]$

[stoic python]

bladexyz17:

[hollow finch]

And those are orthogonal so yeah thats right

The reason I said ew was because I knew the eigenvectors would be a bitch to normalize

[hidden sable]

oh

Then I normalized it

[hollow finch]

The magnitude for the first is 4-2sqrt2 and the second is 4+2sqrt2

So what are your normalized eigenvectors

[hidden sable]

$[\frac{\sqrt{2}+1},{\sqrt{4+2\sqrt{2}}}]$

[stoic python]

bladexyz17:

[hidden sable]

uhhh

[hollow finch]

no comma between

[hidden sable]

$[\frac{\sqrt{2}+1}{\sqrt{4+2\sqrt{2}}}]$

[hollow finch]

$[\frac{\sqrt{2}+1}{\sqrt{4+2\sqrt{2}}}]$

[stoic python]

bladexyz17:

nix:

[hidden sable]

hold on

$\frac{\sqrt{2}+1}{\sqrt{4+2\sqrt{2}}},\frac{1}{\sqrt{4+2\sqrt{2}}}$

[stoic python]

bladexyz17:

[hidden sable]

$\frac{-\sqrt{2}+1}{\sqrt{4-2\sqrt{2}}},\frac{1}{\sqrt{4-2\sqrt{2}}}]$

[stoic python]

bladexyz17:

[hollow finch]

Good

[hidden sable]

?

[hollow finch]

$Q=\begin{bmatrix}\frac{\sqrt{2}+1}{\sqrt{4+2\sqrt{2}}}&\frac{-\sqrt{2}+1}{\sqrt{4-2\sqrt{2}}} \
\frac{1}{\sqrt{4+2\sqrt{2}}} & \frac{1}{\sqrt{4-2\sqrt{2}}}]\end{bmatrix}$

oh god wtf

[hidden sable]

lol

i think i got it

thank you

[hollow finch]

yea

[stoic python]

nix:

[hidden sable]

Just another question

Is the diagonalization of a symmetic matrix always orthogonal?

[hollow finch]

It can always be made orthogonal. If you have multiple eigenvectors for a single eigenvalue you need to make them mutually orthogonal to get an orthogonal matrix of eigenvectors. But yes, the eigenvectors from different eigenspaces are always orthogonal for symmetric matrices.

[hidden sable]

ok thank you

[half ice]

Yus

An invertible matrix is an iso between spaces

And all isos have an invertible matrix

[hollow finch]

someone lost points on a test for saying that all eigenvalues of a matrix A where A^3=0 must be zero. Supposedly, there can be complex eigenvalues, but i cant see how.

[sonic osprey]

do you know the size of the matrix?

[hollow finch]

No

A cubed is the zero matrix. There is no restriction on the size.

[sonic osprey]

okay wait, no that must be true

all eigenvalues of nilpotent matrices must be 0

[hollow finch]

yeah thats what i thought. but the professor said otherwise. im going to email him because i really cant see how it can have a nonzero eigenvalue.

[sonic osprey]

Just send him a proof

https://math.stackexchange.com/questions/1265679/why-are-eigenvalues-of-nilpotent-matrices-equal-to-zero

[hollow finch]

$A^3v=\lambda^3 v$

$0=\lambda^3 v$

Therefore $\lambda=0$ since $v$ is an eigenvector which implies $v\neq 0$

[stoic python]

nix:

[sonic osprey]

There are matrices such that A^3 = 0, but A is not 0

[hollow finch]

That was my thinking

[sonic osprey]

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

[stoic python]

Zopherus:

[sonic osprey]

yeah that's the right idea

[hollow finch]

Are all nilpotent matrices similar to an upper triangular matrix which has all zero diagonal entries?

[sonic osprey]

Yeah, that should be true

[hollow finch]

Yeah then the characteristic equation will just be t^n=0 and all roots will be zero. im wondering if im missing part of the question or something.

[sonic osprey]

all that seems right

[hollow finch]

Hm I seem to remember as well that theres some theorem that states that a plugging in a matrix to its characteristic polynomial also gives the zero matrix. Are there conditions that make this unique? As in, say A is 3x3, are there two or more distinct 3rd degree polynomials such that p(A)=0?

[sonic osprey]

It's called the Cayley-Hamilton theorem

and there could be multiply polynomials yes, even if you require the polynomial to be monic

[hollow finch]

That's right. Thank you very much for the help @sonic osprey 🙂

[wintry steppe]

For isomorphic the kernel does need to be 0, and onto. the thm requires T to be one to one and onto to be isomorphic

to prove isomorphic if in the format T(v)=Av why not prove A is invertible ? if A^-1 is true the transformation is isomorphic. Basically if det(A)=/=0 then T(v)=Av is isomorphic

I think it is much easier to just prove A invertible, I used that to prove isomorphism

there are three conditions, if one is true the rest are true

By proving condition 3. Condition 2 and 1 are true

Not sure if your professor would like you to solve this way. You first learn to prove one to one and onto then you learn of this thm

I would try an example with finding one to one and onto with an invertible A

you will see how using an invertible matrix always yields the same results

[limber sierra]

"invertible" and "one to one & onto" are

literally the same thing

this isnt unique to linear operators either

any bijective function is invertible, and vice versa

[wintry steppe]

yes sorry I bad at explaining

[limber sierra]

see theorems 1 and 2 here http://math.colorado.edu/~kstange/has-inverse-is-bijective.pdf

[wintry steppe]

👑

[limber sierra]

so yes, if you know a linear function is invertible (or, equivalently, you know it has an invertible matrix representation)

you know it's one-to-one and onto

as for where "isomorphisms" come from out of this

the idea behind an isomorphism is it lets us say "two structures behave the same"

this means that the isomorphism should preserve the space's structure (i.e. be a homomorphism), but moreover

that we should be able to use it to "go between" the two vector spaces

and indeed to "go back"

hence, it makes sense to ask for isomorphisms to be invertible

if two things are "the same", we should be able to convert between them in both directions

[wintry steppe]

for 21 I get Range(T)=[-1/4;1/4;1]. The textbook's format is Range(T)= [4a,4b,a-b]. I cant seem to find what I do wrong

RREF(A)=[1 0 1/4;0 1 -1/4]

Letting x3=s I get my answer s[-1/4;1/4;1]

[autumn kraken]

Hi, I have an exercise to show the equivalence relation of f ~ g <=> f(3) = g(3) in the set of all functions from R to R.

Is proving the symmetry here trivial or am I missing something? f(3) = g(3) => g(3) = f(3)

[dusky epoch]

all three properties are pretty trivial to prove

since they boil down to the same properties of equality

[autumn kraken]

Yeah I mean it's obvious that if f(3) = g(3) then g(3) = f(3) is the same thing. So I don't understand what they expect me to write down

[limber sierra]

"real number equality is an equivalence relation, therefore..."

that's what i'd say, at least

[autumn kraken]

Okay thanks

Just wondered because I often assume things are trivial but then they are not at all

[autumn kraken]

What does the ~ with the dash above it mean?

[dusky epoch]

i'm guessing that this funky tilde is just their definition of their relation

like

:<=> is "is by definition equivalent to"

[autumn kraken]

oh I see

by the way is it a mistake to mix those up?

or are they just a tool to show intent

[dusky epoch]

they're a tool to show intent

[autumn kraken]

I must be missing something.. Proving that it's an equivalence relation boils down again to just saying "real number equality is an equivalence relation, therefore..." right?

this is weird

[subtle walrus]

why?

this is kinda supposed to happen if your equivalence relation is defined in terms of "=" for real numbers

[autumn kraken]

because doing the same thing twice for seperate exercises seems weird to me. Anyway xD

a quotient of a set is defined with modulo, and it basically is just a subset of that set? Is that correct?

[gray dust]

@wintry steppe most of what you said is sorta incomprehensible. if you're taking your answer directly from RREF(A) then something definitely went wrong. at most you can use RREF to find the dimension of range(T), or count off the pivots of RREF(A) to determine which and how many of A's columns can be used to build a basis for range(T)

also the question doesn't seem to ask for much so really you can just say range(T)=span{(5,1,1),(-3,1,-1)} and be done

[subtle walrus]

@autumn kraken i assume you didn't cover enough content yet to do interesting examples of equivalence relations, so your prof just wants you to get a feel by doing a few trivial calculations

for your question you would have to tell me what you mean by quotient of a set and by modulo

[wintry steppe]

@gray dust thank you !

[autumn kraken]

@subtle walrus for example Z/~ which we call "Z modulo ~"

or Z/n

I know this would give the remainder if you did this on two simple numbers

[subtle walrus]

this is never a subset of Z though

the elements of Z/~ are subsets of Z

[autumn kraken]

ah I see

and these subsets contain all elements for which ~ works?

[subtle walrus]

i think you mean the correct thing

an element of Z/~ is a subset of Z that contains all equivalent elements(by that equivalence relation)

[autumn kraken]

thanks

[hollow finch]

Is this correct? I'm pretty sure the only eigenvalues are zero.

[dusky epoch]

seems ok to me

[hollow finch]

So there are complex eigenvalues? @dusky epoch

[dusky epoch]

no

[hidden sable]

Is there a generalized eigenvector for this matrix?

[hollow finch]

well we know e1 is going to be one

i mean can you see what both eigenvalues have to be by inspection?

[hidden sable]

6 and 4

[hollow finch]

Yep

So find a basis for the null space of A-4I and A-6I

[hidden sable]

[1/2,1],[1,0]

What is e1?

@hollow finch

Is there a generalized eigenvector for this matrix?

[hollow finch]

$e_i$ is a vector with all zeros except the $i$th row has a 1

[stoic python]

nix:

[hidden sable]

oh

[1/2,1],[1,0]

i got that from the eigenvalues

[hidden sable]

<@&286206848099549185>

[gloomy arrow]

Not too sure about that symbol

[azure owl]

@hidden sable what do you mean generalized eigenvectors?

Are you asking what they are?

[hidden sable]

it is in the null space of (A-lambda(I))^2 and is orthogonal to the eigenvector of (A-lambda(I))

I am asking how to find the generalized eigenvector for this matrix

I thought that a matrix had to be defective in order to find the generalized eigenvector, but this matrix is not defective

[gray dust]

$W^\perp$ denotes the orthogonal complement of $W$, which is the set of vectors that are orthogonal to the vectors in $W$

[stoic python]

RokettoJanpu:

[gray dust]

@gloomy arrow

[pale shell]

How do you calculate the span of a set of vectors?

And by that I mean

Like

The space it spans

[subtle walrus]

what do you expect the result to be?

like it's usually a set of infinitely many vectors, so

[half ice]

Usually there's no great way to express the space

The span itself is the way we normally express it

[pale shell]

Mk

So like

R2

Or sum

[cursive narwhal]

@pale shell Yes, you would express it as a sum. So, let $B = (v_1,v_2,\ldots,v_n)$ be a list of vectors. Then, the linear span of these vectors is the following set:

$L(B) = {v= \sum_{k=1}^{n} \alpha_k v_k | \alpha_k \in \mathbb{F} }$

[stoic python]

Abhijeet Vats:

[cursive narwhal]

The above is a linear combination of the vectors in B

[limber sierra]

a matrix is bijective

if you mean "the linear map the matrix represents"

then yes

the linear map is bijective iff the matrix is invertible

because every matrix has a unique inverse

i'm not sure what the "because" is implying here though

[also, only invertible matrices have a unique inverse - noninvertible matrices don't have any inverse - but i'd assume that's just weird wording]

"i am not a dog, therefore i am a human"

it is true that I am not a dog, and it is also true that, if I were a dog, I would not be a human

it is not true, however, that I am a human because I am not a dog

indeed, it would be possible for me to be, say, a cat if that's all the information you know

just because nonunique inverses would imply not injective does not mean that the matrix is injective because of unique inverses

you're affirming the consequent

the proof is a bit more sophisticated

generally the proof strategy is

bijective linear function <=> invertible linear function <=> invertible matrix representation

proving a bijective function is invertible (and vice versa) is very standard, there's a bunch of proofs online

and proving that a linear function is invertible iff its matrix is is done here https://math.stackexchange.com/questions/2294040/1-1-correspondence-between-texthom-k-u-v-to-set-of-all-m-times-n-matri

[implicitly]

not only if

er wait

nvm misinterpeted that sentence, yeah thats correct

assuming you've proven surjectivity in the first place

well, if it's invertible then it cant be noninjective

that's the whole point here

"bijective" and "invertible" are

literally the same thing

and if there exists an invertible (AKA bijective) map between two spaces, then that map is an isomorphism

[river jasper]

Why are the possible eigenvalues for an orthogonal and symmetric matrix only 1 and -1?

[limber sierra]

note that an orthogonal symmetric matrix is its own inverse

you should be able to prove it from this fact

since $Ax = \lambda x$ and therefore $A^{-1}Ax = A^{-1}\lambda x$ so $x = A\lambda x$

[stoic python]

Namington:

[limber sierra]

hence $x = \lambda A x = \lambda^2 x$ (since $Ax = \lambda x$)

[stoic python]

Namington:

[limber sierra]

rearranging gives $(1-\lambda^2)x = 0$

[stoic python]

Namington:

[limber sierra]

so $\lambda\in {-1, 1}$

[stoic python]

Namington:

[river jasper]

alright thanks, the online explanation were so confusing

[hollow finch]

will a real symmetric matrix ever have an eigenvalue with algebraic multiplicity greater than geometric multiplicity?

[low plank]

So I have a question where
A is an n×n matrix such that A^4=In and M = A^3+A^2+A+In
It wants me to show that if |M| does not equal zero, show that A=In

I know that I should multiply M by A, look at the results and use the inverse of A

But can someone explain how that's done ?

I did that but still no idea

Or am I supposed to.use a different way ?

[tranquil trail]

anybody able to help me with this?

[odd kite]

@tranquil trail forget they are matrices for a second. If I gave you 2 vectors u,v in 4D and said W= span(u,v) would you know how to find a basis for $W^\perp$ ?

[stoic python]

Timon:

[tranquil trail]

ummmm not necessarily

just the $W^\perp$ completely throws me off

[stoic python]

jkonyan:

[odd kite]

It's called the orthogonal complement

$W^\perp$ is the set of all vectors which are orthogonal to every vector in W

[stoic python]

Timon:

[pale shell]

What would a 1x3 matrix mean geometrically as a transformation

[odd kite]

@pale shell projection of a 3d vector onto a 1D subspace (a line)

@tranquil trail in other words for every $w\in W$ and and $q\in W^\perp,;; (w, q) = 0$

[stoic python]

Timon:

[odd kite]

where (w,q) is the bi-linear form in this case $\mathrm {Tr}(B^TA) $

[stoic python]

Timon:

[river jasper]

How do I find a generalized eigenvector of a matrix?

[jagged saffron]

in JCF, the power in the min poly determines the largest sub-block size, but say you char poly is (x-9)^2(x-7)^3, then what happens if your min poly is (x-9)?

does that mean that in JCF the block for 7 is 0?

[dusky epoch]

your minpoly cannot be (x-9) for a matrix with an eigenvalue of 7

[jagged saffron]

I see, so if our char poly is (x-$\lambda_1)^n_1 (x-$\lambda_2)^n_2....$ then our min poly has to be at least contain each factor once?

[limber sierra]

yes

here mu_A is the min poly of A

[idle echo]

I forgot the formula for determining the linear dependence of a matrix given the dimensions

for n x m, doesn't m > n indicate that the columns must be independent?

[hallow cliff]

if there are more columns, than rows then there must be linearly dependant columns

same thing for rows

[ocean sequoia]

the coefficients of a polynomial are uniquely determined by the polynomial.

Can anyone phrase that in a different way im kinda confused as to what that means

this is coming from Linear Algebra the Right Way so i can put it some where else if need be

[late jasper]

I need help with this one I don't understand what formula to use

[limber sierra]

this is not linear algebra

#prealg-and-algebra would be a better fit

@ocean sequoia basically, there's only one "set of coefficients" for each polynomial

if you have a polynomial and you know how it behaves, you know how to write its coefficients

it's impossible for, say, 5x^3 + 2x to be the same polynomial as 6x^4 - x^3 + 2x^2

or whatever

in other words, "polynomials are uniquely determined"

[viscid kernel]

@devout pine yup

Det (A*B) = det(A) * det(B)

[devout pine]

👍
another question: would det(A^5)=32?

[viscid kernel]

Det ( AAAAA ) = det(A) * det(A) * det(A) * det(A) * det(A)

So yes

[sonic osprey]

This isn't linear algebra

like #calculus

[hollow finch]

Can a linear transformation which is injective be invertible (or have an inverse transformation) if its not surjective?

[limber sierra]

no

you can make a new function that "corrects" the codomain to the range

[i.e. force it to be surjective]

in fact, this isnt unique to linear transformations

a function is invertible if and only if it is bijective (injective and surjective)

"invertible" and "bijective" are literally the same thing

[hollow finch]

@limber sierra Could you theoretically define a pseudo inverse?

$T: V \to W$

$T^{-1}: \text{Range}(T) \to V$

[stoic python]

nix:

[hollow finch]

With the condition that T is injective

[limber sierra]

dont call it a pseudoinverse, that term is "reserved"

but sure

in that case you're just defining an inverse for the function $T^*\colon V \to \text{Range}(T)$ though

[stoic python]

Namington:

[limber sierra]

i.e. the function i mentioned that "corrects" the codomain to the range

as far as T^{-1}'s relation to T

let me use f and g here actually for simplicity

suppose $f\colon V\to W$ is injective and we want to define an ``inverse-like object" $g\colon \text{Range}(f)\to V$

[stoic python]

Namington:

[limber sierra]

then $g$ would be a left inverse of $f$, in the sense that $g \circ f = \text{id}$ the identity function

[stoic python]

Namington:

[limber sierra]

but there's no way to make a right-inverse of f unless it's surjective

[in which case g would just be an inverse]

this is captured more generally by the theory of monomorphisms and epimorphisms

and more broadly by elementary category theory as a whole

[though the terminology doesnt quite coincide; it's possible for a monomorphism to not be left-invertible, but it is in "nice enough" contexts]

[but left-invertible functions are always monomorphisms, and therefore injective functions are always monomorphisms]

[cold topaz]

Can we say:
:Since we have a 3x3 matrix, and λ=1 is the only value that is more than 0and less than or equal to 3, then 1 is the possible dim of eigenspace."?

[dusky epoch]

that's uh

no?

correct answer, wonky wording and even wonkier reasoning

[dawn fractal]

ooooh, is that from gallian?

my hardbound copy is at my other house, which i can't get 😢

[hollow finch]

@limber sierra Thank you for the very thorough answer. So if I'm understanding right, essentially sometimes we can find a right/left inverse but not necessarily both
unless it's bijective?

[cold topaz]

I'm looking at the values for λ. and 1 is the only positive value that is not more than 3. that's how I got to that conclusion. Is that a right approach?

[dusky epoch]

uh

what if your polynomial was (λ-10)(λ-20)(λ-30) instead?

what would the same reasoning have led you to?

[cold topaz]

then my n = 3.

[dusky epoch]

what's n?

[cold topaz]

meaning Ann.

[dusky epoch]

what?

[cold topaz]

So it is a sqaure matrix

and

[dusky epoch]

the size of the matrix is 3 in either case, but that's not what i was talking about.

[hollow finch]

@cold topaz The degree of the C.P. is 3 so it's a 3x3. Every eigenvalue has at least one eigenvector, and at most the multiplicity of its root in the C.P. So what is the multiplicity of each root?

also there is no sense in looking at eigenvectors for a nonsquare matrix. that just doesn't make any sense. how do you stay on your span in a different dimension

nor can you even evaluate the determinant

[cold topaz]

from my textbook

[dusky epoch]

yeah but what you said is distorted to the point where your reasoning was impossible to follow

[hollow finch]

thats based on the multiplicity not the eigenvalue

tbh thats pretty shitty and confusing to have your eigenvalues be exactly the same as their multiplicity. no wonder youre having trouble lol

[cold topaz]

my book suck big time!

Ive encountered several issues

[dusky epoch]

this is not one of those issues

i mean

ok yeah no that example was bad

but the wording is fine

[hollow finch]

if the C.P. was $(\lambda-3)^1 (\lambda+1)^2$ then the dimension of the eigenspace for $\lambda=3$ is at most 1 because its $(\lambda-3)^1$, and the dimension of the eigenspace for $\lambda=-1$ is at most 2 because its $(\lambda+1)^2$

[stoic python]

nix:

[cold topaz]

so the power is the hint?

[hollow finch]

we also know each eigenspace has a dimension of at least 1. so we definitely have an eigenvector for each. whether or not -1 has a second eigenvector is not guaranteed.

the power is everything in terms of the dimension of the eigenspace

the number of eigenvectors you get for a given eigenvalue has nothing to do with the actual eigenvalue itself. only its multiplicity in the C.P.

[wintry steppe]

How is C. Onto if range=2 and n=3

or would it be onto because it matches R^m

[dusky epoch]

dim(range) = 2 and dim(R^2) = 2

[wintry steppe]

im confused on which value do we base n of rank(T)+Nullity(T)=n on isnt it R^n->R^m

or can you use the dimensions of m

[dusky epoch]

rank + nullity = dimension of domain

[wintry steppe]

ok I see now, much clearer answer than textbook

[distant granite]

is I_{n} = I_{m} for an identity matrix ?

[pallid rampart]

Well

if n≠m, then they're not even the same size

can you compare two matrices with different size?

[distant granite]

oh yes makes sense

this is what made me ask the question

[pallid rampart]

Well

One of the first things you learn when learning matrix multiplication is that

AC=BC does not mean A=B

Where A,B,C are matrices

[distant granite]

oh but in an identity matrix that works

thanks a lot

[pallid rampart]

Well AC=BC doesn't imply A=B means that there are matrices A,B,C such that A≠B and AC=BC

It doesn't mean that AC=BC doesn't imply A=B is always false

But ok

[distant granite]

i get the idea now i thought identity matrices do not need to have same number of rows and columns

but that is not the case

[pallid rampart]

Well there is an identity matrix for each n

[heady dagger]

Hey, I'm a little confused. When I first encountered the determinant, it was defined as a mapping from $\mathbb{R}^{n\cross n}$ to $\mathbb{R}$. Under this definition, non-invertible matrices simply have zero determinant. However, in group theory I came across the definition det: $GL_n(\mathbb{R})\rightarrow \mathbb{R}^*$. The determinant isn't a group homomorphism under the first definiton, right? Then why is the determinant usually referred to as one without mentioning the domain and codomain that we choose? Also, under the second definition, what happens when we have a non-invertible matrix, i.e. one that is not an element of $GL_n(\mathbb{R})$? Do we just not consider them? But if so, then why define it this way, considering that we often have to deal with non-invertible matrices?

[stoic python]

Paddy:

[broken hawk]

I mean it can’t be a group homomorphism in the first case simply because $\R^{n \times n}$ is not a group under matrix multiplication

[stoic python]

Sascha Baer:

[broken hawk]

(it is a ring where matrix multiplication is the ring multiplication, but that is not useful here since det does not respect addition)

I guess the sensible way to think about it is:
\begin{itemize}
\item det is a mulitlinear map $(\R^n)^n \to \R$
\item restricted to invertible matrices, it is also a group homomorphism between $(\mathrm{GL_n}, \cdot)$ and $(\R^*, \cdot)$.
\end{itemize}

[stoic python]

Sascha Baer:

[heady dagger]

Oh, I see! So there aren't two separate definitions - the domain and codomain restrictions are just used to be able to use/study the properties of groups. That's where I got confused. Thanks!

[eager kestrel]

Ping if anyone can help me do this transformation! 🙏🙏

Basically rotate the coord. Sys. In pic to match regular one, using Householder and angle rotation!

[void minnow]

https://cdn.discordapp.com/attachments/359052604149465088/707258699768659978/unknown.png

You guys got any clue about this one? ;-;

[distant granite]

which question a or b ?

@void minnow

we've done the proof to the same problem in my uni today

have u tried to do it on your own yet ?

[void minnow]

both of them

yeah, i've been pretty much looking at it trying to like do anything for the past two hours now

absolutely no clue to it :c

[severe magnet]

what do u think those $(a_i)$ represent?

[stoic python]

Nabil:

[severe magnet]

I mean it should be obvious looking at the linear combination

if u understood that I think it should be clear how to construct the proof

[void minnow]

ai is the i-th column of the matrix A?

[distant granite]

yes

but i don't see how could that help @severe magnet

in the exercise we've done we didn't take the same approach

they gave us the solution form

and they told us to prove it's that one

which is this form

[severe magnet]

well what I wanted to get at was something else

have u done linear transformations?

that proof is basically showing that every linear transformation can be represented by matrix multiplication (speaking of finite vector spaces)

[distant granite]

nope

i see

i can't answer the questions then

[severe magnet]

u can start by showing that Ae_1 = a1e_1 right

then u know that every vector x can be written as a linear combination of basis vectors e_i in V

since th elinear combination is unique

u can use that to show that the linear combinations of the columns is unique proving (a)

[distant granite]

can i ask you a quick question real quick ?

[severe magnet]

alright

[distant granite]

the form of every solution is this right ?

of a System of linear equations

https://media.discordapp.net/attachments/540211747613704221/707267437195952138/unknown.png

[severe magnet]

they want u to show that every linear system is a affine subspace of a homgenous linear system + x_o right

[distant granite]

yes exqctly

exactly*

[severe magnet]

hm quite odd that u have done linear systems without having defined linear maps

since there is a theorem u can use to prove this

[distant granite]

i can send u a pic of the proof

i don't really get it

or do you have a source on the internet

?

[severe magnet]

I could tell u how to prove it in terms of linear transformations

nope I dont rn

[distant granite]

can i send you the proof and u tell me if this proof uses linear transfomations ?

[severe magnet]

sure

[distant granite]

becquse i m hqving the course in german

[severe magnet]

i am german

[distant granite]

haha nice

this is the first part

and this is the second

the Aufgabestellung

i forgot about it

one second

[severe magnet]

what exactly confuses you tehre?

[distant granite]

i don't know what is $x_{0}$

does latex work ?

[severe magnet]

u need to put two dollar signs

one at the beginning and one at the end

[stoic python]

Kasadraf:

[severe magnet]

alright so x_0 is the solution corresponding to the linear system Ax_0 = b

for b as a non zero vector

[distant granite]

so it's any solution

?

[severe magnet]

well the solution to that specific system

that is

v is the solution to the homgenous system

think about it this way

a linear system is Ax=b whereas x is the solution to the linear system

which is denoted here by x_0

nothing more

[distant granite]

oh thanks i see now how that works

[severe magnet]

no problem

once u introduce linear transformations it should be clear why this holds

[distant granite]

so they didn't use it here right ?

cuz we didn't define it

[severe magnet]

no

they just used the definition of linear systems

[distant granite]

oh ok thanks again

[severe magnet]

anytime

[hollow finch]

So my geometric argument for part a would be that P1x+P2x us just writing x as the sum of orthogonal vectors, which is always possible. But I'm struggling to come up with a concise proof of it. The one I came up with wasn't very convincing imo.

$(P_1+P_2)q_1=q_1q_1^Tq_1+q_2q_2^Tq_1=q_1$

$(P_1+P_2)q_2=q_1q_1^Tq_2+q_2q_2^Tq_2=q_2$

[stoic python]

nix:

[hollow finch]

So the sum has two independent eigenvectors of eigenvalue 1

Which I'm pretty sure can only be the identity for a 2x2

But I dislike that reasoning

If anyone could nudge me in a better direction, I'd appreciate it

[hollow finch]

<@&286206848099549185>

[gleaming marsh]

yo anyone has a second?

tiny doubt

Let W be the vector space of all polynomials in t. Moreover, for each k ≥ 1,
let fk(t) = 1 + t + · · · + t_k.
For any n ≥ 1, let Hn be the subspace of W spanned by the polynomials f1(t), f2(t), · · · , fn(t), i.e., Hn = Span{f1(t), f2(t), · · · , fn(t)}. Find the dimension of Hn.

do I have to prove that f1(t), f2(t) ... fn(t) are linearly independent?

[pallid rampart]

You don't necessarily need to prove that they're linearly independent

You can do it that way but it's obviously not the only way

But if you prove that they're linearly independent then you're basically done

[gleaming marsh]

since f1t is [1,0,0,...], f2t is [1,1,0,0,...], so on till fnt is [1,1,1,1,...,1]. This makes them linearly independent right @pallid rampart?

[pallid rampart]

Well you need to prove that those are linearly independent

[severe magnet]

anyone who has a helping hand proving this?

if I use G on both sides I would get $G(w) = \sum_{i=1}^k \alpha_i G(v_i)$ and would need to show that the vectors on the right side span the eigenspace of G

[stoic python]

Nabil:

[severe magnet]

that's the only attempt I have

[severe magnet]

<@&286206848099549185> anyone who can help?

[severe magnet]

its a good attempt to always start with a linear combination

when talking linear independence

[viscid kernel]

What is the difference between a columnvector and a rowvector. I always thought that only the columns of a matrix represent the vectors.

@real plaza actually thats what made me confused even more. So the columns of a matrix dont always represent the vectors ?

Ahhhhh

So its a matter of interpretation, you could also consider your rows as the vectors or the columns

@real plaza dont mind my question, I was gonna delete it xd

[severe magnet]

well linear dependence means that there is a nontrivial linear combination a_1v_1+....+a_nv_n=0 right

[distant granite]

what is the name of such a matrix ?

[wintry steppe]

Augmented matrix

I think

[distant granite]

yeah but there is a 4 th variable

that has nothing to do witrh the others

[wintry steppe]

Ok but that 4th variable isn't like the other 3

[severe magnet]

well I said linear dependence means there is a nontrivial solution meaning v_i != 0 and a_i!=0 to some extend

[distant granite]

@wintry steppe yes

[wintry steppe]

The other 3 are being multiplied by the 4th variable

[distant granite]

a coefficient

[severe magnet]

you want to check whether u find a nontrivial solution to the linear combination

[distant granite]

r is coefficient

[cold topaz]

Are the answers to these types of problems unique? Ignore about the part that says "using the Example x method..."

[wintry steppe]

yes r is a coefficent

[distant granite]

so how do i proceed ?

[wintry steppe]

x y z aren't

well thats just a system of eqautions. its not even a matrix

if you want it to be a matrix you have to put it in matrix form

[distant granite]

i know but like after doing that i need to solve the matrix according to the value of r

but i can't seem to find another apparent value of r rather than 0

[wintry steppe]

r cant be 0

cuz youre multiplying every variable (x, y, z) with r

and if you multiply a variable with 0 u get 0

r is 1

and one of x, y, z is 1, while the other are 0

[distant granite]

r is a real number it can take any value in R unless you mean when r is zero the matrix has no solution right ?

choosing r as 1 seems like a coincidence

i need to know how do i find these apparent values

i need to know how do i find these apparent values

anyone able to help please ?

<@&286206848099549185>

[wintry steppe]

if r = 0, then all the answers to the equations will be 0

[distant granite]

ok what about every other possible value of R ?

[wintry steppe]

for every other value of R thats not 1, you'll get an equation where r*(either x,y,or,z) equals a number greater than 1

[severe magnet]

@real plaza vectors in a linear combination are linear independent if the coefficients are 0 that's what u probably mean by only one solution (which is equivilent to what I said above)

well u should index them differently

two different eigenvalues corresponding to the eigenvectors

[wintry steppe]

say x = 1, and y = z = 0
for the first equation, you need a value r, such that r*x + y + z = 1
what would your value of r be in this case?

[distant granite]

r needs to be 1

[wintry steppe]

exactly

now apply that to the 2nd and 3rd equations

x + r*y + z = 1; if r = 1, then you get 1 + 1(0) + 0 = ?

fill in the ?

[distant granite]

it's 1

but here you chose r = 1 i need to solve this in a formal way

[wintry steppe]

yup

[distant granite]

i ll send you a screenshot

so that you get what i mean,

[wintry steppe]

choose a different value of r and see if my method works

hint: it won't work

[distant granite]

i know it works but i need to write a formal solution wait a sec i ll show you

look here

so this one obviously does not have any solution

for those fixed values of a

it's apparent

in the matrix i ve given it's not that clear what the values of r should be

if i could write it in a formal way or i reduce the matrix

it will be easier to see which values are possible for r

but thanks a lot anyway

[wintry steppe]

shouldnt your matrix have 3 rows

[distant granite]

no i just gave another example

[severe magnet]

um you want to check v and w for linear independence right?

might wanna edit that abit

a little thing to point out is also that you work with different eigenvalues

so rather $c_1 \lambda_1 v + c_2 \lambda_2 w = 0$

[stoic python]

Nabil:

[distant granite]

how do i write a formal solution for this equation ?

$x+y+z=1$

[stoic python]

Kasadraf:

[limber sierra]

not sure what you mean, there are infinitely many solutions and that equation expresses them all

do you want, like, a basis for the solution space? where solutions are vectors of the form $\begin{pmatrix}x\y\z\end{pmatrix}$

[stoic python]

Namington:

[quiet moss]

Is this channel in use? I just want to have someone look at one of my answers on an already marked test I am 99% sure my prof marked it incorrectly but just don’t want to look stupid if I fight it and am wrong.

[limber sierra]

go ahead

[quiet moss]

Actually taking a second look it is definitely wrong how can t=6 give you infinite and no solutions

Whoops that’s does not =6

[limber sierra]

does the system have infinitely many solutions for t = 1?

er

let me rephrase

how many solutions does that system have for t = 1?

[arctic shard]

Sorry for interruption but can someone (eventually) help with this question. Let $V$ be the volume of the parallelepiped spanned by the basis $e_1,e_2,e_3$ and let $V'$ be the volume of the parallelepiped spanned by the reciprocal basis $e^1,e^2,e^3$. Prove $VV' = 1$

[stoic python]

Joshua S:

[limber sierra]

anyway, the point is paolo

you only addressed 3 possible values of t

0, 3, and 6

rather than, you know, all possible values of t

you are correct that there are only infinitely many solutions when t = 6

in every other case, there's no solutions

so your answer for no solutions should be $t \neq 6$

[stoic python]

Namington:

[quiet moss]

No I see it now

Thank you

[cold topaz]

is this a correct way of writing eigenvector using math language?
(λI - A) x = 0

[wintry steppe]

yah

actually, its (A- lambda(I))x = 0

[cold topaz]

my book has it the other way around.

[wintry steppe]

both ways are correct

i just googled it and what i said came up first but either way is not wrong

[severe magnet]

https://cdn.discordapp.com/attachments/540211747613704221/707317121553793074/ajsdfbgliasjdhfoiasduf.PNG anyone here who could help me on this?

[cold topaz]

when λ= 1, -2, -2 , do we say the eigenvalues are 1, -2, -2, or just 1 and -2?

[wintry steppe]

uh i think the first thing

[cold topaz]

somebody said the latter in a different server lol

[severe magnet]

U would say the latter

Eigenvalues span open corresponding eigenspaces which contain vectors depending on the eigenvalue

Vectors with the same eigenvalue are linearly dependent

[viscid kernel]

What is a direct sum

[arctic shard]

Sorry for posting this again but I think it’s getting lost in the chat. Can someone help me with this question? Let $V$ be the volume of the parallelepiped spanned by the basis $e_1,e_2,e_3$ and let $V'$ be the volume of the parallelepiped spanned by the reciprocal basis $e^1,e^2,e^3$. Prove $VV' = 1$

[stoic python]

Joshua S:

[hollow finch]

If we know a 2x2 matrix has two eigenvectors with eigenvalue 1, does it have to be the identity? Does it change if we know it is symmetric?

[wintry steppe]

Yah it's the identity matrix

And what do u mean with your end question

[hollow finch]

I wasnt sure if being symmetric or not decides the question. I just need to figure out how to prove it now.

[wintry steppe]

I think it decides the question, if I'm understanding correctly

[meager bolt]

would at^6 be in the subspace of P^6? I think it’s yes because it contains the 0 vector (t=0) and you can add and multiply and it would still be a 6th degree polynomial right?

[wintry steppe]

@real plaza you should have (1 -1) as your first row

even still, you should have (1 -1) as your top row

because the first entry of the matrix should pretty much always be 1 if possible

in this case it's possible so you should make it (1 -1)

so your left with 1x -y = 0, which equals x - y = 0

so x = y

so your first eigenvector should be
[1
1]

@real plaza

it's [ 1 1] because when you reduce x and y, it becomes [1 1]

for example, if you set x = y = 6, you'll get [6 6]

but [6 6] can reduce into [1 1]

no that's not wrong

you can pick an arbitrary value but you have to reduce it after

so its just easier to choose [ 1 1]

it would but its a rule of thumb that when you can reduce, you should reduce

[meager bolt]

yo @wintry steppe would at^6 be in the subspace of P^6? I think it’s yes because it contains the 0 vector (t=0) and you can add and multiply and it would still be a 6th degree polynomial right?

[wintry steppe]

i think thats correct

[wintry steppe]

yes

thats your change of basis matrix

[gray dust]

@meager bolt the proper way to phrase that question is "is the set S={at^6 | a is any scalar} a subspace of P_6?" yes it is but your reasoning doesn't support any of it. S contains the 0 vector because "a is any scalar" covers the case when a=0, giving the 0 polynomial, and so S is nonempty. you can easily show that the sum of any 2 vectors in S as well as any scalar multiple of a vector in S is of the form scalar*t^6 and therefore S is closed under addition & scalar multiplication. THEN you can say S is a subspace of P_6

[prisma drift]

So I'm supposed to be finding all values of d which make the matrix A positive definite. I know matrices are positive definite when their eigenvalues are also positive. The issue I'm running into is that I have this gross cubic equation. I think I might have to use the cubic discriminant... Can you guys think of any alternate route I could take

[eternal finch]

Did you learn about the relationship between the pivot elements and the positive definiteness of the matrix?

[severe magnet]

If I have two diagonisable, linear maps $F,G: V \to V$ with $F \circ G = G \circ F$ and pairwise distinct eigenvalues $\lambda_1,...,\lambda_k$ and $\mu_1,...,\mu_j$ and I've already proven that $F(\text{Eig}(G,\mu_j)) \subset \text{Eig}(G,\mu_j) $, what can I say about $F|_{\text{Eig }}:\text{Eig}(G,\mu_j) \to \text{Eig}(G,\mu_j)$ denoting the linear map restricted to the domain of all the eigenvectors of G with eigenvalue $\mu_j$? Can I say the map is injective/surjective?

[stoic python]

Nabil:

[severe magnet]

since that would help for a proof alot

but I feel like it's too fast saying that considering we only have a subset relation

[eager burrow]

Without any more information, F could be zero on some eigenspace of G, so nah, neither injective nor surjective

e.g. V one-dimensional, F = 0 and G = id

[severe magnet]

hm damn ur right

quite hard to do the proof to the exercise but maybe I'm missing something

mainly this

[severe magnet]

<@&286206848099549185> maybe

[severe magnet]

anyone?

[wintry steppe]

Hello, my textbook asks "What does x^2 + y^2 + z^2 <= 3 represent?"

When I try to graph the equation in GeoGebra, nothing shows up, why is that ?

[quartz compass]

can geogebra graph 3d graphs?

[wintry steppe]

https://www.geogebra.org/3d?lang=en

It works fine, up until I add the z^2, for some reason.

I just started linear algebra, so I'm not sure what's up.

[quartz compass]

just graph it with = instead of <=

<= would be graphing like a solid object