linear-algebra

[lavish jewel]

this tells you A spans R^m

[heavy crown]

oh wow I think I understand yes

[lavish jewel]

then one could probably make a statement about the dimensions based on the rank

[heavy crown]

then how'd you show it's square?

if the rank doesn't change

[lavish jewel]

that the solutions are unique means that the matrix is full column rank, so rank(A) = n, and m >= n

let's see

[heavy crown]

how'd you know that m >= n? and is it important to say that?

oh I think I see it now

[lavish jewel]

lemme think for a bit. this part should be easy, but i'm sleepy and i suck at math

right, the idea would be that for unique solutions, the null space should have only the 0 vector

[heavy crown]

that's because
after you said A spans R^m
means C(A)=R^M
so dimC(A) = dimR^M = m

and we also know that dimC(A) = rank(a) = n

so n=m

nice

[heavy crown]

you're very good no worries (:

[lavish jewel]

that seems ok to me, i'm sure someone else will pop up with a correction if we made a mistake somewhere

[heavy crown]

thank you for your help! 🙂 appreciate it

[misty storm]

Is "vector in terms of basis" different from "alpha of the linear combination of basis to find vector"?

[twilit anvil]

@misty storm not sure what an alpha of a linear combination is, could you give more context or an example?

[feral iron]

not really sure how i'm supposed to approach this problem, I just got introduced to vector spaces and subspaces. My best guess is C right now. Do I write it as a matrix? How do I test for additive closure and all that from its current form

[twilit anvil]

i think a 'vector in terms of a basis' means: for a vector x in your vector space, say R^n, choose a basis v1,...,vn. then your x can be written uniquely as a linear combination of v1,...,vn.
say x = c1v1 + ... + cnvn for some scalars c1,...,cn.
writing x as the vector (c1 ... cn) is expressing x in terms of the basis v1 , ... , vn.

[twilit anvil]

you believe it's C. is that because you are able to write a matrix that has its nullspace defined by the set specified in the question?

or, have you tried doing that, but failed?

[feral iron]

no its literally just a hunch, but I did try to write a matrix

[twilit anvil]

have you computed the nullspace of a matrix before, by RREF?

[feral iron]

I'm not sure if I understand whhat I'm even supposed to do I just started making a matrix and I moved 4(x3) to one side to set it equal to 0

I think so but I don't really get what nullspace is

[twilit anvil]

my recommendation would be to study some examples of how you compute the nullspace of a matrix. this, you probably did before. then, from understanding how nullspaces are computed, you can tackle your problem of creating a matrix with a specific nullspace.

[feral iron]

so is a nullspace just the set of all the vectors that multiply by the matrix and equal zero?

[twilit anvil]

yes, in fact, the nullspace of the matrix A is an example of a subspace

it would be a good exercise to try and prove that, if you haven't already

[feral iron]

"W is a subspace because it has a zero element." what does this mean for it to have a zero element

does that mean an entry in the matrix is zero? or a row? or neither

or does that mean if every xn = 0 Ax = 0

[nocturne jewel]

$0\in W$

[twilit anvil]

they mean that the zero vector is in W

[stoic python]

Mosh

[twilit anvil]

W is a set of vectors, not a matrix

[feral iron]

okay so W has a zero element because

0-2(0) = 4(0) etc?

[nocturne jewel]

yes, x1=...=x4=0 satisfy the constraint on R4 to make W

[twilit anvil]

i have to take my leave, good luck with your problem, hope i was of help to you

[feral iron]

thank you

[misty storm]

one of the practice exercises my teacher issued mentions "spaces generated by vectors"

but vectors would only generate a subspace, technically, right?

or is it the same thing basically?

[stable kindle]

subspaces are spaces

[misty storm]

neat

the spaces generated by a given vector always match their vectors' dimensions?

[north hedge]

dont think so

[misty storm]

how would I go about figuring out wether or not that was the case?

[north hedge]

see how many linearly independent vectors there are that generate it ig

[misty storm]

0

[north hedge]

how can there be 0

ig if you have just 0 vector

[misty storm]

well

having 3 linerarly dependent vectors

ahhh

but I see where you're going

[north hedge]

that still counts as having 1 linearly independent one

[misty storm]

sensible

[north hedge]

e.g. (1 2 3), (2 4 6) still generates 1D subspace

(a line)

[misty storm]

and if I have, say 2 or 3?

that would generate a subspace of dimensions 2 or 3, accordingly?

[north hedge]

yes

so 2 of them gives a plane

etc

[misty storm]

ah

[heavy crown]

any clue anyone?
Let A be a real matrix non-square of order m x n, while m ≠ n.
Show that at least one of these matrices is non-invertible: A * transpose(A) , transpose(A) * A.

[stray granite]

what does constrain mean in matrices

I have a matrix in non-RREF, and it is asking "Write down all the constrains on the variables xi
, i = 1, . . . , 6 from the system"

[heavy crown]

Is it correct to do rank(A* transpose(A)) = rank(A)² ?

[novel jetty]

no

[heavy crown]

what is it equal to?

[novel jetty]

it is equal to rank(A)

[heavy crown]

why's that?

we know that rank(transpose(A)) = rank(A)

then rank(A* transpose(A)) = rank(A) too?

[novel jetty]

as a quick reason for why it's not rank A squared, if A=I and the dimension was n, the dimension on the left is n and the dimension on the right in your equation is n^2

[novel jetty]

yeah

you can show that they have the same null spaces

[heavy crown]

hmm I'm trying to see it

[novel jetty]

It's easier to see if you replace the right part with rank (transpose(A))

if x is in your nullspace and transpose(A)x=0, then multiplying both sides by A gives one part

if x is in the nullspace of A * transpose(A), then A*transpose(A)x=0 (vector) and so transpose(x)A * transpose(A)x = 0 (number)

but the lefthand side of that equation can be written as a dot product

[heavy crown]

oo I think I understand now

thanks for the detailed explanation 🙂 I see it now @novel jetty

[novel jetty]

np. If you're interested in applications of this, it comes up a lot in doing least squares estimates

[heavy crown]

good to know)

[misty storm]

guys

real quick

this is a change o' coordination matrix

is it from a->b

or b->a?

[novel jetty]

that notation isn't standard, so it depends on what book or notes you're using

[umbral birch]

hey could anyone help with this?

[solid hedge]

where are u stuck?

@umbral birch

[umbral birch]

im assuming part a is just 4x5 matrix, but for the rest i have no clue @solid hedge

[flint jackal]

That's incorrect dimensions, I'm pretty sure

[solid hedge]

what did u learn about linear systems?

[umbral birch]

how come?

[novel jetty]

it's an augmented matrix, not a normal one

4x5 is correct

[umbral birch]

i thought so too

[solid hedge]

u know gauss elimination?

[umbral birch]

i do but im not too confident with it

[novel jetty]

having no solutions is equivalent to the right column being independent of the others

[solid hedge]

but u know the theory behind it?

[umbral birch]

not really

im having a hard time visualizing possible matrices for each scenario and its connection in terms of ranks

we would assume the matrix would be in RREF right?

[novel jetty]

you can assume that if you want

[umbral birch]

they dont give specific numbers right, so it could be anything, but i know rank has to do with leading 1s within a matrix no?

[novel jetty]

I find it easier to think of things in terms of linearly independent columns, but you could also look at 1's in RREF

[umbral birch]

ah i see

[solid hedge]

https://www.youtube.com/watch?v=5HJKYefV1Ik

[flint jackal]

Do you know the term full rank?

[novel jetty]

did your class cover span, basis, and linear independence?

[umbral birch]

going over span rn

ln independance is something im still learning

so im a bit confused about this question

[umbral birch]

no i dont recall hearing that term

[novel jetty]

for the purposes of this problem, the right column being in the span of the left columns means that there is a way to write it as a linear combination of the left columns

the left columns being linearly independent means that there is at most one way to get any possible answer

being a basis can be important to parts of this problem, but knowledge of that isn't needed

[umbral birch]

so what would i put for each rank

im still a bit confused ;-;

[novel jetty]

that would be too much of a hint

finding variables to solve the system of equations is equivalent to finding a linear combination of the left columns that is equal to the right one

[novel jetty]

using what I said here, you could consider whether it is possible for this to occur at any number of linearly independent left columns

I think it will be hard for you to do things this way if you're struggling with spans and linear independence, but I can't think of a simpler way to do this

[umbral birch]

ah okay

[twilit anvil]

i was having trouble understanding this passage in lax's linear algebra book , about the transpose of a matrix:

namely the sentence '...we see the matrix T acting from the right on row vectors is the same as the transpose of the matrix T acting from the left on column vectors'. isnt the transpose of the matrix T acting from the left on the row vector l?

[feral iron]

is something like
[1 0 0;
0 0 0;
0 1 0;
0 0 1 ]
considered not in echelon form because of the row of zeros not being on the bottom?

[limber sierra]

correct, it is not in row echelon form.

but its very close

just need a couple swaps

[radiant yarrow]

What does an object mean here?

[dusky epoch]

nothing

[lavish jewel]

a thing

[dusky epoch]

they're introducing a new notation here

[lavish jewel]

it says "you run into this thing, and it looks like this, and you go 'oh, that's a tuple' "

[fickle citrus]

Long story short tuples are equal if all components are equal

tuples are ordered pairs/triples/quadrutuple/quintuple/n-tuple(see the reason why they are called tuples now?), so (0,1) and (1,0) are not the same

[lapis fern]

Hi #linear-algebra I’m kinda stuck on this problem.

This is what I have so far, just some rando thought that I’ve not managed to string together cohesively.

1. Since $\mathbb{U}_i $ is finite for all i, then each one has a finite basis. Call each basis $\mathcal{B}_i$ respectively.

2. If I take the Union of the bases, I will also get a finite set, (which I can’t prove, but MSE say this is covered in elementary set theory classes.)

$\bigcup^{m}_{i=1}\mathcal{B}_i \text{ is finite. }$

3. It would be convenient at this point to take the span of the Union because that equals the subspace sum somehow?? I can prove this...

$span\bigg(\bigcup^{m}_{i=1}\mathcal{B}i \bigg)\stackrel{?}{=} \sum^{m}{i=1}\mathbb{U}_i $

And I’m stuck. I don’t know if this is the right plan of attack.....

Hi #linear-algebra I’m kinda stuck on this problem.

This is what I have so far, just some rando thought that I’ve not managed to string together cohesively.

1. Since $\mathbb{U}_i $ is finite for all i, then each one has a finite basis. Call each basis $\mathcal{B}_i$ respectively.

2. If I take the Union of the bases, I will also get a finite set, (which I can’t prove, but MSE say this is covered in elementary set theory classes.)

$\bigcup^{m}_{i=1}\mathcal{B}_i \text{ is finite. }$

3. It would be convenient at this point to take the span of the Union because that equals the subspace sum somehow?? I can prove this...

$span\bigg(\bigcup^{m}_{i=1}\mathcal{B}i \bigg)\stackrel{?}{=} \sum^{m}{i=1}\mathbb{U}_i $

And I’m stuck. I don’t know if this is the right plan of attack.....

[native rampart]

Finite union of finite things is finite

[stoic python]

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

[native rampart]

Yes, That's the right angle of attack

[lapis fern]

I believe it, but I can’t prove it. There’s a proof in Halmos Elementary set theory

[native rampart]

The maximum number of elements in a finite union is when the sets are distinct

In that case,you just add the cardinalities

And finite sum of finite numbers is finite

[lapis fern]

How do I prove point 3?

$span\bigg(\bigcup^{m}_{i=1}\mathcal{B}i \bigg)\stackrel{?}{=} \sum^{m}{i=1}\mathbb{U}_i $

$$span\bigg(\bigcup^{m}_{i=1}\mathcal{B}i \bigg)\stackrel{?}{=} \sum^{m}{i=1}\mathbb{U}_i $$

[stoic python]

Videlicet

[native rampart]

You know span(UB_i) is a subset of \sum U_i?

[lapis fern]

i was trying to do double inclusion....

[lapis fern]

So this was first step....

Span(U B_i) is just the set all linear combos of all the basis vectors....

[native rampart]

Now each a_i b_i is in \sum U_i

[lapis fern]

Yeah. Yeah. Okay. I see that

[native rampart]

So,Any Element in that set is in \sum U_i

[lapis fern]

okay.....

i can almost see it, lol.

Yes...okay. I see it

[native rampart]

I mean not exactly U_1,U_2 and so on

[radiant yarrow]

Won't representing a set of n-tuples from a field F make sense to be showed as $F_{n}$ instead of $F^n$?

[stoic python]

Researcher in Pre-algebra

[native rampart]

No

[fickle citrus]

The superscript notation is inspired from how the tuples are constructed from some product

[radiant yarrow]

Ohh

[fickle citrus]

And well, any notation works but ^n is common notation

[radiant yarrow]

Okay

[native rampart]

Also,You can see them as functions from {1,2,3...n} to F

[lapis fern]

b_i \in B_i, means that b_i is just a vector in U_i.....i see it.

[native rampart]

b_i could be in U_j where i!=j

[radiant yarrow]

Okay I didn't get that far yet

I'll keep reading

[lapis fern]

Yes.....of course cuz these subspace could be equal for all we know.....but for sure b_i is in U_i cuz b_i is a vector whose can be decompose as a linear combo of B_i elements, which basis vectors

Wait, I’m confused, youre right. Each B_i can be of some any finite length.

[native rampart]

Take U_1=span{b_1,b_2,b_3}
U_2=span{b_3,b_4}

[lapis fern]

yeah.

[native rampart]

b_1 is in U_1,b_2 is in U_1,not U_2

[lapis fern]

Yeah. I see it.

Okay, so that’s one directional inclusion. How do I do the other inclusion?

[native rampart]

For the other direction note that if v=u_1+u_2...u_m, where each u_i is in U_i,then each u_i can be decomposed individually into basis vectors of that space

[native rampart]

Suppose u_1=b_1+b_2 and u_2=b_3

Then u_1+u_2 can be written as (b_1+b_2)+(b_3)

[lapis fern]

so v= b_1 + b_2 + b_3

Okay. Okay. Cool. So the two subspace are equal.

[dusky epoch]

how come you mentioned being unable to prove that the finite union of finite sets is finite

[lapis fern]

cuz I don’t know how to do it. But i believe it

[lapis fern]

https://math.stackexchange.com/a/1159890/102443

[native rampart]

Now,do you know |U A_i|<=\sum{|A_i|}?

[lapis fern]

No?

I don’t know what that even means?

Are those cardinalities?

[dusky epoch]

yes

[lapis fern]

so “lengths” of sets...

[dusky epoch]

the union of finitely many sets has at most as many elements as the total of the numbers of elements in each set

[native rampart]

Cardinality of union is less than sum of cardinalities

[dusky epoch]

if you take the union of five sets with 1, 10, 100, 420 and 69 elements, the union has cardinality at most 600

[lapis fern]

sure. I believe it. Ive seen a lot of finite sets in my life.

[native rampart]

Well,You can show that fact inductively from |A U B| <= |A|+|B|

[lapis fern]

So this is some kind of elementary set theory proof?

[native rampart]

Suppose A and B have a common element,then |A|+|B| will have atleast one more element compared to |A U B| because you will be counting that one element twice in |A|+|B|

Yes

[dusky epoch]

yes, this is basic set theory

[wintry steppe]

by Induction?

[dusky epoch]

it's mostly bookkeeping

[wintry steppe]

yeap.

[native rampart]

|AUBUC|<=|AUB|+|C|<=|A|+|B|+|C|

Repeat similarly

There is also a thing called principle of Inclusion and Exclusion

$|A \cup B|+|A \cap B|=|A|+|B|$

[limber sierra]

uh

[lapis fern]

if its basic set theory at this point and it involves induction, i will try it solo. Lol, wish me luck.

[limber sierra]

first of all, \cup

but also

[stoic python]

Buncho Dragons

[native rampart]

For finite sets

[limber sierra]

that is the most unintuitive form lmao

[lapis fern]

Thanks for the help folks! Y’all gave me a lot to work with.... this is all I had before:

[lapis fern]

It double counts the intersection....it seems intuitive....

[limber sierra]

eh i guess

i always see it presented as |A U B| = |A| + |B| - |A cap B|

which feels more intuitive to me personally

[lapis fern]

Oh, lol. That one I have seen. Yes. That’s the form Axler shows in the text.

[dawn fractal]

how would i obtain the basis for the column spaces of A and AB then

it's now known that the column vectors of A span Ax,

but are they necessarily LI?

bases*

[lavish jewel]

no, they are a spanning set

[dawn fractal]

as of now i have found out that C(AB) <= C(A)

but to relate it to the rank, i am not sure

[lavish jewel]

show me the task again, i forget what we're doing

[dawn fractal]

i used linear transformations to prove that im of composition is a subspace of im of the other

by im i mean range of the corresponding linear transformations

that i used

so that i could prove C(AB) <= C(A)

[lavish jewel]

ok, so

let's say A is size m x n

[dawn fractal]

obviously these are both finite dimensional, but

[lavish jewel]

and B is size n x p

[dawn fractal]

(btw AB is of size mxp and A is of size mxn)

A in C^{mxn}, B in C^{nxp}

[dawn fractal]

but idk if that directly implies rank AB <= rank A

together with C(AB) <= C(A)

wait no, lemme rephrase that

idk if C(AB) <= C(A) and C(AB),C(A) being finite dimensional directly imply rank AB <= rank A

[lavish jewel]

the image of A has dimensionality equal to the rank of A, yeah?

similarly for B

[dawn fractal]

we didnt define the image of a matrix

its why i used linear transformations first

[lavish jewel]

it's the column space

[dawn fractal]

i see

[lavish jewel]

the col space has dim = rank

[dawn fractal]

yes indeed

[lavish jewel]

without thinking too hard about it, consider what happens when the column space of B matches the row space of A

[dawn fractal]

wdym

when rank B = rank A?

[lavish jewel]

you know how the row space is the orthogonal complement of the null space

oof

[dawn fractal]

i mean idk what u mean by "matches"

[lavish jewel]

is equal to

[dawn fractal]

somehow i can imagine that in my head, but i don't directly know it

so i have to do casework?

[lavish jewel]

that's the easiest way imo. lots of grunt work without thinking much about it

[dawn fractal]

for sure that would imply rank A = rank B

but in my proof i assume WLOG that rank A <= rank B

so that i need to prove rank AB <= min(rank A, rank B) = rank A

[lavish jewel]

you can split it into a handful of cases

[dawn fractal]

i.e. rank AB < rank A and rank AB = rank A?

[lavish jewel]

i was more thinking about the null and row space of A

[dawn fractal]

or that R(A) = C(B) and otherwise

[lavish jewel]

yeah, like so

[dawn fractal]

what about the null space

[lavish jewel]

like, is there anything other than the zero vector in the null space?

if there is, then the column space of B may or may not have components in the null space of A

[dawn fractal]

im not sure what u mean

[lavish jewel]

i'm not sure i can help you with this proof cuz i would do it based on what the dimensions and rank imply about the null space of A

maybe someone else has a different idea

[dawn fractal]

N(A) contains all the nx1 vectors that produce 0 when mulitplied to A

[lavish jewel]

sure

and that depends on the shape of A and its rank

[dawn fractal]

C(B) contains all the nx1 columns of B

mhmmmm

[lavish jewel]

so the question is

[dawn fractal]

interesting

[lavish jewel]

since rank(A) <= rank(B)

in which ways can B have a higher rank than A? how about the same rank?

i.e. which shapes of A and B, and are they full column/row rank? rank defficient?

[dawn fractal]

1. if A is not full rank, then sure
2. if they are both full rank, perhaps

[lavish jewel]

there aren't that many options, since the inner shapes must match (cols in A, rows in B)

[dawn fractal]

i was considering that from the start but i guess i thought it was too many to handle

[lavish jewel]

you could also think of what A is doing to B

it's taking vectors from B and using them as coordinates for the vectors in A

the result is a set of vectors in the span of the columns of A

so it cannot be higher dim than col A

i guess that's faster

show that AB is a set of vectors, each of which is a linear comb. of the columns of A

then if A has a nontrivial null space, the rank of AB can be smaller than the rank of A, or of B has very few cols too

[dawn fractal]

as in the cols of AB are l.c.'s of the cols of A

[lavish jewel]

mhm

[dawn fractal]

yeah i just said what u said lmao

[dawn fractal]

interesting!

[dawn fractal]

as in the rows of B that when multiplied to A produces 0?

wait wut

no that sounds wrong lmao

perhaps the transposes of those rows?

[lavish jewel]

columns of B

can't multiply the rows from the right

[dawn fractal]

right

ok i will give this thought

ty

[dusky epoch]

if or iff?

[lavish jewel]

can you share the question exactly how it was originally written?

and S is an operator, a matrix, or what?

[dusky epoch]

are we allowed to view it as a matrix

okay

do you know what it means for an operator to be diagonalizable?

you meant eigenvectors

in any case

i'd rather you would have said that S can be written as PDP^-1 where P is invertible and D is diagonal

and now,

$S^2 = PDP^{-1}PDP^{-1}$

[stoic python]

Ann

[lavish jewel]

you good from there?

[stable kindle]

for 21, can i just do W = {(0, 0, a, b, c)}, and for 22 can i just do: W1 = {(0, 0, a, 0, 0)}, W2 = {(0, 0, 0, b, 0)}, W3 = {(0, 0, 0, 0, c)}

[native rampart]

Yea

[stable kindle]

ez

[dusky epoch]

are we assuming char(F) ≠ 2

[stable kindle]

oh my gods

axler says F is either R or C

so booyah

easy mode

[dusky epoch]

ah

[dawn fractal]

wait, how is this necessary for columns of A to span C(AB)?

[lavish jewel]

do you know of a way to, for example, span a subspace of dim 5 using 4 linearly independent vectors?

i think you are not ready for this proof, you should go review the basics once more

[dawn fractal]

nope

[lavish jewel]

because there isn't

[dawn fractal]

bases are minimal spanning sets

C(A) is spanned by the cols of A corresponding to cols in the RREF with a leading 1

my problem lies in how exactly i can determine those columns in the RREF with a leading 1

especially in AB

so is this where determining the nullspace of A, whether trivial or nontrivial, comes in?

[rose umbra]

if T(u) =0 does it means T(t*u)=0? (t= IR)

[lavish jewel]

if T is a linear transformation, yeah

since you can take the scalar t outside

and yes, that's where the nullspace comes in

[dawn fractal]

oh that's what u meant

[rose umbra]

ty eppa

[dawn fractal]

then if the kth column of B is in N(A), then the kth column of AB is 0_{m,1}

so no leading 1 in the kth column of the RREF of AB and therefore wouldn't be included in the basis of C(AB)

[lavish jewel]

a quick google fu shows that the easiest way is as i said earlier, to write AB as a set of vectors, each of which is a column of B multiplied by A

and then use these rank arguments

[dawn fractal]

google fu?

[lavish jewel]

i'll just send you that, you can go from there using the definition of rank and C(A)

[rose umbra]

how to solve such a question?

[native rampart]

Your conditions will be T(2,2,1)=(0,3,1), T(1,4,0)=(0,0,0),T(1,1,1)=(0,0,0)

Try to think how you get these conditions

[rose umbra]

yes i got to here but not sure how to continue

[native rampart]

That's all

Now,you know {(2,2,1),(1,4,0),(1,1,1)} forms a basis

of R^3

[rose umbra]

right

[native rampart]

A linear transform is uniquely defined by how it acts on some basis

[rose umbra]

what do you mean

[native rampart]

Suppose you have some basis {e_1,e_2,e_3}

Given any x,T(x) can be expressed in terms of T(e_1),T(e_2) and T(e_3)(T(c_1e_1+c_2e_2+c_3e_3)=c_1 T(e_1)+ c_2 T(e_2) +c_3 T(e_3))

So,Your transform is uniquely determined by those values

[lavish jewel]

are they asking you to build the transformation or is it just true/false?

[rose umbra]

true/false

but i need to build it at other questions

[rose umbra]

is this theorem?

[lavish jewel]

what buncho wrote always applies

[rose umbra]

how does it called?

[lavish jewel]

i don't think it has a name. buncho just expressed the vector x in terms of the canonical basis, and then used the linearity of T

[native rampart]

If you want a name,you can call it the canonical theorem of linear algebra

[rose umbra]

i cant understand this, it should be seperated ?

[native rampart]

Yea, different lines

mb,I assumed it was on different lines for everyone

[rose umbra]

why i need that then?

i can just say the form basis and that all (?)

[native rampart]

Yes

You also have to add that condition tells you what the images of the basis elements are

[rose umbra]

what if i told to find the linear transformtion?

[native rampart]

This is your linear transform

I have a linear transform T, which satisfies T(2,2,1)=(0,3,1), T(1,4,0)=(0,0,0), T(1,1,1)=(0,0,0) and any value of T(x) can be computed knowing this

[rose umbra]

i mean how to find the build of T

[lavish jewel]

you can take the expressions buncho wrote and notice that it's basically a system of 9 variables, 9 equations

[rose umbra]

can someone give me example of this transfomation?

what does it means z+i?

[lavish jewel]

say z = 1 + 1i

T(z) = (1 + 2i, 1)

i had a typo, sorry

[rose umbra]

yea didnt got it lol

thx

[dusky epoch]

z+i means z+i...

it means the sum of z and i

[umbral birch]

would it be rank 1 and nullity 0

[lapis fern]

what page in Axler is this?

[stable kindle]

it's exercise 1C

[lapis fern]

oh. i had not done 21 and 22.

[wintry steppe]

are you asking how to do it or how to make sense of it

put backslashes before *'s or _'s

that prevents discord from making them vanish

[drowsy sleet]

det(A*) = det(A)*
Is this true?

[drowsy sleet]

Thxx

[wintry steppe]

yes it's true

one way to see it is to just write out the huge formula for the determinant in terms of the entries and see what happens to the complex conjugates

[drowsy sleet]

Hmm i see
It didnt seemed to be intuitive to me

[stoic python]

ττερρα

[snow jetty]

Hello sirs, could you suggest me some video on Legendre polynomials? They are given as an example of an orthonormal basis... but I don't really get their sense and meaning

[wintry steppe]

this is for 2x2 matrices. you can do a similar computation for matrices of arbitrary dimension

maybe with some induction if you want to be 1000% careful and rigorous

[neon turret]

What is this question exactly asking for? I thought its asking for whether the eigenvalues for this matrix belongs to one of those sets.

Like what are the steps i need to do to get the answer here. Idc about the answer just wanna know what the question is asking for and what to do.

[wintry steppe]

yes, it's asking for the eigenvalues

compute the roots of the polynomial det(matrix - tI) in Z_5

[drowsy sleet]

Thx so much
Its actually from a physics problem, so i wouldnt have to be 1000% rigorous luckily

[rose umbra]

is T(U)-T(V) = T(U-V) ?

if T is linear transformation

[nocturne jewel]

if T is linear, then yes

[wintry steppe]

[nocturne jewel]

and U and V are from the input space obviously

[rose umbra]

is this obvious and I cant see it ? because every where I see has only the +

[nocturne jewel]

$T(au+bv)=aT(u)+bT(v)$

[stoic python]

Mosh

[nocturne jewel]

a=1 b=-1 gets T(u-v)=T(u)-T(v)

assuming -1 is in the scalar field

[rose umbra]

ah right

thx

[drowsy sleet]

I have some questions on this problem
My attempt:
Part 1) by using the property of unitary and determinant, I can show that det(U)=1 and
|det(U)| =1
With that, i can then link to the euler’s formula and for some θ, e^iθ = 1 ?
How would this help part 2 tho

[quartz compass]

but det(U) is not always 1

[drowsy sleet]

Even for unitary ones??

[quartz compass]

they explicitly say it's what you are to prove

it can be written as e^{i theta}

which isn't always 1

it's not even true for orthogonal matrices, a reflection can have determinant -1

(all orthogonal matrices are unitary matrices)

[drowsy sleet]

I see where my logic is wrong.
Btw does anyone have any recommendations on materials that can help brush up on some linalg concepts such as inner products, eigenvalues, hermitian etc

Im currently taking a quantum mechanics course and the theory is closely tied to linear algebra and yet, linalg2 course was not a prerequisite

Im struggling quite a bit

[rose umbra]

what i did here wrong ?

[nocturne jewel]

looks right

[rose umbra]

but 2x3 matix make no sence

[nocturne jewel]

how so?

[rose umbra]

it should be R3->R2

[nocturne jewel]

$T[v]=\begin{bmatrix}0&0&0\ -1&0&1\end{bmatrix}v$

[stoic python]

Mosh

[rose umbra]

where v is (a,b,c)

[nocturne jewel]

needs to be 2x3 to be able to multiply by a R3 vector and get a R2 vector out

[rose umbra]

it should be 3x2 (?)

[nocturne jewel]

how do you multiply a 3x2 by a 3x1?

[rose umbra]

its regular

wait

[nocturne jewel]

If $L: V\to W$ and dim(V)=n and dim(W)=m, then $(L)_\alpha ^\beta \in \mathbb{K}^{m\times n}$

[stoic python]

Mosh

[rose umbra]

i multping 1x3 by 3x2

wait is it always Av and not vA

oh dam i got confused

[nocturne jewel]

yes, it's always T[v]=Av

[rose umbra]

@nocturne jewelwait a second its still make no sence to muliply 2x3 by 1x3

[nocturne jewel]

yeah, cause it's a column vector not a row vector

[rose umbra]

aha right

thanks

[solar tendon]

can someone help with this problem

If a matrix A is not invertible, then A has an eigenvector.

would it be false?

[wintry steppe]

A is not invertible if and only if there is a non-zero vector v such that Av = 0.

[solar tendon]

so it would be false based on that right

[wintry steppe]

why?

[solar tendon]

because there is at least 1 matrix

[wintry steppe]

0 = 0 * v

[solar tendon]

where there is no non zero vector that can make the matrix 0

[wintry steppe]

i don't understand what you mean

the claim is true

[solar tendon]

oh

could you explain it to me

a little bit more

[wintry steppe]

so if A is not invertible, then its null space is non-trivial, right?

so you can find a non-zero vector v such that Av = 0

[solar tendon]

right

[wintry steppe]

but that can be written as Av = 0v

and since v is non zero, that says that v is an eigenvector of A

[solar tendon]

oh i see

i got it now

thanks

i have 1 more

All 2x2 real matrices have eigenvectors.

i think it is true

[wintry steppe]

do all second degree polynomials have roots?

since the roots of the characteristic polynomial of your matrix are the eigenvalues, it's basically the same question

[solar tendon]

yes

[wintry steppe]

are u sure

roots in R, i mean

[solar tendon]

some have irrationals

and some are not rela

*real

[wintry steppe]

right, that's what i mean

if you can find a matrix whose characteristic polynomial has no real roots then you're done

[solar tendon]

ok i got it

[wintry steppe]

what's the simplest quadratic polynomial with no real roots?

ok

[solar tendon]

0 -1

1 0

[wintry steppe]

right

[solar tendon]

thanks man

[wintry steppe]

this thing rotates vectors by 90 degrees, so it's impossible for any vector to be on the same line as its image

[solar tendon]

oh i see

[wintry steppe]

which is a nice geometric way to think about it having no eigenvectors

also, by the same logic, you can prove that any real matrix of odd dimension has an eigenvector

which is nice to know

[rustic moon]

hey guys, i have a troubling friend. he's trying to tell me Axler's convention to use 0 to represent every 0 scalar, vector and matrix is bad but this is sacrilegious because clearly axler is a deity

how do i send him to hell

[verbal pivot]

do not worry, in the afterlife, axler personally sends everyone of those sinners to hell

[limber sierra]

bet him $10 that he cant find a situation where it creates a meaningful ambiguity in axlers text

sometimes it creates an ambiguity that doesnt matter

(like 0 scalar or 0 matrix would both work)

or its clear from context that only the 0 vector makes sense in a given use

or whatever

(like if youre adding a matrix to it, it must be a matrix)

unless you go out of your way to make it a problem (eg use one-element matrices interchangably with scalars or something dumb), it isnt one

and you win $10

[old dirge]

Having a hard time with this problem

Isn't this -4? I'm arguing with my friend since he thinks its -14

[fickle citrus]

Unless the 0 vector is represented by 1 I don't see an issue with using just the symbol '0' and overloading it

,w det {{2, -1, -2}, {0, -3, 0}, {6, 0, -6}}

[stoic python]

Results provided by WolframAlpha

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

Requested by ShatteredSunlight#7267

[old dirge]

that's what I fucking thought

[fickle citrus]

You could have typed that yourself

[old dirge]

Sorry I don't know wolfram Alpha that well. I appreciate the help man

[fickle citrus]

Uhh you don't need wolfram specifically, but do try to learn a few computational things

For example numpy
https://numpy.org/doc/stable/reference/generated/numpy.linalg.det.html
Or matlab
https://www.mathworks.com/help/matlab/ref/det.html
These will help you verify computational things

[viral flint]

How much Axiom of Choice is needed to prove the following:

given linearly independent vectors v_ 0, (v_i : i in I) in a vector space V, there is a linear map f : V -> basefield such that f(v_i) = 0 but f(v_0) is non-zero?

[lavish jewel]

this is probably a better fit for the computing software channel

[night tapir]

ok

[lavish jewel]

i can possibly answer your question there

yes, it's mainly for linalg, but computationally. this channel is more for theory stuff

[night tapir]

humm..

[weak needle]

I have a feeling no choice is needed here but this is a bit of an infirmal argument, you probably have to look closer to see if i’m using choice somehow. defining S=span(v0){0}={v| there is a in k st v=ak}{0} we can define a relation on Sxk by vRa iff v=av0. Now this relation is a function as if vRa and vRb then av0=bv0 which implies (a-b)v0=0 and if a-b is nonzero we multiply by the inverse to get v0=0 a contradiction. Now we call this function f and extend it to all of V by saying g(v)=0 if v is not in S and g(v) =f(v) otherwise. Do you think this works @viral flint ?

[weak needle]

There is a weird thing discord is doing with the backslashes. Wherever you see {0} nest to some set you should remove it from that set

[viral flint]

It's treating them as escapes.

Whenever \ is typed in front of a letter: \a, it doesn't get hidden. But \ in front of certain characters (mostly everything except letters and numbers and space) escapes it and disappears. It's to remove the formatting effects of *,_ (italics/bold/underline) and `.

[weak needle]

I see

[sage ibex]

do \\ to get \

[viral flint]

Yes

Also
The relation -> function part is indeed, not a big deal

You're basically defining f(kv_0) = k, right?
The challenge is to extend that to the whole space so that it is linear.

[weak needle]

Yes, i just did it that way to be safe

[viral flint]

Your extension is not linear, as can be seen if we consider
k = g(kv) = g((kv + v_1) + (-v_1)) =/= 0 + 0 = g(kv + v_1) + g(-v_1)

where v_1 is a non-zero vector not in S.

[weak needle]

Ah shit you’re right

[sage ibex]

[viral flint]

There is a pretty "standard" way to prove this, but it requires constructing a basis for V, which relies on the Axiom of Choice.
Which is why I asked whether you need the full strength of AoC.

[weak needle]

Yes i know the standard proof, i was hoping that i could circumvent it aomehow

[unkempt light]

Hey guys, why is it that if
B is the set of vectors
dim(B) = dim (V) (the vector space)
If it's linearly independent then B spans V
and vice versa?

[lavish jewel]

B is a set of vectors in V?

[unkempt light]

Yes, here's the full example

[lavish jewel]

if you have a set of linearly independent vectors taken from V, then you can do elementary transformations to get back to the canonical basis, i.e. rref a matrix with those vectors as its columns into an identity matrix

so you could express the canonical basis as a linear combination of the vectors in B (which were taken from V), and then use those canonical basis vectors to generate any vector in V

(this is just an example, not a proof)

[wary storm]

What is the basis of vector space R(R)

[nocturne jewel]

what is R(R)?

[unkempt light]

Maybe they meant Rn?

[rose umbra]

if im asked to prove/dis that in the euclidean space : does it refers to standard multiplication as the inner product?

[nocturne jewel]

Well im waiting for them to respond, not someone with a wild guess

[nocturne jewel]

My guess is you can still just use a general IP

[rose umbra]

does it matters?

[lavish jewel]

which vector space are u and v from?

[rose umbra]

any*

[nocturne jewel]

any complex inner product space?

[rose umbra]

( any u and v )

[lavish jewel]

...

[nocturne jewel]

that's not what Edd asked

[rose umbra]

i mean from any vector space

[nocturne jewel]

Ok so if it's any complex IP space, then just a general IP

[rose umbra]

is it usually like that when the question doesn't refer to a specific IP?

[nocturne jewel]

yes

$\norm{u-v}^2=\langle u-v,u-v\rangle$

[stoic python]

Mosh

[lavish jewel]

yeah, can't say anything about it, other than presumably they mean this $\Vert u \Vert^2$ norm is induced by the inner product $\langle u,u \rangle$

[stoic python]

Edd

[lavish jewel]

which is precisely what mosh just said

[rose umbra]

got it

[nocturne jewel]

The one thing is that the proof should be done in a complex IP space, so taking a real component makes sense

[rose umbra]

yea question asked for both

[nocturne jewel]

but yeah what I tex'd is the start, then it's just use linearity / conjugate linearity

[lavish jewel]

something like $\langle u, v \rangle = \overline{ \langle v, u \rangle }$ is gonna be needed

[stoic python]

Edd

[rose umbra]

what to do after that?

[nocturne jewel]

.

[rose umbra]

what that means?

like to get the result of the < > ?

[nocturne jewel]

expand the IP using linearity and conjugate linearity

[rose umbra]

by which space of R?

[nocturne jewel]

?

[rose umbra]

in order to exapnd the IP I need to know U and V dimension

[nocturne jewel]

no??

$\langle u-v,u-v\rangle = \langle u,u-v\rangle - \langle v,u-v\rangle$ by linearity

[stoic python]

Mosh

[nocturne jewel]

missing a lot of steps

given it's a proof

[rose umbra]

yea its just what i got by now

[nocturne jewel]

<u-v,u-v>=<u,u-v>-<v,u-v>=<u,u>-<u,v>-<v,u>+<v,v>

[rose umbra]

i think i got something wrong

[nocturne jewel]

yeah

[lavish jewel]

what to the slashes mean

nothing in that expression cancels out

[rose umbra]

why not

[lavish jewel]

why would it

[rose umbra]

those are scalars

[lavish jewel]

what two things are equal and have opposite signs

[rose umbra]

0

xD

[rose umbra]

what are wrong with the cancels?

i got this now and it seems fine

[lavish jewel]

what are you even trying to do

is this the expansion of <u-v, u-v>?

[rose umbra]

left is the expansion i did on left

[lavish jewel]

the RHS of this is not the RHS of the original problem

[rose umbra]

-2Re<u,v> =? -<u,v>-<u,v>

[lavish jewel]

no

that's only true if <u,v> is real, which isn't in general the case

[rose umbra]

but that is the euclidean space

[lavish jewel]

what?

u and v are in general complex, and so is <u,v>

you can't just remove the Re

[lavish jewel]

you're gonna need this

[gray dust]

we're in a general (complex) inner product space V

such an inner product is a map $V\times V\to\bC$

[stoic python]

RoκερραJαnpu

[rose umbra]

question asked to solve it both in euclidiean space

and unitary space

individually

[gray dust]

do it for unitary spaces 1st

[lavish jewel]

if you do it right for the complex one, it applies ot the real one too

[gray dust]

the result simplifies in real inner product spaces

[rose umbra]

what should be assumed in complex ?

[rose umbra]

and is it wrong for euclidiean space?

[gray dust]

use the properties of inner products on unitary spaces

[lavish jewel]

why are you removing the Re?

please read what i wrote

don't remove it

[lavish jewel]

and use this

[rose umbra]

ah i think i got it

thanks

[wintry steppe]

Which topics in linear algebra is important if I am going to learn tensors

[lavish jewel]

bases, the 4 fundamental subspaces, svds, diagonalization

you can always reshape tensors into matrices, and the idea is to use tensors as mappings between vector spaces

though to represent stuff directly as multidimensional arrays, multilinear maps and forms are handy

[split hull]

What is a-8(6+)m=

[wintry steppe]

it's a - 48m.

[lavish jewel]

[unique forge]

Is it considered bad practice to use round brackets for vectors and square brackets for matrices?

[opaque heath]

dos anyone know how to do this

[nocturne jewel]

$m\cdot n =0$

[stoic python]

Mosh

[wintry steppe]

do you know how to find the value of inner product? @opaque heath you should've learnt that

if you already solved it just let us know by saying all g ok?

[opaque heath]

yes i know how to find that

i still havent solved the probelm though

[wintry steppe]

i already solved it but i won't give you answer otherwise ann will hate me

[wintry steppe]

ok so where are you stuck at?

[twilit anvil]

for handing in assignments, i think its fine.. just be consistent

[ruby wagon]

Would anybody mind helping me out with this

I've been looking at it for a while now and I'm not sure where to start. The question should be self explanatory enough to not need much background clarification.

[quartz compass]

start by writing ABv=lambda v

now it'd be nice if we had BA there instead of AB

we can nearly get it if we multiply on the left by B

BABv = lambda Bv

so we should just define the new vector u=Bv

BAu = lambda u

oh hey, u is apparently the eigenvector of BA with eigenvalue lambda now

[lavish jewel]

byotiful

[ruby wagon]

thank you i feel stupid now

[quartz compass]

kinda just 50% writing down the eigenvalue/eigenvector thing and 50% just trying random stuff and hoping something sticks

to make an omelette you gotta break some eggs etc etc haha

you're welcome

[ruby wagon]

thank

[opaque heath]

did you get 1,2,3,4,5,6,7 as possible values?

[stark saffron]

What is abstract algebra? since I can't post in that channel.

[limber sierra]

the study of algebraic structures and their morphisms

groups, rings and ideals, fields, modules, algebras, etc

not a perfect definition but

good enough for a rough idea

[twilit anvil]

i think youre allowed to post, you can give yourself the role if you look at the #get-advanced-access channel

[cold harbor]

how do i solve this?

[slow scroll]

N_A = N_{rref A}, and finding dim(N_{rref A}) is just a matter of counting the number of free columns (columns that don't have pivots).

A is just the coefficient matrix corresponding to this system btw

[blissful vault]

Can someone look over this proof? My approach was to first prove that a certain list is a basis with a contradiction argument. Then use this basis to show the properties of a direct sum. "F" in linear algebra done right is the real or complex numbers. For the proof, I wanted in either case for the dimension of F to be 1.

The part where I consider the F to be dimension 1 is where I'm a concerned about.

[verbal pivot]

I think that assuming that null(T) has a basis is not a correct step (at least in the context of Axler's book)

[blissful vault]

I believe we proved null(T) to be a subspace in all cases, so shouldn't that mean it has a basis?

[verbal pivot]

Yes, that is correct, null(T) is always a subspace, but I'm pretty sure, that doesn't necessarily means that it will be finite dimensional

You may consider the linear function that takes any polynomial (so that the domain is a infinite dimensional space) to the 0 scalar, the null space is the same as the space of all the polynomials

[blissful vault]

Ah, darn.

I see

Ya, the problem doesn't say anything about V being finite-dimensional. Thanks for that. Does the proof work if it is finite-dimensional?

[verbal pivot]

mmm

in the last part, where you justify that the intersection is {0}, you should show that a*u doesnt belong to the null space (although this a pretty direct)

the part where you show that R spans range(T) seems kinda weird to me, but this doesn't really mean that it's wrong too lol,

I don't think that saying that dim F = 1 is wrong (at least when we talk about real and complex fields)

Im pretty sure that Axler says that it is of dimension 1

(maybe in the chapter of duality ?)

I would inspect the proof a bit more but I have some things to do 😦 sorry

[twilit anvil]

it was proved earlier in Axler's book that if U is a subspace of V, then there is another subspace W so that V is the direct sum of U and W. I believe that in the finite dimensional case, having U = null T can simplify your proof. i dont see right away how you might use this fact for the general case though.

edit: after some thought i think this fact can be used nicely in the general (not finite dimensional) case. please message if you would like to talk more about it

[wintry steppe]

no, these are not possible values
so please reread the question completely
you missed one statement

sorry i just fell asleep

[opaque heath]

this is my work i dont see anythign wrong with it though

[wintry steppe]

i saw it

can you guess which statement you missed?

https://cdn.discordapp.com/attachments/540211747613704221/847904418506080306/unknown.png

let's read this again

missing what i mean is not to satisfy 1 statement

@opaque heath

[opaque heath]

uhh

i dont see it

[wintry steppe]

@opaque heath wysi

i emphasized it

[opaque heath]

the values i got were integers tho right

huhh

[wintry steppe]

not only n_y, m_x, and n_x @opaque heath

how about the possible values for m_y?

[opaque heath]

ohh

i see now

[wintry steppe]

all g?

[opaque heath]

yesyes

[wintry steppe]

よかたああ。。

[slow scroll]

@blissful vault You don't need to reference bases for this proof. I can walk u through it if you'd like. Just ping me. EDIT: actually you kind of do need bases, but in an almost trivial way i guess.

[wispy panther]

Hi I have a question about SVD / PCA
are there any caveats when you cannot use it?
not when you should not use it
but you numerically / algebraically cannot use it?
(from #help-7｜zen1thxyz )

[lavish jewel]

i don't see a reason why one couldn't calculate it

other than issues with numerical stability and how long it takes for huge matrices

i.e. the singular values might be very small but nonzero, and the computer does not have enough precision to get nonzero values. or backwards, small errors make singular values that should be 0 turn into small values

the latter can anyways be handled by using low rank approx

[wispy panther]

thanks for the reply!
sorry I missed this

[lavish jewel]

hmm?

[wispy panther]

I'm already using the low_rank packages, but it still give an error

[lavish jewel]

what error are you getting

[wispy panther]

RuntimeError: svd_cuda: For batch 0: U(65,65) is zero, singular U.

[lavish jewel]

i've never used pytorch, i couldn't say what the problem is

you'll have to trace the error back to see what exactly is the issue. all matrices have an svd

[wispy panther]

File "/usr/local/lib/python3.7/dist-packages/torch/_lowrank.py", line 170, in _svd_lowrank
U, S, V = torch.svd(B_t)
RuntimeError: svd_cuda: For batch 0: U(65,65) is zero, singular U.

I think it is somewhere deep inside their package

okay so, theoretically, all matrix have an SVD

that's good to know

[wispy panther]

thanks!

[wintry steppe]

i am so confused

does anyone know what to do here?

[marble lance]

Yes

But what are you confused about

Do you not understand the question or not understand how to do it?

[urban spear]

What I've got here is a matrix in some set of coordinates (s_1,s_2,s_3) and we're making a change of coordinates s -> t such that the matrix becomes constant. Does anyone know how they found that exact change of coordinates to make it work?

actually I'm not even sure I understand how to get the new matrix given that change of coordinates

[sharp wave]

Hey 🙂 I've got a question concerning vectors. (It's been quite some time since I learned about them in uni)

For context: I'm implementing a 3D visualization application.
I've got a camera object in a 3D coordinate system. It's position would be my vector A.
I've got a point where my camera is looking at. This would be my vector B.

Now I want to calculate the straight line that is in a 90 degree angle from the line AB (Red Line in the picture).
Or to be more precise: Let's say I want to move the camera on the red line 1 unit to the left (which would move vector B also 1 unit to the left). What is my camera position now?

(Keep in mind: It has to be in 3D)

[lavish jewel]

do you have the red line segment?

[sharp wave]

I only have Vector A and B.

I want to move A on the red line.

[lavish jewel]

except for the degenerate case where a and b lie on the same ray with source at the origin, a and b should span a plane

you could then do gram schmidt on b-a and a by keeping b-a as is

the second vector that gram schmidt yields will be in the direction of the red line segment

you would then want to move along the ray a + (second vector from g.s.)

if a and b are linearly dependent though, you will need more information

otherwise what you drew here as a red line segment would be a plane

[sharp wave]

okay. thank you very much for the help. there is still some stuff which I don't understand (yet). but now i know what I have to google and look into (gram-schmidt etc.)

[snow jetty]

maths 🤯 .

Where could I find a proof or explanation of equivalence of definitions of a bilinear non degenerate form: the first one is this crazy abstract thing

The second one is that the associated Gram matrix is inversible?

[limber sierra]

only proves equivalence between the second sentence of your image and the matrix property

(because the first sentence is actually more general)

(applies to the infinite dim case as well)

actually https://www.mat.uniroma2.it/~sorrenti/MAT204Fall06_files/Lecture5.pdf is a better source

proof of prop 4

[snow jetty]

Strange that they even consider** ii and iii** as different things. The roles are interchangeable

[lavish jewel]

these are general bilinear forms

the form need not be symmetric

b(u,v) =/= b(v,u)

if you wanna write this as a matrix (since they said n-dimensional), this means that the col and row spaces might be different

and so the null and left null spaces might also be different

[snow jetty]

but still, we just change the order of input; we prove just ii for example and then we say that to prove iii we just consider a function defined by the Gram matrix transposed and then that there is a bijection...

[lavish jewel]

i'm not sure i follow

where'd the gram matrix come from?

gram matrices are (hermitian) symmetric and define inner products

those are only a special case of bilinear forms

and also, if you just transpose, it's still the same vector u_0, for example, just now on the other side of the transposed matrix

the way this is written makes it clear that the left and right null spaces might be different

i.e. in general, u_0 =/= v_0

[snow jetty]

Haha I think that they decided to call every associated matrix Gram matrix in my course 🤪 ... This is how they illustrate Sylvester's theorem

[lavish jewel]

doesn't it just say the Gram matrix is a bilinear form?

not that all bilinear forms are gram matrices

[snow jetty]

They say "the Gram matrix of the associated bilinear form looks like this ☝️ "

But yes indeed, probably all that in the context od symmetric bilinear applications... the Sylvester's theorem makes no sense otherwise?

[lavish jewel]

wouldn't that just mean they took the bilinear form B and did B^T B?

idk which of all the sylvester theorems you mean

[snow jetty]

Yes like they gave two definitions of this theorem, the first one like this

"If Q is a quadratic form over a real vector space V of dimension "n" and if Q(x)= Σai*xi for a basis {v1,...vn} then the numbers p, q do not depend from the basis {v1,...vn}"

[lavish jewel]

mhm. quadratic forms are symmetric

[snow jetty]

then they say that "in terms of matrices"...

but obviously they mean symmetric forms indeed

[snow jetty]

so if I understand correctly this one can be generalised for all symmetric forms in some way?

I think they would call it Sylvester's law of inertia https://en.wikipedia.org/wiki/Sylvester's_law_of_inertia

[lavish jewel]

yep that's what i was looking at

and you can see that they immediately require the matrix to be symmetric

[snow jetty]

Seems that a symmetric matrix always defines a quadratic form and in the other sense

[elfin ether]

Does anyone know what a minimal polynomial of a square matrix can tell about it's Jordan form? (for example here m_A is the minimal, and xhi_A is the char poly)

[snow jetty]

the largest multiplicity of the minimal polynomial is the largest jordan block

(for each eigenvalue)

[elfin ether]

can this largest block appear multiple times?

[snow jetty]

yes, take for example the identity matrix

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

go to the "Minimal polynomial" paragraph

[elfin ether]

huh, a pretty weird feature

thanks!

[snow jetty]

I understand it intuitively like this

There are several "Jordan chains" and each Jordan block corresponds to an separate "chain"

That's why it's logical that the sum of the geometrical multiplicities is the number of blocks

[elfin ether]

well i'm quite confused about this jordan thing

couldn't find much resources on youtube

[snow jetty]

The sum of algebraic multiplicities is the sum of "sizes" of blocks

[snow jetty]

There is one

[elfin ether]

yeah that's like diagonalizable matrices

[snow jetty]

as a particular case

https://www.youtube.com/watch?v=GVixvieNnyc

there are 5 parts

[elfin ether]

i saw all parts of this, it didn't give me intuitions

he gives the "chains" thing as a given

[snow jetty]

it follows from the primary decomposition theorem

[elfin ether]

and where can i learn it?

my uni professor isn't great 😦

[snow jetty]

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

see paragraph

Invariant subspace decompositions

so if you have a split characteristic polynomial (for example in complex numbers) then you can split your vector space in invariant subspaces in such a way that they will be generalized eigenspaces

but don't worry we were learning this stuff 3 months though our section is "physics"

[elfin ether]

wut

all i know is that every eigenspace is an invariant subspace

[snow jetty]

true, but the thing is, we also can define a generalized eigenspace like this

(a simple eigenspace would be a particular case for m=1)

[elfin ether]

why do we take m to be the multiplicity in the min poly?

a larger m will give a larger kernel i believe

so normally eigenspaces don't sum to the vectorspace, but generalized ones do?

[snow jetty]

yes, if the characteristic polynomial is split

[elfin ether]

wdym split?

[snow jetty]

for example x^2+1 is not split, but (x+1)^2 is

[elfin ether]

ah ok

didn't know the english term

[snow jetty]

(But in complex numbers all of them are split)

This is the fundamental theorem of algebra

[snow jetty]

For some time, but since a certain "m" the increasing stops 😄

[elfin ether]

you use all this in physics?

pretty cool

[snow jetty]

I didn't remember using Jordan forms (yet)

But diagonalisation for example in mechanics you diagonalise the inertia tensor to find an axis such that the moment of inertia L would be colinear to the rotation vector ω

But I am in the first year so it will come soon I guess

[elfin ether]

I don't like Jordan form, too much calculations involved :/

but it's cool to have a general form of diagonalization

[snow jetty]

However that's one of the major subjects

[frosty vapor]

jordan form is horrifying

i am learning it now too

and man the computations suck

[wintry steppe]

jordan form is nice

[frosty vapor]

yeah when u dont have to do it urself

arithmetic is hard

[neat peak]

is it true that $\tr_{V^}\left( \rho \left( g^{-1} \right)^ \right)=\tr_{V} \rho \left( g^{-1} \right)$?

[stoic python]

ProphetX

[elfin ether]

can someone help find the jordan form of this?

it's alphas on the main diagonal, and ones on a diagonal which is 2 above the main

the characteristic polynomial of this is (x - a)^n

i'm not sure what the process is now

[dusky epoch]

do we know if the order of this matrix is even or odd

[elfin ether]

no, but we can separate into these cases

[dusky epoch]

this is a simple permutation away from being in jnf

it'll have 2 blocks and both blocks will have alphas down the diagonal

and their sizes are half the size of the whole matrix

[elfin ether]

why though

[dusky epoch]

with the obvious interpretation for odd orders (ie order 2k+1 gives a block of size k and a block of size k+1)

the jordan basis is {e_1, e_3, e_5, ..., e_2, e_4, e_6, ...}

like. if you reorder your basis vectors so that all the odd ones come first and then all the even ones

[elfin ether]

how do you know all this?

[dusky epoch]

by looking ig

[elfin ether]

umm how would i derive all this

[dusky epoch]

dunno

fuck around find out

[elfin ether]

lol

[dusky epoch]

this just happened to jump out at me

[frosty vapor]

maybe if you were wearing pants you would know

[spark dome]

Given a vector v, is there a name for the space of linear operators such that v is an eigenvector?

And more generally for a set of vectors, the space of operators for which each vector is an eigenvector, so the intersection of these spaces over the elements in the set

[blissful vault]

Saw this late. I appreciate it. I found a solution for this exercise.

[slow scroll]

This is how I would do it:

Let $u \in V \setminus \ker f$ and suppose WLOG that $f(u) = 1$. Now, let $v \in V$. Then there is $c \in F$ such that $$f(v) = c = cf(u) = f(cu).$$ Therefore, $v - cu \in \ker f$

[stoic python]

kxrider

[blissful vault]

What's the "V\ ker f"? I haven't seen the slash notation before.

[slow scroll]

"V - ker f." Just the set of V not in ker f

so $v-cu = x$ for some $x \in \ker f$, and $v = x + cu$. Hmm, yesterday I thought it was obvious that this is a unique way to write $v$ as sum of an element of the kernel and an element of $span{u}$, but now im not so sure actually.

[stoic python]

kxrider

[slow scroll]

yea yea it is. If x' + c'u = v = x + cu then x'-x = (c-c')u and therefore 0 = f(x'-x) = (c-c')f(u) = c-c' so c = c'.

and since c is uniquely determined, x is also uniquely determined.

yea, i guess to make the idea slightly more intuitive, v uniquely determines the "cu" as cu = f(v)u.

@blissful vault does that make sense? I'm using the characterization of the direct sum that goes like V = U \oplus W if each vector in V can be written uniquely as a sum of an element of U and an element of W

[blissful vault]

Yes! That looks vastly simplified from what I was doing with basis.

[slow scroll]

ye, and this way you sidestep all of the dimensionality issues.

[stable kindle]

sanity check: most two-element lists are linearly independent in C as a vector space over R, but linearly dependent in C as a vector space over C?

[slow scroll]

yeah, all two element sets of vectors in C are linearly dependent over C.

[elfin ether]

"Given an orthogonal transformation, prove that the complement of an invariant subspace is also invariant"

sounds easy but I can't figure it out, help?

[viral flint]

Let T be orthogonal (thus <Tv, Tw> = <v, w> for all v, w) and V be an invariant subspace. Let W be its complement.
Then you want to show that for any w ∈ W, Tw ∈ W. Being ∈ W is equivalent to being orthogonal to every element of V, i.e. x ∈ W iff <x, v> = 0 for every v ∈ V.
So assuming w ∈ W, can you show that Tw ∈ W?

[dire thunder]

@viral flint assume Tw is not in W, that is Tw in V. Then if v is in V we should have 0 = <w, v> = <Tw, Tv> forcing Tw to be in W, contradiction

[viral flint]

Tw not in W and Tw in V are not the same

Wait

[viral flint]

You are talking about the orthogonal complement of a subspace and not the complement of a subset, right?

[dire thunder]

i assumed orthogonal complement

but even if it would be usual complement

[dire thunder]

this would be false still

given set X and Y in X

you take Y^c

for each x either x is in Y or in Y^c

the same for orthogonal complement

well technically you should treat also zero vector but it is easy

[viral flint]

It doesn't seem like you did, which is why I asked in the first place.

[viral flint]

This part is wrong.

[dire thunder]

Tw is not in W, that is Tw in V. Then if v is in V we should have 0 = <w, v> = <Tw, Tv> forcing

[viral flint]

Consider a plane and its normal line in R^3. There are plenty of points which are neither in the plane nor in the line.

[dire thunder]

yes but man

not only one line is in orthogonal complement for plane

oh

i got your point

uh

i would have cleared up my proof then

but gtg

sorry

[dawn fractal]

Let $A\in\C^{n\times n}$ be normal. Show that $A$ is Hermitian if and only if all eigenvalues of $A$ are real. Formulate and prove similar statements for skew Hermitian and unitary matrices.

[stoic python]

!superficialsicko

[dawn fractal]

what does "similar statements" mean?

as in $A$ is skew Hermitian and unitary iff all eigenvalues of $A$ are real?

[stoic python]

!superficialsicko

[dusky epoch]

no

simiar does not mean copy-pasted

A is skew-Hermitian iff all eigenvalues of A are _____. A is unitary iff all eigenvalues of A are _____.

they ask you to fill in the blanks here

[dawn fractal]

figured

[stoic python]

!superficialsicko

!superficialsicko

!superficialsicko

!superficialsicko

!superficialsicko

[dawn fractal]

nvm i got it lmaoo

[dawn fractal]

any hint as to how i prove that A is unitary only if all eigenvalues of A have modulus 1?

given that A is normal in C^{nxn}

[lavish jewel]

if you already know that a unitary matrix is diagonalizable, you can use that

[dawn fractal]

not really
but i do know that A is normal iff it's unitary diagonalizable
and any A, whether or not it's normal, is unitary similar to an upper triangular matrix

[weak needle]

unitary matrices are also normal, hence unitarily diagonalisable

[dawn fractal]

i see

that's what they meant

ty

ill think about it

since A is unitary diagonalizable

U^*AU = D for some unitary U, diagonal D

then
\begin{align*}
D^* &= ((U^A)U)^\
&= U^(U^A)^\
&= U^A^(U^)^*\
&= U^*A^U
\end{align}

[stoic python]

!superficialsicko

[dawn fractal]

but $U^*AU = D$

[stoic python]

!superficialsicko

[dawn fractal]

so is D* always equal to D?

[weak needle]

No, look carefully

A* is different from A

Do you get what to do now?

[dawn fractal]

oh right

i was so confused when i thought D* = D

it seems like D*D needs to be equal to 1

i mean I

[weak needle]

Yes

[dawn fractal]

and from there im lost