linear-algebra

[lavish jewel]

and what eigenvalues and eigenvectors are

[wintry steppe]

yes

[lavish jewel]

what's the definition

[wintry steppe]

@lavish jewel if you can find values to build a vector from others

[lavish jewel]

you're gonna have to write it down as an equation

[wintry steppe]

like a_1x_1 + ... + a_nx_n = 0 with a_1 = ... = a_n = 0

[lavish jewel]

right, that means that the x_i are linearly independent

[wintry steppe]

yes

[gray dust]

"with" should be implies

[lavish jewel]

now notice that that thing you wrote is equivalent to the matrix-vector product $[x_1 x_2 \dots x_n]\begin{bmatrix} a_1 \\ a_2 \\ \vdots \\ a_n\end{bmatrix}$

[stoic python]

Edd

[lavish jewel]

where the x_i are vectors

[wintry steppe]

yes

[lavish jewel]

if the columns are linearly independent, then the system Mv = 0 only has the solution v = 0

if the columns are linearly dependent, then Mv = 0 has nontrivial solutions for v

you could then do elimination and find that the solution is parametrized in terms of free variables

those solutions are a basis for the null space of the matrix M

and they have the property that Mw = 0, for any such vector w in the null space

which is the same as saying Mw = 0w

or in other words, the vector w has an associated singular value of 0

all as a consequence of M having linearly dependent columns

and trivially, Mw = sigma w, with sigma nonzero, means w is a singular vector of M with a singular value sigma

beyond that, i'll just direct you to the rank-nullity theorem

[wintry steppe]

@lavish jewel i mean you have that the rank is the dim of M - dim ker f

[lavish jewel]

now use that and what i just explained

[wintry steppe]

you remove the number of vectors that build Mw = 0

[lavish jewel]

i don't understand what you mean. remove them from what?

[wintry steppe]

i don't get this part

why having Mw = 0w => M having linearly dependant columns

[lavish jewel]

you told me yourself

that's the definition of linear dependence

if w is nonzero and we can still get Mw = 0 = 0w, then M must have linearly dependent columns

since Mw = 0 is the same as the definition of linear dependence that you gave me

[lavish jewel]

here

[lavish jewel]

same as this

just call M = [x1, x2, ..., xn]

[wintry steppe]

wait i'm lost

I said that but for saying x_i were independant

not dependant

[lavish jewel]

if the a_i can be nonzero, then that means the x_i are linearly dependent

i mean

if it's not linearly independent, then surely it is linearly dependent

i don't think i will be able to explain this to you, i would strongly recommend you just start over from the beginning

or maybe someone else can explain it here

[wintry steppe]

ok got it

[wintry steppe]

until that it's ok

so based on the nullity theorem, rank f = dim E - dim ker f => rank f = number of independant vectors => Mw = sigmaw with sigma != 0

@lavish jewel so if rank(f) = n, you can find n sigma and w to have Mw = sigmaw with sigma != 0 ?

[lavish jewel]

yes

[stable urchin]

k for the sake of conceptual understanding

take a columnspace of some matrix A

say the rref of A is equal to some matrix B

so is it true that the basis of the rowspace of A is equal to the basis of the rowspace of B

since the rowspace of B is equal to the rowspace of A

[lavish jewel]

no

the same subspace can have many bases

as a simple example, [1,0] and [0,1] span R^2, and so do [1,0] and [1,1]

[marble lance]

Being a basis for the one is the same as being a basis for the other, since they are the same. The main issue is the use of "the basis" since there is no one basis.

[lavish jewel]

the operations you do in RREF are the same as what i did there, just adding (scaled versions of) the basis vectors together in some way

for the common cases of R^n and C^n, any subspace has infinitely many bases

except {0}

[marble lance]

My snarky {0} reply was sniped

[stable urchin]

i know about the infinite basis thing i just forgot about that when asking the thing

i meant

are the sets of bases for row(A) and row(B) equal

[marble lance]

You started off talking about a column space, so btw, this is not true for column spaces since col(A) ≠ col(B) in general

[lavish jewel]

wdym sets of bases

[marble lance]

row(A) and row(B) are the same thing. Everything about them is the same.

[stable urchin]

yeah i know

[lavish jewel]

it's the same subspace, you just found two different bases for the same subspace

[stable urchin]

okay cool

[wintry steppe]

In any vector space ax=ay implies that x=y . Is This statement True ?

[marble lance]

@wintry steppe no

[wintry steppe]

Why ?

[marble lance]

0x = 0y for any x and y

If a is nonzero, then yes

[wintry steppe]

Okaay ty !

[stable urchin]

How do I find a linear transformation $T:\mathbb{R}^5\rightarrow\mathbb{R}^3$ such that
$$
\text{nul}(T)=\text{span}\left{\begin{pmatrix} 1 \ 1 \ 0 \ 0 \ 1 \end{pmatrix},\begin{pmatrix} 1 \ 0 \ 0 \ 0 \ 1 \end{pmatrix}\right}
$$

[stoic python]

timmmm

[stable urchin]

I'm just finding it hard to work from a nullspace

like I understand that from here you can say that {x1+x2+x5=0,x1+x5=0}

but like I'm tryna think of how to get a 3x5 coefficient matrix from that

[marble lance]

You can uniquely define a linear transformation by describing what happens to the basis vectors of your domain. So you can extend those two vectors to a basis of R^5, send those two vectors to 0 and the others to something nonzero.

[stable urchin]

alrighty so uhm

it may be nearly 5 am and things are having a hard time penetrating my brain

but is there a way in which you could elaborate on that just a bit more

[marble lance]

Can you give me a basis of R^5 that contains those two vectors?

[lavish jewel]

note that you were not given enough info to write a unique transformation T, you'll have to make some stuff up

[stable urchin]

{e1,e2,e5}

[marble lance]

That's not a basis of R^5

A basis should have 5 vectors

[stable urchin]

ah okay so a basis would be {e1,e2,e3,e4,e5}

[marble lance]

Yes, but I want a basis that has as two of its basis vectors, those two vectors

[stable urchin]

okay, {<1,1,0,0,1>,<1,0,0,0,1>,e3,e4,e5} i think

oh wait

[marble lance]

Yeah that works

[stable urchin]

ooh fun

[marble lance]

Now because I'm lazy I will call those b1 b2 b3 b4 b5 in the same order you listed them

So you can define T(bi) = ci for any ci in R^3 and that will define a linear transformation, do you agree?

[stable urchin]

yes

[marble lance]

We need T(b1) = T(b2) = 0 so that b1 and b2 are in the nullspace

So now it's just about sending the other three to something else so that no linear combination of them can map to 0

Except the zero lin comb

So 0 ≠ T(a3 b3 + a4b4 + a5b5) = a3 T(b3) + a4 T(b4) + a5 T(b5) unless a3 = a4 = a5

And that's the same as saying T(b3), T(b4) and T(b5) should be lin independent

[stable urchin]

yep

[marble lance]

So you can map them to any three linearly independent vectors

And then that transformation is what you want

[stable urchin]

oh okay I'm just gonna take a sec to work with that

thanks for the guidance

[marble lance]

👍

[stable urchin]

so would this new basis of dimension 5 be the range of the transformation?

[marble lance]

The basis is the basis for the domain

So it is unrelated to the range

[stable urchin]

oh right that makes sense

[marble lance]

The basis is in R^5, the range is in R^3

[stable urchin]

right so I came up with the transformation
$$
T\begin{pmatrix} x_1 \ x_2 \ x_3 \ x_4 \ x_5 \end{pmatrix}=\begin{pmatrix} x_1+x_2+x_5 \ x_1+x_5 \ x_1+x_5 \end{pmatrix}
$$

[stoic python]

timmmm

[marble lance]

But T(1,0,0,0,1) = (2,2,2)

[stable urchin]

oh shoot

[marble lance]

Look over your work again, maybe you made a mistake when converting it to this form

[storm fractal]

how do they calculate this? (is this the determinate or something different)

[stable kindle]

determinant

[storm fractal]

nvm

[lavish jewel]

looks like they found the cofactors for the first row

since one has to multiply the minors by the corresponding entry in the matrix, this immediately means you're adding 1*(some minor) + 0 + 0

and the minor should be (-1)^2 det([-1, 4; -2, 8])

which comes out to 1*(-8 - -8)

[winged prairie]

does anybody understand why F^n can be seen as a function, this is from linear algebra done right

like i thought F^n is just a set

[small vigil]

Could you explain the S-lemma? What is it?
Are there any intuitive explanations?

[stable kindle]

F^n is basically the same as the set of all functions from n to F

for example, take a function f from {1, 2, 3} to R

then:
f(1) might be 3
f(2) might be 5.5
f(3) might be -1

this function f corresponds to an element in R^3

the element (3, 5.5, -1)

[winged prairie]

ok makes sense

ty

[terse urchin]

For simplex method, how does slack variables transform inequalities into equalities? it makes like no sense to me

[lavish jewel]

if, say, x < y, then one could add some c so that x + c = y

and this c needs to be >= 0

[floral thistle]

Good morning

[pseudo elm]

Yo I need help

[lavish jewel]

what do you need help with

[pseudo elm]

How is that a2 = [a][a] = [9 8 9] ..

[lavish jewel]

do you know how matrix multiplication works?

[pseudo elm]

Explain it to me

[pseudo elm]

;)

Come on edddddddd

Explain it to me

○_○

M waiting

[lavish jewel]

this image is the only thing i'll provide you with

google is a thing, and you presumably have a book and/or course material as well

[pseudo elm]

Ok done

I understood

Thanks

[lavish jewel]

otherwise, go to the help channels instead

[pseudo elm]

I got the ans

Arigato

[heavy crown]

I don't understand it but I believe it should be simple, help ?

[winter harbor]

Let $u \in T^{\perp}$. Then, $\forall v \in T$ we have that $\langle u,v \rangle = 0$. In particular, since $S \subseteq T$, we have that $\forall v \in S$ it is the case that $\langle u,v \rangle = 0$, thus $u \in S^{\perp}$ and $T^{\perp} \subseteq S^{\perp}$.

[stoic python]

MISTERSYSTEM

[winter harbor]

Was just a matter of using the definition of orthogonal complement.

[heavy crown]

Your proof and explanation made it clear for me to understand it, thank you very much (:

[winter harbor]

Np!

[fickle citrus]

Hello all, suppose if I have
$$Av_{1}=v_{1}$$
and
$$Av_{2}=v_{2}$$
then what conditions do I need to say
$$v_{1}=c\cdot v_{2}$$
for $c\in\mathbb{R}$? In other words, Is there a way I would know they share an eigenvalue and indeed the multiplicity of this eigenvalue is hopefully $1$

[stoic python]

ShatteredSunlight

[fickle citrus]

Hmm, I think in general it's not possible?

[teal grotto]

one sufficient condition is that eigen space corresponding to eval 1 is one dimensional

this guarantees that the geometric multiplicity of eval 1 is 1, but not necessarily the algebraic multiplicity

[fickle citrus]

Yeah, indeed, I require the geometric multiplicity to be 1

I'm not sure if I can show that, so I will have to do my problem using high-power theorems I don't comprehend again...

[teal grotto]

this one shouldnt need any other heavy machinery

[lavish jewel]

the original question is a bit weirdly formulated

since v1 = c v2 has nothing to do with the matrix

you could simply normalize the vectors and then use cauchy-schwarz to show whether they are parallel

[fickle citrus]

FWIW I'm doing the doubly stochastic matrix proof of stationary distribution (yet another proof)

A lot of the online proofs basically 'consider solution' and say it is the solution by uniqueness.

For me I don't like how it comes about, so I wanted to see how much I can push it.

The only additional thing I could add is that the vector of all 1s is linearly dependent to the final stationary distribution, given by
$$\mathbf{1}^{T}P=\mathbf{1}^{T}$$
under bistochasticity and
$$\pi^{T}P=\pi^{T}$$
under existence of stationarity

Then Perron-Frobenius says the there is only one eigenvector with all positive entries, so I finally get
$$\pi = c * \mathbf{1}$$

[stoic python]

ShatteredSunlight

[fickle citrus]

In essence I'd like a more algebraic a priori result of the stationary distribution rather than 'guess and check'

And no Perron-Frobenius if I can help it, since I didn't prove PF

[ionic laurel]

can anyone explain what the minimum rank of any matrix can be?

[hard drum]

Will need more context lol but the true minimum across all possible matrices of any size is 0 since dimensions of vector spaces are nonnegative

[ionic laurel]

well

in the context of an nxn, mxn, nxm i mean to say

like the maximum rank of a matrix cannot extend beyond the # of columns and # of rows

on the other spectrum

what would the minimum rank be

would it be dependent on the type of matrix?

[ionic laurel]

i see

[teal grotto]

not dependent on the size of the matrix. any n by m matrix with every entry zero has rank zero. it may depend on the properties of the matrix. for example, minimum rank of all symmetric matrices is still 0, but minimum rank of n by n invertible matrices is n

[ionic laurel]

ok

[teal grotto]

?

y is a function of x. there is no fixed value for y

y = 2x is saying that given x, y is going to be twice the value of x

x is not by default 0, whatever that means

should be y = 6 if y = 2x

this is not a linear algebra question btw. but still not clear what your question is.

[teal grotto]

yes, it’s redundant to write y = 4x + 0

[nocturne jewel]

btw, this isnt linear algebra

[stable kindle]

google matrix inverse

[prisma sail]

oh

[nocturne jewel]

Im gonna guess it's wrong

[prisma sail]

im dumb alright

[nocturne jewel]

since that matrix isnt invertible

[prisma sail]

huh

[nocturne jewel]

6 6
1 1

This isn't invertible

[prisma sail]

right

[nocturne jewel]

so its inverse doesnt exist

[prisma sail]

not sure how to go about this problem

the diagonal is right but not the rest

nvm nvm got it

[wooden iris]

can someone simplify this explanation to why the matrix is linearly independent

[prisma sail]

columns aren't scalar multiples of each other

and you can see if augment that matrix with the 0 vector
there will be a pivot in each non augmented column thus has only the trivial solution

note that it can have atmost rank 3 because of the number of columns

so that 0 row is ignored there are no free variables

hopefully that helps im not too good at linear algebra

[wooden iris]

thank you no it definitely helps!

[solid cobalt]

I need to find the the matrix T, B2_T_B1

I know how to perform the change of base with real numbers but how do I compute this since it involves complex numbers?
Is it the same process as performing the Gram Schmidt process on complex space?

I have already shown that B2 is an orthonormal basis for C2
although I do not understand why we were asked to show that B2 is an orthonormal basis for C2 in the first place

any explanation will be greatly appreciated

[winter harbor]

What do they mean by T_B2, T_B1 ? Is it asking us to find the matrix of T with respect to the basis B2 and B1?

[solid cobalt]

Yes, this is what they mean

Transforming from one basis to another

I don't know if I phrased that correctly

hopefully you understand what I'm trying to say

[winter harbor]

Alright

So, the first column of the matrix T_ B2,T_B1 will be given as follows

I suppose (1,2) is written in the standard basis of C^2

So what we have to do

Is find those coordinates

With respect to basis B2

This means that we must find constants a,b

For which:
$$
\begin{bmatrix}
1 \
2
\end{bmatrix}

a
\cdot
\begin{bmatrix}
1/\sqrt{2} \

• i / \sqrt{2}
  \end{bmatrix}

b \cdot
\begin{bmatrix}
(1+i)/2 \
(-1+i)/2
\end{bmatrix}
$$

[stoic python]

MISTERSYSTEM

[winter harbor]

So we basically want to write (1,2) = a* w_1 + b *w_2

For some constants a,b

These constants

Are the coordinates of (1,2)

With respect to basis B_2

These coordinates

Will be the first column of the matrix we are trying to find

Are you fine with this?

[solid cobalt]

I see
do I do the same process for the other columns?

[winter harbor]

Yup!

Now

You will see why B2 being orthonormal is so useful

Notice that, a priori, we would have to solve 3 different systems of linear equations

But since B2 is orthonormal

We don't have to

Notice the following

Take the inner product with respect to w_1 on both sides

We will be able to simplify things down as follows

$
T(v_{1}) = a w_{1} + b w_{2}
$
This implies that:
$$
\langle T(v_{1}), w_{1} \rangle = a \langle w_{1}, w_{1} \rangle + b \langle w_{2}, w_{1} \rangle
$$
But $B_{2}$ is orthogonal, meaning that $\langle w_{1}, w_{1} \rangle = 1$ and $\langle w_{2}, w_{1} \rangle = 0$. So in fact we have that:
$$
\langle T(v_{1}), w_{1} \rangle = a
$$

[stoic python]

MISTERSYSTEM

[winter harbor]

So in fact, what we found out is that in order to calculate the first coordinate of T(v_{1}) with respect to B2

All we have to do is take the inner product of T(v_1) with w_1

We can do the same calculation to find the second coordinate of T(v_1)

Just take the inner product with respect to w_2 on both sides

[solid cobalt]

mindblown
Thank you so so much for that explanation, It was spot on

[winter harbor]

Np

Just do the same thing to find the other coordinates

Take the respective inner products and so on

[solid cobalt]

will do
thank you so much again
hope you have a great day!

[worldly bear]

why are x4 and x5 0 here

do you take like -13/6 * x5 = 0 or?

[wintry steppe]

I get that the eigenvectors of a matrix are the vectors that don't get rotated, only get scaled so they're always on their span and the eigenvalues of that matrix are the magnitude of the eigenvectors

but

how do I compute them for a given matrix?

and eigenvalues an eigenvectors should exist for non-square matricies right?

[worldly bear]

eigenvalues are only computable for square matrices

If A is a n x n matrix, you can find its eigenvalues by solving the det(A - λI) = 0

where I is the identity matrix

[wintry steppe]

what's lambda?

[worldly bear]

just a variable

eigenvalues are typically denoted by lambda

[wintry steppe]

ah so you solve for lambda to get the eigenvalue

[worldly bear]

exactly

and those are the values for which the matrix A is scaled on its span

[winter harbor]

First of all, it doesn't make sense to talk about eigenvalues and eigenvectors for non square matrices, because notice that given a $n \times n$ square matrix $A$ over a field $\mathbb{K}$ we define a non zero vector $x \in \mathbb{K}^{n}$ to be an eigenvector of $A$ if $\exists \lambda \in \mathbb{K}$ for which:
$$
Ax = \lambda x
$$
Notice that this makes sense, because since $A$ is a square matrix, it sends vectors with $n$ entries to vectors with $n$ entries as well, so that asking $Ax = \lambda x$ makes sense. Notice, however, that if $A$ is an $m \times n$ matrix with $m \neq n$, i.e it is a non square matrix, we have that $A$ sends a vector $x$ with $n$ entries to a vector with $m$ entries, so that $Ax$ is a vector with $m$ entries and $\lambda x$ is a vector with $n$ entries, so asking for:
$$
Ax = \lambda x
$$
Doesn't make sense.

[wintry steppe]

ahh okay got it

thanks

[stoic python]

MISTERSYSTEM

[wintry steppe]

so once you find lambda, would you like do

(A - λI)V = 0

to get the eigenvectors?

[winter harbor]

Yeah, once you have λ you basically have to solve the system of equations Ax = λ x where x is unknown.

[wintry steppe]

ah right

[worldly bear]

can a set of vectors give a basis for a certain space if the set of vectors is < the dimension of the space?

they can right, but not if there are more vectors than the dimension of the space?

[wintry steppe]

I think it's the other way around

[blissful trout]

I have been struggling with my math homework, so basically I have to solve systems of equations using the Gaussian Elimination Methtod but it’s been making my head hurt trying to do it

Can I send a picture of it?

[worldly bear]

[winter harbor]

No lmao

Like

The dimension of a vector space is defined as the cardinality of a basis set

The reason this concept is well defined

is because the cardinality of any basis you take is the same

It is unique

[worldly bear]

so for a set of vectors to be a basis, we have to have the same amount of vectors as the dimension

[winter harbor]

that's one of the requirements, yes.

[worldly bear]

and even if we do, it may or may not be a basis

yes

[winter harbor]

But they also need to be linearly independent.

[worldly bear]

yes i understand all that

lin independent and span

so its not worth checking if a set of vectors form a basis if i have less than the dimension of the space, how would i denote that to my professor

[winter harbor]

What's the exercise, exactly?

[worldly bear]

,rotate

[worldly bear]

a is not a basis bc it is linear dependent

b has more vectors than the space so it’s also not a basis

c has less so i thought i had to check it for linear independence and spanning but you say i don’t have to since it can’t be a basis

and d is a basis since it’s linear independent, don’t need to check spanning since we have 4 vectors in R^4, so spanning is given

[winter harbor]

You can explicitly show that at least some vector in R^4 can't be written as a linear combination of vectors in the set described by (c)

If you don't want to use the fact that dimension is well defined

also

I am not sure if you professor has discussed the definition of dimension and proved that it is well defined + every vector space has a basis

So checking things out using the definition would be a good exercise

[worldly bear]

well defined?

[winter harbor]

By well defined I mean the following

Given a vector space

We define its dimension as the cardinality of a basis set.

Now

There's two things we have to be careful here

First of all

What guarantees us that every vector space has a basis?

Maybe there are vector spaces which do not admit basis, so we'd have to verify that.

Turns out every vector space has a basis, so we are good for now.

But there's also another problem we have to check

Because even if every vector space has a basis

What guarantees us that a vector space might admit two basis with different cardinalities? It's not obvious at first, and you explicitly made a question regarding this.

And turns out that the cardiniality of any two different basis sets of a vector space are the same.

So we can, first, guarantee that every vector space has a basis and the cardinality of any two basis are the same. Thus talking about the dimension of a vector space makes sense, or in mathematical jargon, it is well defined.

[worldly bear]

why are the cardinality of any two basis the same?

[winter harbor]

It is not straightforward

For finitely generated vector spaces

[worldly bear]

just useful to know?

[winter harbor]

This is a consequence of what is called the steinitz exchange lemma

And the proof is quite nice

It doesn't use anything fancy

You can even prove this result using stuff from system of linear equations

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

The wikipedia article has a proof of this

[worldly bear]

i’m gonna look into it after i finish these review problems i think

[winter harbor]

For vector spaces which are not finitely generated

We need more tools to prove this

It involves something called Zorn's Lemma (Or you can be fancier and use something weaker like the Ultrafilter Lemma but shhh)

Which usually isn't covered in a first year course

And is a tool from set theory/order theory

[winter harbor]

Yeah, it is definitely useful to know these more conceptual facts about linear algebra

[worldly bear]

i took differential equations before linear algebra and i’ve already seen a lot of the conceptual relationships between them

[winter harbor]

Differential equations before linear algebra ?

How's that a thing

[worldly bear]

dude i have no idea

i kinda struggled in it too bc he kept alluding to linear algebra and how stuff works but i didnt know what was going on

like for systems of diff eqs i was just mindlessly doing the computations bc i didnt understand what the point of finding these eigenvalues was

[winter harbor]

You have to do a lot of diagonalization and jordan canonical form machinery in an introductory differential equations course

a background in LA helps a lot

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

[worldly bear]

yeah it seemed like it would while i was taking it

[winter harbor]

Btw, here's the general theorem

This article has a proof for the infinite dimensional case

while the other one describes the finite dimensional case

[worldly bear]

im scared to even dip my toes in that

the infinite case

[winter harbor]

Are you doing engineering?

If you are doing physics

At some point you will have to

[worldly bear]

i am not

just a general mathematics major with a minor in comp sci

[winter harbor]

Oh nice

At some point you will have to get yourself used to this zorn's lemma machinery and so on

It's not as hard as it sounds

Plus, infinite dimensional vector spaces are as useful as finite dimensional ones

[worldly bear]

im sure ill cross that bridge at some point

[winter harbor]

they appear everywhere

[worldly bear]

yeah they seem more real life based then finite dimensional spaces

maybe im wrong though

[winter harbor]

Idk what real life based mean

But it's just that a lot of the vector spaces we care about in mathematics, in particular analysis, are spaces of functions which are in a major part infinite dimensional

Like the L^2 space, which is the main framework where one does fourier analysis

Or the space of real valued smooth functions on a manifold

or the space of continuous functions on the real line

Or the space of entire functions on the complex plane

and so much more

[worldly bear]

i appreciate your help and insight

[winter harbor]

[worldly bear]

@winter harbor i think i found the proof lol

[winter harbor]

What theorem is it referring to when it says "the above theorem"?

[worldly bear]

not sure, maybe one on a previous page

[winter harbor]

I hope so

It's prolly a version of Steinitz Exchange Lemma

[ionic laurel]

Is it possible for a finite set to span an infinite-dimensional vector space?

i am confused on that

there is a true or false that says

and this is false

but i am unable to understand how that is possible

if anyone could help that would be greatly appreciated

[teal grotto]

R^2 is spanned by all of its vectors, an infinite spanning set, but R^2 is not infinite dimensional as a vector space over R

[teal grotto]

no, but this is not the same question as the one in the picture you posted. a finite set spans a vector space with dimension the same as the number of linearly independent vectors in that set

which is always finite

[flint cloak]

@teal grotto hey i was wondering if u could help me on this problem

I pinged helpers multiple times but i think there's no one available rn

[teal grotto]

i can try. what problem

[flint cloak]

there's 2 problems but they're very similar

[ionic laurel]

Oh

Thank you for clarifying

[flint cloak]

do you know what the answer should be, im running out of time almost xD

i know the idea that for 1-1 u gotta get 2 polynomials to equal the 02x2 matrix but idk how to figure it out because the matrix given to us has all variables in each slot

so u cant even isolate for a degree

@teal grotto do you know how to do it

[teal grotto]

wow this is brutal

set a = -b and c = d = 0

just choose some random a, like a = 1

[flint cloak]

for 1-1?

[teal grotto]

ye

[flint cloak]

so <1,0,0> = [0,0,0,0]?

[teal grotto]

no, consider the polynomial x^3 - x^2

what is its image under T

[flint cloak]

0

if x = 1

[teal grotto]

what?

no, what is T(1*x^3 + (-1)*x^2 + 0x + 0) ?

[flint cloak]

idk

my brain honestly fried after 50 attempts

of different maps

r^3 p2 m2x2

lol

[teal grotto]

just set a = 1, b = -1, c = d = 0 and compute brah

[flint cloak]

so whats the other polynomial

that = [0,0,0,0]

[teal grotto]

the zero polynomial. all its coefficents are 0

[flint cloak]

what

so just T(0)?

[teal grotto]

yea

[flint cloak]

like this ^

[teal grotto]

yes

[flint cloak]

ah ok

and for onto

do u know if its onto or not

[teal grotto]

well, it wont be

[flint cloak]

and any y that satisfies

in matrix form

[teal grotto]

uhhh idk

[flint cloak]

would 1,1,1,1 work?

[teal grotto]

idk try it

[flint cloak]

i dont want to get it wrong because itll put me on a diff question then

lol

yolo

[teal grotto]

wait

[flint cloak]

alright

[wintry steppe]

is e^x the eigenvector of the derivative and the integral operator?

e^nx to be a bit more general

[flint cloak]

im going to just do {3,1,0,0}

worked

[teal grotto]

wow

actually insane

[teal grotto]

was that a guess

[flint cloak]

it was an educated guess

because

i just looked at row echelon

and just went w putting non zero on dependant basis vectors

just need help with this now

^

[teal grotto]

for injectivity, check if the matrices are linearly independent. if not, then its not injective.
for surjectivity, check if the matrices are spanning. if not, then its not surjective.
this is a general way to do this

[flint cloak]

yeah im just pressed for time rn otherwise i'd do it the normal way

[teal grotto]

use symbol lab or something and put the matrices in as column vectors to check linear independence. like, put them in a four by four matrix. if it has zero determinant, then you know the matrices are linearly dependent. otherwise, theyre linearly independent and your matrix is injective and surjective

[flint cloak]

yh just did that

this thing has rank 3 so its lin dependant

just need to find an example of not 1-1 now

lol

thats the hard part

and an example of not onto

do you know an example that could work

[winter harbor]

Yup, it is! Notice that if we have an open subset $U \subseteq \mathbb{R}$, we can think about the space:
$$
C^{1}(U) := {f : U \rightarrow \mathbb{R} , \vert , \text{f is differentiable with continuous derivatice} , }
$$
Notice then that the derivative can be seen as the operator:
\begin{align*}
D : & C^{1}(U) \rightarrow C^{1}(U) \
f & \mapsto \dfrac{d f}{dt}
\end{align*}
Which is a linear operator. Now, $C^{1}(U)$ is an infinite dimensional vector space. In any case, the definition of eigenvalues and eigenvectors stays the same, i.e we define a function $f \in C^{1}(U)$ to be an eigenvector of $D$ if $\exists \lambda \in \mathbb{R}$ for which:
$$
D(f) = \dfrac{d f}{dt} = \lambda f
$$
And as you can see, the function $ f(t) = e^{\lambda t}$ satisfies this since:
$$
D(f) = D(e^{\lambda t}) = \lambda e^{\lambda t} = \lambda \cdot f
$$

[flint cloak]

@teal grotto do u know of any that would work

[wintry steppe]

ahh okay awesome

[winter harbor]

Yeah, one thing with operators on infinite dimensional vector spaces tho

[wintry steppe]

how does C^1(U) work? like I have never encountered that notation before

that text going off the screen lol

[teal grotto]

well, you should be able to figure this out now that you know some more information. i cant think of any examples on spot, but you have enough

[winter harbor]

Noticed this now

UGH

[lavish jewel]

i like my functions continuou

[stoic python]

MISTERSYSTEM

[winter harbor]

It's still going offscreen

[lavish jewel]

must now interject with "derivatice"

[wintry steppe]

Lol

[winter harbor]

Anyways

If we take an open of the real line

[wintry steppe]

$$C^1(U) := {f\colon U \to \bR : f \text{ differentiable, } f' \in C(U)}$$

[stoic python]

TTerra

[wintry steppe]

oh so it's basically a set of functions which map U to R with continuous derivatives?

[winter harbor]

Yup

Differentiable functions with continuous derivative

[wintry steppe]

does the 1 indicate that it only has 1 continuous derivative?

[winter harbor]

We can talk about C^k functions too

Which are functions wich are k times differentiable

[wintry steppe]

ahh

[winter harbor]

And all its derivatives are continuous

Naturally

We can also talk about C infinity functions

Which are functions whose all derivatives exist

And all of them continuous (trivially in this case)

[wintry steppe]

would the set of all polynomials be a good example?

[winter harbor]

Yup

They are one of the nicest examples

The exponential function is also C^infinity

Sine and cosine too

In fact, they are analytic

[lavish jewel]

the poly mention ties nicely into taylor's theorem

[winter harbor]

Which is stronger

[wintry steppe]

does that mean it has a complex derivative everywhere?

[lavish jewel]

there's real analytic too, which is weaker

rn everything was real

[winter harbor]

In this case, I mean that they have a Taylor series expansion everywhere

[wintry steppe]

ahh okay

[winter harbor]

Btw

I was about to talk about the spectrum of an operator

[wintry steppe]

this?

[winter harbor]

Which is a detail we have to care about when we are dealing with operators on infinite dimensional vectors

[winter harbor]

Yeah

[wintry steppe]

ohh

[winter harbor]

Basically

In infinite dimensional vector spaces, some of the stuff we know about eigenvalues breaks down

[wintry steppe]

oh how?

[winter harbor]

In the following sense

In infinite dimensional vector spaces

Not every linear operator is continuous

We usually call continuous linear operators bounded operators.

[wintry steppe]

oh that's strange

[winter harbor]

Yeah, and turns out because of this

And some other technical details

It is more fruitful to talk about the spectrum of an operator

Rather than its eigenvalues

Its basically a generalization of what an eigenvalue is

In the finite dimensional case we don't have to care about this

[winter harbor]

A generalization in the sense that every eigenvalue of an operator is in the spectrum of its operator, but not necessarily the converse is true.

This is the detail I wanted to mention

[wintry steppe]

wow functional analysis is so cool

[winter harbor]

But this is basically the only difference between dealing with eigenvalues on an infinite dimensional vector space

The definition is the same

But turns out that

We can talk about more general things in the context of infinite dimensional vectors

Which comes with the spectrum and so on

Its quite nice

https://en.m.wikipedia.org/wiki/Spectrum_(functional_analysis)

[wintry steppe]

so the spectrum of on operator works even if the operator is not continuous?

[winter harbor]

Yeah, we can talk about the spectrum of a bounded operator in a Banach Space.

[wintry steppe]

can k be real or complex? like using the fractional derivative? so can we talk about something like C^(5.4 + 2i)

[pallid yarrow]

hello i am having trouble computing jordan normal forms

[winter harbor]

Never really messed around with fractional derivatives

So I don't know

[wintry steppe]

me neither, I was just curious

[pallid yarrow]

i know how to find algebraic and geometric multiplicity. so i can find the jordan normal form up to reordering of the jordan blocks on the diagonal. is this sufficient to find a matrix's normal form?

i.e. are all jordan normal forms of a given dimension similar.

[winter harbor]

Yup, the Jordan Normal Form is unique up to ordering.

So this should not be a trouble.

[pallid yarrow]

i thought this just means that if a matrix has two jordan forms, they are the same but reordered on the diagonal. how does that show that the two jordan forms are similar?

like is there a matrix P such that PJP^-1 reorders the blocks or something?

[winter harbor]

If you have two Jordan normal forms J_1 and J_2 for a matrix A with, possibly distinctic orders, then act on J_1 via a permutation matrix to get J_2.

Permutation matrices are invertible

And I think you can work out the details

[pallid yarrow]

hm. my textbook (artin's) says unique up to reordering of the jordan blocks on the diagonal. i dont think permutation matrices alone can reorder jordan blocks of different dimensions on the diagonal.

[lavish jewel]

i'm pretty sure you can set up block permutation matrices

[pallid yarrow]

would u mind showing me a quick example on a three by three jordan form with say, dimension 2 and dimension 1 blocks? i tried for a bit and got kinda stuck

[lavish jewel]

smth like this for a MWE

and of you call these T

you'd get smth like PTT^-1JT^T T^T,-1P^-1

terrible notation lol

$P T T^{-1} J T^\text{T} (T^{-1})^\text{T} P^{-1}$

[stoic python]

Edd

[pallid yarrow]

whoa lol

[lavish jewel]

where, for clarity, the Ts on the right of the J can be different from the ones on the left

but that shouldn't happen here since you wanna move blocks along the block diagonal

then just associate as needed to get an equivalent form

if you're more familiar with it, perhaps, it's the same as why singular value decompositions are unique up to rotations

[pallid yarrow]

im not familiar with those

[pallid yarrow]

i really dont understand what you mean here. so i take T to be a matrix of the form you drew??

[lavish jewel]

yeah

[pallid yarrow]

what is P?

[lavish jewel]

oh, i guess i mixed up your notation

one second

[pallid yarrow]

yep np

[lavish jewel]

let's say there is some matrix A whose jordan normal form is PJP^-1

then another equivalent jordan normal form is the one i gave above

what i did was construct a similarity transformation between some J1 and another J2

[pallid yarrow]

ah thats what i thought

[lavish jewel]

some of the transposes and inverses can be removed from what i wrote, thanks to the properties of permutation matrices

yeah, you can remove the transposes on the right of J. then, if desired, replace the inverses with transposes (for the T only)

[pallid yarrow]

but why would there be transposes? J' being similar to J just means there is X such that XJ'X^(-1)=J

[lavish jewel]

because permutation matrices are orthogonal

so X^-1 = X^T

i think i put in the original transposes by mistake tho, to be honest

$P T T^{-1} J T (T^{-1}) P^{-1}$ so this should do

[stoic python]

Edd

[lavish jewel]

and you can replace T^-1 by T^T

there you can see that there's a similarity transformation acting on J, and that the resulting P' = PT matrices are the same as the original, but with shuffled columns

[pallid yarrow]

im not sure if we are talking past each other or not

im trying to show that any two jordan normal forms which are the same except the blocks on the diagonals are shuffled are similar

[lavish jewel]

that's exactly what is written up there?

[pallid yarrow]

TT^-1 J TT^-1 is just J lol. thats not a jordan normal form with different ordered diagonal blocks

[lavish jewel]

bruh

[pallid yarrow]

maybe im just being slow.

[lavish jewel]

$(P T) (T^{-1} J T) (T^{-1} P^{-1})$ how about now

[stoic python]

Edd

[lavish jewel]

now let $(P T) (T^{-1} J T) (T^{-1} P^{-1}) = P_2 J_2 P_2^{-1} = A = P J P^{-1}$

[stoic python]

Edd

[lavish jewel]

then you can solve for J in terms of J_2

[pallid yarrow]

oh! lmao

[lavish jewel]

and the result will be T^-1 J T = J_2

give it a shot

[errant mist]

I want to write that the abs. value of eigenvalue is less than or equal to the sum of each row in a matrix in mathematical notation . Would I just notate this as .: $|\lambda | \leq \sum_{i=1}^n A_{ij}$? Or does this sum over the entire matrix what I have written?

[stoic python]

Fredrikpiano

[lavish jewel]

what you wrote adds the elements in a column

idk if that's what you meant

pick a column and add up all the entries in it

and if you wanted to say that this should work for any column, i would toss in a \vorall 1 <= j <= no. of cols.

[errant mist]

Its related to Gergorin's disc theorem where I have a symmetric matrix and I want to give a bound for the eigenvalues. If I understand correctly then the eigenvalue should satisfy being less than or equal to each row in the matrix as well as for each column.

This is what I'm trying to convey in mathematical notation

[lavish jewel]

sure

well, fixing j and iterating over i in the sum adds up the entries of a column

on the other hand, fixing i and iterating over j adds up the entries of a row

[serene solstice]

Find det(A). I'm thinking of partitioning, but how?

[lavish jewel]

so if you take what you wrote and add \forall j, this would say that the absolute value of an eigenvalue would be larger than the sum of the entries of any column of the matrix

i'm guessing you also have to specify an eigenvalue though

unless you mean this holds for all of them

[errant mist]

so, I add a $\forall j = (1,2,\dots, n)$ columns and $\forall i = (1,2,\dots, n)$ rows right?

[stoic python]

Fredrikpiano

[lavish jewel]

one or the other, depending on what you're adding up (entries of columns or entries of rows)

[errant mist]

yes, thank you

means I will write the expression twice. Once for adding columns and once for adding rows

[teal grotto]

this matrix is equal to (a-b)I + bJ where I is the 4 by 4 identity matrx and J is the 4 by 4 matrix with all entries set to one. try figuring out the eigenvalues of J and how many times they occur

[serene solstice]

Hmm, haven't studied about eigenvalues yet.

[teal grotto]

they’re really helpful here : (

well, i guess it’s small enough to do laplace expansion

[near crescent]

Hello,
how can i show that
R3 ⊆ {(2, 1, 0), (-1, 2, 1), (0, 3, 0)}

where R3 means real numbers in 3 dimensions

[dusky epoch]

this is very clearly false as stated

did you by any chance mean to ask how to show R^3 ⊆ span {(2, 1, 0), (-1, 2, 1), (0, 3, 0)} ?

[near crescent]

yes

@dusky epoch sorry, i should have been more clearly

[marble lance]

Let (x,y,z) be an element of R^3. We want to show that (x,y,z) is in the span. So find a,b,c such that
(x,y,z) = a(2,1,0) + b(-1,2,1) + c(0,3,0)

[near crescent]

is it (0, 0, 0)?

[marble lance]

Is what (0,0,0)?

[near crescent]

(x, y, z) = (0, 0, 0) when a = b = c = 0

[marble lance]

That's true

But (x,y,z) has to be an arbitrary vector of R^3

Because you want to show that every element of R^3 is in the span

Now you have to find a b and c in terms of x y and z

[near crescent]

So that its not 0,0,0?

[marble lance]

No

It's an arbitrary vector

[near crescent]

you mean like this?: (2a - b, a + 2b + 3c, b)

[marble lance]

You cannot give it specific values

Well, if you simplify the RHS, that is what you get

So (x,y,z) = (2a-b, a+2b+3c, b)

Now you need to find what a b and c makes that true

[near crescent]

I have no clue beside from a = b = c = 0

[marble lance]

You are not understanding what you have to do

If you want to show that a set A is a subset of another set B, then you have to show that every element of A is also an element of B.

We now want to show that R^3 is a subset of the span of those 3 vectors

So you must show that every element is also an element of the span

The elements of R^3 look like (x,y,z) where x y and z are real numbers

The elements of the span look like a(2,1,0) + b(-1,2,1) + c(0,3,0)

So you have to take an ARBITRARY element (x,y,z) and show it belongs to the span

And the way to do that is to write (x,y,z) in the form a(2,1,0) + b(-1,2,1) + c(0,3,0) where a, b and c depend on what x y and z are

So you are not working with specific values for x y and z

So you cannot just say it is 0

[near crescent]

Im not sure how to insert it right

[marble lance]

I will do an example

Let's say we wanted to show that (x,y,z) is in span {(1,2,0), (0,3,0), (0,0,1)}

Then (x,y,z) = a(1,2,0) + b(0,3,0) + c(0,0,1) = (a, 2a+3b, c)

So we get
a = x
2a+3b = y
c = z

Subbing a = x into 2a + 3b = y, we get
b = y/3 - 2x/3.

So,
(x,y,z) = x(1,2,0) + (y/3 - 2x/3) (0,3,0) + z (0,0,1)

This shows that (x,y,z) belongs to the span since we have written it as a linear combination of the three vectors.

[near crescent]

thank you for the great example, ill try it on my own

[marble lance]

👍

[near crescent]

2a - b = x
a + 2b +3c = y
b = z

Subbing b = z into 2a - b = x, we get
a = 2/z + x

Subbing a = 2/z + x into a + 2b + 3c = y, we get
c = 1/3( -(2/z + x) - 2z + y)

So,
(x, y, z) = 2/z + x, 1/3( -(2/z + x) - 2z +y), z

[runic snow]

If (3x-1) is a factor of x^2+17x+k what is the numerical value of k?

[marble lance]

You did your substitutions wrong, and your conclusion is wrong too

(x,y,z) = the linear combination of the vectors with those coefficients

(x,y,z) is not (a,b,c)

[marble lance]

This is not linear algebr, but if 3x-1 is a factor then we can write
x^2 + 17x + k = (3x-1)(ax+b) for some a and b, multiply out and compare coefficient to get the values of a and b, and then you can get the value of k

[near crescent]

@marble lance can you tell me where my mistake slies in the 1st part?

Subbing b = z into 2a - b = x, we get
a = 2/z + x

or was the method right and i just twisted some numbers?

[runic snow]

how about the ac test? ac=mn and b=m+n where m and n are factors

[marble lance]

You can use other methods yeen

2a - z = x does not lead to a = 2/z + x

[near crescent]

a = 1/2(x + z) whoops, my bad, thank you

[wintry steppe]

why is the 0 vector not considered the eigenvector of every matrix?

[marble lance]

Because it doesn't have a single associated eigenvalue

[wintry steppe]

thanks

[marble lance]

It belongs to every eigenspace

👍

[chilly shard]

this proposition is in artins. but for the matrix $A=\begin{bmatrix}0 &1\ 1&0 \end{bmatrix}$, i think this is false(???). the eigenvectors of A are exactly those on the diagonal, which is a 1d subspace. but A is similar to a diagonal matrix. how can there be a basis of $F^n$ of eigenvectors?

dunno where i'm going wrong with this.

[stoic python]

shortcut

[chilly shard]

(to be clear A is a matrix over the reals)

[hard drum]

The proposition is certainly true

For starters saying the eigenvectors are on the diagonal must be a mistake, do you mean eigenvalues? And that's unfortunately incorrect too as that's true only for triangular matrices, which this is not

The eigenvalues of A are 1 and - 1 as the characteristic polynomial is x^2 - 1

[chilly shard]

omg lol.

im very silly.

[hard drum]

(you might be confusing this diagonal (ig the antidiagonal) with the main diagonal)

[chilly shard]

||or tired||

[hard drum]

dw

[pallid rampart]

How do I show this?

Number 13

Preferably just a hint

[zinc timber]

||Orthogonalize||

[wintry steppe]

I am trying to understand any vectors v in the vector space V can be uniquely defined by its coefficients in the decomposition. I know if V has a basis then every v can be expressed as v = $\sum a_kv_k$.

[stoic python]

Outlander

[quasi vale]

suppose that a vector can be represented in two(or more ways) by different coefficients, then try to come up with a contradiction

[sick sandal]

do you understand what the basis does?

[wintry steppe]

yeah. The basis lets us express every represent every vector v in V as a unique linear combination.

[sick sandal]

each vector can be represented by its coefficients in an ordered basis
these are just namely its "coordinates" wrt the basis

and its convenient to use

[quasi vale]

@pallid rampart I did this recently, xd. Try to use the Riez Representation Theorem

[sick sandal]

are you familiar with ordered basis?

[wintry steppe]

Not really.

[sick sandal]

its a basis with more information provided namely which vector comes 1st and 2nd etc so if you have some basis B={v1,v2...,vn} and you have a vector v=c1v1+c2v2+..cnvn the scalars c1,..,cn are the coordinates of this vector
and you can identify each vector using them only when the basis is ordered which should be clear to see why this would not apply to any basis
or in other words we can say there is a 1 to 1 correspondance between each vector v-->(x1,..,xn) to all of n tuples in F^n

[wintry steppe]

I see, it still a confusing but I understand why we can represent a vector with its coordinates with respect to a basis. Thanks.

[sick sandal]

i was correcting something and i had a blackout 🤦
what i meant was each ordered basis makes a 1 to 1 correspondance between all vector v in V and all n-tuple (x1,..,xn) in F^n
v-->(x1,...,xn)

or creates/determine is the proper word i suppose

the n-tuples will naturally be unique by this

you can ofcourse prove that anw

[wintry steppe]

by 1 to 1 correspondance you mean for each vector v we have a n-tuple (x_1,...,x_n) in F^n. So a basis allows us to have that.

but ordered basis? why does it only work for it.

[sick sandal]

take some unordered basis and try to assign some coordinates

lets say you get
2v1+3v2 and 2v2+3v1

are these the same?

[wintry steppe]

we don't know.

[sick sandal]

can you identify a vector with its coordinates

[sick sandal]

just take some example and you'll see they are not

[wintry steppe]

ok.

[sick sandal]

moreover its not unique

this only works when you know for a fact

the 1st vector is v1

the 2nd is v2

so on

else the whole concepts fails

beauty of this is each n tuple assigns a vector in V for some ordered basis

and you can do addition and scalar multiplication normally as mentioned

[sick sandal]

this might be a bad example but imagine the coordinate system and you want to assign a point (1,2,3)
would this work if the coordinate system wasnt ordered as (x, y, z) in that order?

[wintry steppe]

oh I see it would be confusing. Ok let me see I get it. For a basis B is a just a collection of vectors. There could be different linear combination of B for example if B = {e,b,d}, so for any v in V can be a_1e + a_2b + a_3d or a_1b + a_2d + a_3e etc. But if B was ordered we know which comes first.

Meaning a ordered basis is just a n-tuple of vectors. Am I getting it right?

[sick sandal]

eh sounds about right

should be clear why order is important to represent vectors as coordinates

[wintry steppe]

yeah. thanks for much.

[wintry steppe]

Ok I want to deeper insight why a system of vectors v_1, v_2, ..., v_n in V is linear independent if only the linear combination of the zero of V is trivial? Also what should I be doing when I encounter definitions that I understand but I do not understand why that definition was considered?. Sorry for the questions.

it's another way of saying that you can't solve for any of the vectors in terms of the others as a linear combination

Also what should I be doing when I encounter definitions that I understand but I do not understand why that definition was considered
try to understand through examples

definitions aren't made out of thin air

there's always some reason something's defined

ok thank you.

[autumn kraken]

Hello

I have a simple function R -> R, x -> x², how can I test if it's surjective?

I tried writing down f(x) = x² but got no further

dunno how to start

[stable kindle]

iff it's not surjective, there's some y such that no x in R satisfies x^2 = y

[autumn kraken]

oh

so any negative y 😄

[stable kindle]

mhm

[autumn kraken]

if it was surjective, how would you start there?

[stable kindle]

well it's not

but just explicitly find an element

like if it were x^3

show that cbrt(y) works

[autumn kraken]

cbrt?

x->x³ is actually the 2nd exercise lul

I will try

[stable kindle]

cube root

[autumn kraken]

okay

I also need to prove that the functions are well-defined

for the x->x² function, I just said that if x in R, then x² in R

is that enough? 😅

[winter harbor]

First of all

A function has 3 informations

A domain, a codomain and an assignment

For the x->x^2 function, I assume you are taking as domain and codomain R

So we have f : R->R which maps x->x^2

To check this function is well defined

We have to verify that for every element in the domain

I can assign f(x) to it

And we can do that in this case

For every real number we can assign its square

As a counter example

Consider g : R -> R given by x->√x

This is not well defined

Because we can't assign a real square root to a negative real number.

If we changed the domain say working with the non negative reals R+, then yes, g : R+ -> R which maps x->√x is well defined.

After checking if the domain makes sense

We have to check that the codomain makes sense

Namely

We have to check that the function indeed assigns an element in the domain to an element in the codomain and not another different set

For instance

Consider the negative reals R-

And the "function" f : R -> R- which maps x->x^2

This is not well defined

Since the square of a real number is always non negative

Or the "function" f : R -> [0,1], where x->x^2

This is not well defined

Because the square of a real number might be bigger than 1

Ofc, f : R -> R x->x^2 makes sense, since the square of a real number is always a real number

And similarly, f : R->R+ x->x^2 is well defined since the square of a real number is always non negative.

We have so far only checked that the domain and the codomain make sense.

Now we have to check if the assignment makes sense

Namely, if there's no ambiguity

For instance, the assignment f : R -> R where x->x^2 makes sense

Because the square of a real number is always unique

An assignment that has ambiguities is the following one

[autumn kraken]

Is it enough for me to say that every x² is unique?

I don't need to prove that right

I mean for ambiguity

[winter harbor]

You don't need to, it's just straightforward.

But consider this one

\begin{align*}
f : & \mathbb{Q} \rightarrow \mathbb{Q} \
\dfrac{a}{b}& \mapsto a + b
\end{align*}

[stoic python]

MISTERSYSTEM

[winter harbor]

This assignment is not well defined

[wintry steppe]

what's the connection between the dual of a vector space and the inner product defined in that vector space?

[winter harbor]

Why? Well, it seems that the domain and codomain make sense

But notice that 1/2=2/4

While f(1/2) = 1+2 = 3

And f(2/4) = 2+4=6

Which are different

A function must assign a unique value to every element in the domain

So we must check that

And in this case, it doesn't.

But that's basically all the technicalities you must be careful about @autumn kraken

[winter harbor]

There's tons of

[wintry steppe]

oh nice

[winter harbor]

Namely

Something called "Riesz Representation"

Let's work over a finite dimensional vector space, for simplicity

[autumn kraken]

thank you @winter harbor

for surjectivity of x->x³, I just said that for every x³ in R, there is cbrt(x³) in R, which means every element in R has at least 1 archetype/prototype (??? I don't know the English term, in German it's Urbild). Is this correct or not sufficient?

[stable kindle]

no

cbrt

pls

it makes sense

[autumn kraken]

yeah square root to the power of 3 is kinda dumb ngl lol

[winter harbor]

Let $V$ be a finite dimensional vector space over $\mathbb{R}$ or $\mathbb{C}$ where $V$ is endowed with an inner product $\langle \cdot, \cdot \rangle$. Notice that we can define a function:
\begin{align*}
\flat : V& \rightarrow V^{\ast} \
v& \mapsto v^{\flat}
\end{align*}
Where $v^{\flat}(u) = \langle u,v \rangle, \forall u \in V$.
\
\
Notice that since $\langle \cdot, \cdot \rangle$ is positive-definite, in particular it is non degenerate and so if $\exists v \in V$ for which $v^{\flat} \equiv 0$, this means that $\forall u \in V$ we have:
$$
v^{\flat}(u) = \langle u, v \rangle = 0
$$
Thus, $v = 0$ (by non-degeneracy) and as such $\flat$ is an injective linear map. Since $\text{dim}(V) = \text{dim}(V^{\ast})$ and both $V$ and $V^{\ast}$ are finite dimensional, this implies (by rank-nullity) that $\flat$ is an isomorphism. So notice that an inner product on $V$ induces a canonical isomorphism (that does not depend on a basis) between $V$ and its dual.

[stoic python]

MISTERSYSTEM

[winter harbor]

This is a baby version of what we call the Riez Representation Theorem for Hillbert Spaces

Turns out that in the infinite dimensional case

A Hillbert Space is also canonically isomorphic to its continuous dual (i.e the space of continuous linear functionals on V) via the musical isomorphism $\flat$

[stoic python]

MISTERSYSTEM

[winter harbor]

Yes

We call $\flat$ a musical isomorphism, its inverse is called $\sharp$.

[stoic python]

MISTERSYSTEM

[wintry steppe]

oh wow really?

nice

[winter harbor]

These are the so called musical isomorphisms

Yup

[wintry steppe]

wait I'm a bit confused by that notation of v^bflat

[wintry steppe]

here

[winter harbor]

What is confusing?

[wintry steppe]

there's a function bflat mapping stuff from V to it's dual and what does v^bflat denote?

[winter harbor]

Ah

v^flat denotes a linear functional

What it does

Is that it takes an element in u

And what this functional does

Is take the inner product of u with v

v^flat(u) = <u,v>

This is what v^flat does as a linear functional

[wintry steppe]

ah okay

[winter harbor]

And flat maps V into its dual via v -> v^flat

So it indeed sends vectors to linear functionals

And we can show that this map is injective, and thus an isomorphism

[winter harbor]

That fact that injective => isomorphism uses heavily the fact that we are working over finite dimensional vector spaces here.

[wintry steppe]

shouldn't it be bijective for it to be an isomorphism?

[winter harbor]

Yes, but an injective map between finite dimensional vector spaces of the same dimension is always an isomorphism.

So we don't even have to explicitly write down an inverse.

[wintry steppe]

I see

[winter harbor]

Just for completeness

The inverse of flat does the following

\begin{align*}
\sharp : V^{\ast}& \rightarrow V \
f \mapsto f^{\sharp}
\end{align*}
Where $\f^{\sharp}$ is the unique vector in $V$ such that $\forall u \in V$:
$$
f(u) = \langle u, f^{\sharp} \rangle
$$

[stoic python]

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

[winter harbor]

So in some sense, this tells us that every linear functional on a finite dimensional vector space

Can be represented by a unique vector

Where the linear functional is just given by taking the inner product with this vector.

[wintry steppe]

ah

well thanks

[winter harbor]

Np, you can search up "musical isomorphisms" or "Riesz Representation" for more information.

[wintry steppe]

I do not know how a) and b) can be done but I know that c) and d) follows from a and b.

[winter harbor]

What is the 0 vector in the vector space of polynomials?

[lavish jewel]

use the definition of the kernel and the 0 element in the codomain

or if it helps you, you can write a larger matrix that transforms a vectorized form of M_22 into a vector isomorphic to a poly in P1

[wintry steppe]

[(a+b); (c+d)] = [0;0].
I tried this.

and I got { a = -b, c = -d, b and d are free in R }

[wintry steppe]

How is this?

[lavish jewel]

$\begin{bmatrix} a + b \\ c + d \end{bmatrix} = \begin{bmatrix} 1 && 1 && 0 && 0 \\ 0 && 0 && 1 && 1 \end{bmatrix} \begin{bmatrix} a \\ b \\ c \\ d \end{bmatrix} $

[stoic python]

Edd

[lavish jewel]

this represents the same thing

and it's possibly easier to find the kernel of

at least easier in the sense that you already have the intuition for it

[wintry steppe]

then I'll reduce that to rref then check for linearly independent columns?

[lavish jewel]

if you don't see why this is equivalent, though, i strongly recommend you stick to the original formulation and follow mistersystem's tip

[wintry steppe]

I got dim(ker(L)) = 2.

[autumn kraken]

f: R->R, x²->x
this function is not well defined, is that correct?

because (-3)² != -3 for example?

[winter harbor]

It is not well defined because if you take 9

Then 3^2 = 9

And so f(9) = 3

But (-3)^2 = 9

So f(9) = -3

The same input gets mapped into different values

[autumn kraken]

oh

I thought the x has to be the same

but you can do f(3^2) = 9 ?

I don't understand

[wintry steppe]

it's not one-to-one then?

[winter harbor]

No

That's not what f does

[wintry steppe]

now, i'll have to go with basis(Im(L)).

how do i do that?

[autumn kraken]

let x=-3, then f((-3)^2) = -3 <=> f(9) = -3, which is fine, but also for x=3 we get f(3^2) = 3 <=> f(9) = 3

ahh I see

thanks

[winter harbor]

Yup

Also

This is not really linear algebra

I guess this is more suitable for #prealg-and-algebra I guess

[autumn kraken]

dunno my module is called linear algebra 1 lul

[wintry steppe]

so I got dim([a,b; c, d]) = 4.

[winter harbor]

Yeah, but this topic is specific seems to be mostly just a bit of HS algebra review.

[winter harbor]

You can try using rank-nullity to find a basis for the image of L if you have seen that already.

[winter harbor]

Or just use this.

[ionic laurel]

I have a small question

the dimensions of the column space is the same as the dimensions of the transpose of the column space correct?

and this same concept applies to the row space

[marble lance]

What is the transpose of the column space? Do you mean the column space of the transpose?

Because that would just be the row space of the original matrix

[ionic laurel]

yeah my bad

i mean to say like

dim(ColA) = dim(ColA)^T

but yeah that makes sense

the column space of the transpose is the row space

and the dimensions of row space and column space are the same

[marble lance]

Yes

[autumn kraken]

Can I ask something about surjectivity here or should I go to #prealg-and-algebra ?

[winter harbor]

If it is related to vector spaces and linear maps somehow

[ionic laurel]

Sort of a dumb question but does anyone know whether this is true or false?

The dimension of the vector space of signals, §, is 10

It's a true or false question in my textbook and I have not read anything about vector spaces of signals but yeah if this is sort of irrelevant then feel free to tell me 😅

[lavish jewel]

depends which signals lol

[winter harbor]

What is a space of signals

[lavish jewel]

same as usual. can be a vector space of some specifically shaped vectors in R^n, or something like that

signal usually only means it has some property or another that makes it more or less easy to fourier transform

[ionic laurel]

yea idk lmao thats literally just the question

its all good

i doubt that type of specificity will show up on a exam

[lavish jewel]

there needs to be more context or it makes no sense

if that is all there is, the correct answer is "what the fuck"

[lavish jewel]

what is \mathbb{S}

[fickle citrus]

Update on this, I have considered alternative algebraic proofs which use 1-pi, and later on use the normalisation equation and distribution uniqueness later on to solve for a unique pi. There is no use of Perron-Frobenius, but rather on Kolmogorov Axioms/Probability axioms. In a way, this is nice in showing the resulting stationary distribution is a uniform discrete distribution.

However, I have decided to use Perron-Frobenius due to what it allows me to say: the linear dependence on the stationary distribution vector with the vector of all 1s, as achieved for every doubly stochastic matrix, means having a uniform discrete distribution as the stationary distribution happens only for the the family of all doubly stochastic matrices. In essence, there is a converse result achieveable with use of Perron-Frobenius which I think is a good addition.

In other words, by probability axioms and uniqueness of stationary distribution,
$$\text{Doubly Stochastic}\implies\text{stationary distribution is uniform discrete}$$

But Perron-Frobenius additionally says
$$\text{Doubly Stochastic}\iff\text{stationary distribution is uniform discrete}$$

[stoic python]

ShatteredSunlight

[fickle citrus]

Probability axioms cannot do the converse for it says nothing of the dimensionality of the eigenspace of Perron root of ergodic transition matrices, and so cannot comment too much on the linear dependence of vector of 1s with the final stationary distribution

That said I really should look at a proof of Perron, and see if I can really 'get it'

[pallid rampart]

I figured out a way in the end, but I’m interested to see how you did it with riesz representation

[winter harbor]

Let $\lambda_{1}, \cdots, \lambda_{m} \in \mathbb{R}$ be distinct positive real numbers and $V$ a Hillbert Space.
\
\
Let $f \in V^{\ast}$ be such that $\forall i \in {1, \cdots, m}$ we have:
$$
f(v_{j}) = \lambda_{j}
$$
And whose support is the closure of $\text{span}(v_{1}, \cdots, v_{m})$.
\
\
Since $v_{1}, \cdots, v_{m}$ are linearly independent vectors in $V$, this in fact induces a unique continuous linear functional on $V$ and thus $f$ is well defined. By Riesz Representation Theorem/Musical Isomorphism $\exists ! f^{\sharp} \in V$ for which $\forall v \in V$ we have:
$$
f(v) = \langle f^{\sharp}, v \rangle
$$
In particular, $\forall j \in {1, \cdots,m}$ we have that:
$$
f(v_{j}) = \langle f^{\sharp}, v_{j} \rangle = \lambda_{j} > 0
$$
Thus, by taking $w = f^{\sharp}$ the result follows ; $\square$

[stoic python]

MISTERSYSTEM

[winter harbor]

I didn't take as hypothesis that V is a finite dimensional vector space. Which makes things harder because we have to guarantee f is continuous in order to apply the Riesz representation theorem.

And I think I was a bit handwavy there when making f vanish outside of the closure of the span of v_1, ...,v_m.

Because idk if it is straightforward if, in this way, we make f continuous.

But I suppose we can do something along these lines.

The finite dimensional case is more straightforward

[noble swan]

Hey, just making sure I understand basis' correctly. A certain set is the basis if the determinant != 0, right?

[silent dune]

How do I find two orthogonal vectors that are orthogonal to themselves and the plane? I can't figure any out that work

[lavish jewel]

if you put the vectors as columns of a matrix, this works. vectors need not be tuples of numbers, though, so this is too restrictive

[dusky epoch]

are you sure you phrased correctly what you are looking for?

'two vectors orthogonal to themselves and the plane' sounds sus

even if you replace 'themselves' with 'each other'

[silent dune]

ah maybe I wrote it wrong but the dot product of the two vectors has to be 0, and they both must be orthogonal to the plane

[lavish jewel]

they must be orthogonal to the plane's normal

not to the plane itself. they must be ON the plane

[silent dune]

right

but still I can't figure out two vectors that are

[lavish jewel]

you only need one. you could find the second using a cross product, which will by definition satisfy the conditions you're looking for

constructing just one is simple enough

[silent dune]

cross product?

[lavish jewel]

say (-3, -1, 1)

[silent dune]

ya

[lavish jewel]

at any rate, you can write is as a system of equations and use the techniques you already know