Mathematics - 2022-02-10

PenAndPaperMathematics

00:02

@robjohn wouldn't interest you

Jakobian

00:52

I'm unstuck

Jakobian

01:32

I've finally finished all the exercises from the appendix

it was pretty fun

Koro

04:02

Why is it said that transpose of a Jordan block also a Jordan block?

A $3\times 3$ Jordan Block by definition looks like:
$\begin{pmatrix} a&1&0\\0&a&1\\0&0&a\end{pmatrix}$

leslie townes

do people say that?

a jordan block is similar to its transpose in a simple way, and this is one way that people generally prove that a matrix is similar to its transpose.

maybe that's where it's coming from?

i'm not sure that i've seen any textbook call the transpose of a (non-diagonal) jordan block a "jordan block"

Koro

it was written in class notes of my coaching class.

That was probably a mistake as by definition of JB, we consider 'super-diagonal' viz. the diagonal just above the main diagonal.

leslie townes

that's certainly the definition i'm used to. although you'd get the same theorems if you used the other ones. :D

Ted Shifrin

04:40

They’re similar. That doesn’t make them both JB.

Koro

05:22

brilliant.org/wiki/jordan-canonical-form

I don't understand why answer (a) in try yourself question is not correct.

The matrix given in option a) has dimension of eigenspace of 1 equal to 1.

But for the other options, the dimension is $2$.

That is, geometric multiplicity of 1 in 1 in option a) and 2 in the others.

leslie townes

mm, the page won't load for me. but maybe it's not correct?

it wouldn't be the first time someone was wrong on the internet

Koro

So JCF of option a) shall have one block for eigenvalue 1 in option a) and two blocks for the other options.

Ted Shifrin

@leslietownes shocking

Koro

Leslie: I've shared the screenshot of the page.

Ted Shifrin

Isn’t it 2 in 2 of the others?

leslie townes

05:31

flips over the sign as to read: 00 DAYS SINCE SOMEONE WAS WRONG ON THE INTERNET

Koro

I think that it's 2 in all 3 of the others.

leslie townes

yes koro

the JCF of the first is a single 3x3 block and the others are 1x1 + 2x2

Ted Shifrin

Yeah, you’re right.

leslie townes

can you have quotes in a domain name? they should call it "brilliant".org, like they're being sarcastic

Koro

haha

My answer was right :).

now that I think about it, it happens to me often.

Even in tests, I mark something but the app shows incorrect sometimes even if I am right.

Rithaniel

05:35

Alright, I need a quick sanity check: Is there a distinction between $E(X\mid Y)$ and $E(X\mid Y=y)$ aside from the fact that the latter uses a lowercase $y$?

Ted Shifrin

Yes.

Koro

I used to know that. I seem to have forgotten this.

Rithaniel

Okay, then I have failed my sanity check, because I don't remember any distinction between the two

Ted Shifrin

Event or RV Y may have many possible outcomes.

Rithaniel

Ah, wait, so if a question asks for $E(X\mid Y)$ I should answer by describing a distribution (and that description could possibly involve a description of its pmf/pdf). If it asks for $E(X\mid Y=y)$, then I can just give a function in terms of $y$ (which is likely to be the pmf/pdf from the previous)?

(or am I still crazy?)

Koro

05:50

what's the use of JCFs?

1. showing that two matrices are similar.
2. computing powers of a matrix

leslie townes

those are the big ones. if some matrix property of interest is invariant under similarity you can prove it for all matrices by considering jordan blocks only.

in my own life, their main use is linear algebra problems on exams. if you are heavily into matrix calculations it is just one of a zillion other canonical forms and you probably aren't computing them by hand.

Koro

I see. Like showing a matrix is similar to its own transpose

leslie townes

it's helpful to have invariants that characterize a matrix up to some notion of equivalence, and JCF happens to be one of them, with similarity the notion of equivalence.

Koro

I was just in the middle of writing 'similarity being one of them' in response to your message but you already included that :).

copper.hat

@Koro The Jordan form is never used for numerical work.

leslie townes

05:55

unrestricted similarity, i should say. it's important for JCF that the similarity transformations can be any invertible matrix. sometimes you want to be more restrictive than that and you get different notions of equivalence.

that are nevertheless rooted in similarity.

Rithaniel

I like that one problem where you show that a nilpotent matrix has to have all-zero eigenvectors using the JCF

Or, rather, I like to approach that problem with JCF

copper.hat

@Rithaniel That follows from just an eigenvector.

Rithaniel

@copper.hat Yeah, but using that approach would make it not relevant to the discussion

leslie townes

koro: it's useful if for some reason you know, say, the characteristic polynomial of a matrix, but want information about eigenspaces. you can't generally shake that information out of the characteristic polynomial, but JCF at least lets you see what the possibilities are.

Koro

using the fact that minimal polynomial's roots are same as eigenvalues, we can conclude that nilpotent matrix N has only zero eigenvalues.

@leslietownes One question: doesn't that tell exactly the dimension of the eigenspace?

number of blocks of corresponding to $\lambda$ is same as dimension of eigenspace of $\lambda$.

leslie townes

06:04

what's "that" in "doesn't that"

Koro

that is JCF.
I mean suppose JCF is given.

leslie townes

yes, but i'm assuming it's not given. only the characteristic polynomial is given.

that's not enough to tell you dimensions of eigenspaces, but if you're familiar with JCF you know what the various possibilities are.

when i said "but JCF at least lets you see" i should have said "but familiarity with the concept of the JCF at least lets you see."

not familiarity with the actual JCF of the matrix.

Koro

@leslietownes I see. Thanks a lot :).

copper.hat

anyone need an nft installed in their driveway, good deal going

Rithaniel

How exactly would you install an nft at all?

copper.hat

06:12

ongoing joke

in absence of any mathematical contribution, i offer diversion

Rithaniel

Hey, diversion is good

I have been wont to do so in this chat before, too

copper.hat

some might argue that my sort of diversion would not qualify for good.

i still have pssd

post suspension stress disorder

Rithaniel

Oh, that's rough

copper.hat

i can hardly add two numbers now

Rithaniel

I have gssd (graduate student stress disorder)

copper.hat

06:15

thankfully that is far in my past

or maybe not thankfully

Rithaniel

(Somehow I manage to always confuse myself when I'm doing probability stuff)

copper.hat

i always spit when i take expectations

Rithaniel

(I'm trying to figure out what exactly $nY$ is when $Y$ is Uniform. I think it should be Uniform again)

(But with different support, maybe?)

copper.hat

you mean if $Y$ is uniformly distributed on some set then so is nY$?

Rithaniel

Yes?

copper.hat

06:18

Yep,

Koro

Y uniform means pdf of Y is constant?

d stands for density not distribution

copper.hat

well, constant over the range over which it is uniform, assuming that it is continuous

a die roll is uniform

Rithaniel

Yeah, that's right. All points have equal likelihood

copper.hat

my intuition for probability is not great.

Rithaniel

So, if $Y\sim\text{Unif}(0,1)$ would $nY\sim\text{Unif}(0,n)$?

copper.hat

06:21

yep

Rithaniel

I think that's right

Koro

$P(nY\leq y)= P(Y\leq \frac yn)= \int_{-\infty}^{\frac yn} p(x) dx$.

So from here, I think it's possible to show.

here p is pdf of Y and is constant over some [a,b], a<b as you said that Y is uniform.

Rithaniel

That's exactly the stuff I'm typing up now, actually, Koro

Thanks for the sanity check

Koro

I'm glad that I remember at least this :).

copper.hat

We have $P[Y \in A] = \int_0^1 1_A(x)dx$. So, $P[nY \in A] = P[Y \in {1 \over n} A] = \int_0^1 1_A(nx)dx = \int_0^n {1 \over n} 1_A(t)dt$.

Rithaniel

06:39

Alright, got another one: If you know $Y\mid X=x\sim\text{Bin}(n,x)$, then is $E(Y^2\mid X)$ just the second moment of $\text{Bin}(n,x)$?

I feel that this might be a "yes, of course, that's obvious," but, as I've said, I tend to confuse myself when I try to do probability stuff

Rithaniel

06:52

Yep, it's obvious

Koro

07:25

If A is an n by n matrix satisfying $A^k=I_n$, k>1 and it is given that 1 is not an eigenvalue of $A$, then A is diagonalizable.

How do I prove this?

The polynomial $p(x)=x^k-1$ is such that $p(A)=0$.

So minimal polynomial of A is $m_A$ that divides p.

$m_A(1)\ne 0$.

Rithaniel

Look to cyclotomic polynomials. That's probably a first step

Koro

$m_A(x)= \Pi_{r=1}^{k-1}(x-\omega^r)$

where $\omega$ is $k-$th root of unity.

@Rithaniel cyclotomic polynomials? I don't know what those are.

Rithaniel

You just posted a couple. They're polynomials whose roots are the roots of unity

Koro

$A$ has $k-$ distinct eigenvalues. But there is no way to know if sum of dimensions of eigenspaces corresponding to the distinct eigenvalues is equal to dimension of the space that is $n$.

Koro

13:57

(so much silence today)

Astyx

14:43

@Koro You know $\ker A^k-I = \mathbb C^n$

Koro

yes.

Astyx

Can you find a decomposition of the LHS?

Koro

no, I'm afraid.

I know however that $V= \oplus G(\lambda_i, T)$

where G' s are generalized eigenvalues.

Astyx

So the key lemma to prove is that if $P$ and $Q$ are coprime polynomials, then $\ker PQ(A) = \ker Q(A)\oplus \ker P(A)$

Koro

15:03

i'm afraid, I haven't seen the term coprime polynomials yet.

I'll find more about it and then try proving it.

Astyx

do you know about coprime integers?

Koro

@Astyx yes.

Astyx

it's the same idea

the point being that $\mathbb C[X]$, like $\mathbb Z$, is a UFD

Koro

ah, ring theory.

Astyx

Here all you need is the Bezout identity, i.e. there exist polynomials $U,V$ such that $PU+QV=1$

Koro

15:10

isn't there any other way?

:(

Astyx

@Koro not sure what you mean by this

Koro

sorry, I meant generalized eigenspace.

Astyx

which is defined to be..?

Koro

If $\lambda$ is eigenvalue of operator $T\in L(V)$ then $v\in V\setminus \{0\}$ such that $(T-\lambda I)^kv=0$ for some positive integer $k$, is defined as a generalized eigenvector corresponding to $\lambda$.

Astyx

So under which condition does this decomposition give a diagonalization of $A$?

Koro

15:19

and the set of all generalized eigenvectors (corresponding to $\lambda$)of $T$ unioned with the 0 vector is called generalized eigenspace $G(\lambda, T)$.
It can be shown that $G(\lambda, T)=null (T-\lambda I)^{\dim V}$

@Astyx if A has a basis consisting of eigenvectors of $A$.

Astyx

@Koro and in terms of this statement?

Houshou Rattengod

I'm having a hard time remembering/finding the name for a mathematical process. It's part of the process for Collatz Conjecture, where "If you can prove for every number up to 'X', then you only need to prove that the next number falls below 'X' and you can "stop". because you know every number up to 'X' will go down to 1.

I know this process has a name, and the Collatz example is the only way I know to describe it. Does anyone know what that process is called?

Koro

I don't think there's any simple conclusion about diagonalization from generalized eigenspaces.

It may sound strange though considering that they are generalized eigenspaces.

Astyx

@HoushouRattengod strong induction?

Houshou Rattengod

@Astyx See, I'm not sure. I thought the process had an actual name

Koro

15:25

@Astyx Every eigenvector is a generalized eigenvector so $A$ is diagonalizable if $E(\lambda_1, A)+...+E(\lambda_k, A)=C^n$ where + is direct sum.

Astyx

@Koro Under which condition on the symbols you introduced here is $E(\lambda, A) = G(\lambda, A)$

Koro

Here E($\lambda_i, A)=null (A-\lambda_iI)$

Houshou Rattengod

@Astyx Wait... I looked up that term. I think tha tactually is it. Thanks!

Koro

@Astyx one such condition is when all $\lambda$'s are distinct are the number of lambda's equals n (the dimension of $C^n$)

Astyx

What I want you to get to is that you want all generalized eigenvalues to be eigenvalues

this is sufficient to have a diagonalisation: can you see why?

Koro

15:37

This is possible if dimension of generalized eigenspace equals dimension of the eigenspace.

(Algebraic multiplicity =geometric multiplicity)

Astyx

yes. can you relate alg and geo multiplicities to the assumptions of the exercise?

Koro

@Astyx that's what my problem is: I have only $k-1$ eigenvalues (distinct) and dimension of the space is $n$.

Astyx

you do not have k-1 distinct eigenvalues

Koro

We consider the polynomial $p(x)=x^k-1$, then $p(x)=(x-1)(1+x+x^2+...+x^{k-1})$.

x is not 1 so minimal polynomial of A divides the other factor.

minimal polynomial here is $(1+x+...+x^{k-1})=p$, which has $k-1$ roots (distinct).

Astyx

this doesn't imply that $A$ has $k-1$ eigenvalues

Koro

15:47

it does.

because roots of minimal polynomial of A are precisely the eigenvalues of A.

Prithu biswas

23 hours ago, by Prithu biswas

What is the standard definition among mathematicians for the summation and Product function.

Koro

no?

Astyx

the minimal polynomial is not necessarily what you said it is. It could be a divisor of that

@Prithubiswas over the integers? the reals? any ring?

Koro

but $1+x+x^2+...x^{k-1}$ can't be factored further.

Prithu biswas

@Astyx Over the reals.

Astyx

15:48

@Koro not on $\mathbb R$

Koro

Oh wait, I'm wrong.

Astyx

@Prithubiswas formally it's a function $\mathbb R\times \mathbb R \to \mathbb R$, which is induced by multiplication on $\mathbb Q$

Koro

$(x^4-1)=(x-1)(1+x+x^2+x^3)=(x-1)(x+1)(x^2+1)$

Prithu biswas

@Astyx What are the desired properties?

Koro

now I lost any hope of solving this question :(.

Astyx

15:51

@Koro if you try to prove the lemma I stated above it would be easier. It's not too hard

@Prithubiswas continuity

Koro

i'll try that. meanwhile, i'll post this on mse also.

thanks a lot @Astyx.

Astyx

Glad to help

I'm willing to bet there are already similar questions posted on MSE

Prithu biswas

@Astyx Like if we have a sequence of real numbers I would want to take a summation or a product.

Astyx

What I meant is that if there are rational sequences $(x_n)$ and $(y_n)$ approximating reals $x$ and $y$, then $(x_ny_n)$ approximates the real $xy$

Prithu biswas

Lets say I have a sequence g : ℕ → ℝ . Then I would have a function Sum(g,k,m) and Prod(g,k,m) where k,m ∈ ℕ.

Koro

15:56

@Astyx you were right!

3

Q: How to prove that this matrix is diagonalizable?

I am trying assignment questions of Linear algebra and couldn't solve this particular question regarding diagonalizabilty. Let $n \times n$ complex matrix $A$ satisfies $A^k = I$ the $n \times n $ identity matrix, where $k$ is a positive integer $>1$ and let $1$ not be an eigenvalue of $A$. Then...

i found it.

Prithu biswas

@Astyx Like $\sum_{i=k}^{m} g(i)$ or $\prod_{i=k}^{m} g(i)$

Astyx

oh ok, I wasn't talking about that. You define those inductively

Koro

some answers in the post also use JCF :).

Astyx

I didn't know you knew about JCF

Prithu biswas

@Astyx But what is a standard definition for them?

Astyx

16:04

I don't know what you mean by standard definition

Prithu biswas

@Astyx The definition mathematicians commonly use when communicating.

Astyx

Well $\sum_{k=n}^m a_k$ is just $a_n+a_{n+1}+\dots+a_m$

Prithu biswas

@Astyx What if m < n?

Astyx

Usually this is taken to be $0$

Prithu biswas

@Astyx Oh ok. What about the Product function?

Astyx

16:07

1

Prithu biswas

@Astyx Oh ok. Thanks for the clarification =)

Astyx

glad to help

Anthony Saint-Criq

16:29

Hello there! A while ago, I asked a question about why the tubular neighborhood of RP² in CP² has boundary the lens space L(4,1). Now, I know the formal answer, so it's all fine. However, I still struggle to visualize what's going on! If I call M this tubular neighborhod, then it comes with a projection M→RP². What is the pre-image of something on RP² in M? For instance, what is the preimage of a real line? What is the pre-image of a small disk on RP² in M?

Govind75

17:30

How do I prove $\|A\| _2 = \|x\|_2 * \|y\|_2$, where $A = xy^T$

I get that I want the maximal $2$-norm of the vector $(x_1\vec{y}.\vec{v},...,x_n\vec{y}.\vec{v})$

How can I attain the equality above??

Nevermind

I got it

Koro

18:20

If $G=\mathbb Z_4\times \mathbb Z_4$ is written as union of n subgroups (of order 4 each) of G then n can't be 4 or 5.

I don't know any theorem about union of groups except possibly this one: $G_1\cup G_2$ is a subgroup iff one contains the other.

not sure how to prove the statement for $n$.

G has only 5 subgroups of order 4.

So $n$ can't be $4$, fine.

But how to refute the possibility of n=5?

$G=\{(0,0),(0,2),(2,0),(2,2)\}\cup \langle (0,1) \rangle \cup \langle (1,0)\rangle \cup\langle (2,1)\rangle \cup \langle (1,2) \rangle$

Oh wait, where is (3,3)?

nvm, i got that.

Ted Shifrin

19:41

@Koro Really?

Govind75

20:59

I have a weird stats question, for a random variable $X_i$ if I have a probability of $p$ of drawing from one normal distribution $N(\mu_1,\sigma_1^2)$ and a probability of $1-p$ of drawing from another normal distribution $N(\mu_2, \sigma_2^2)$. What's the joint distribution of $X_1,...,X_n$ if they are iid.

I was thinking it could just be the weighted product of the sums of the two different pdfs

$f_X(x) = pf_a(x) + (1-p)f_b(x)$, then take the product for each Xi

robjohn

23:13

@Govind75 Yes, and if my computations are correct (I’m on my phone, so be careful), the mean is $$p\mu_1+(1-p)\mu_2$$ and the variance is $$p\sigma^2_1+(1-p)\sigma^2_2+p(1-p)(\mu_1-\mu_2)^2$$

Clarinetist

0

Q: Finding the distribution of the number of transitions it takes to go from an initial state to an absorbing state

I would like to apologize in advance, as I am extremely unfamiliar with the theory of Markov Chains, aside from the bare basics. So please bear with me. This is not homework. I have a Markov Chain whose transition matrix is given by $$\begin{pmatrix} 3/4 & 1/4 \\ 5/6 & & 1/6 \\ 7/8 & & & 1/8 \\ 9...

probability probability-distributions markov-chains

geocalc33

23:39

is this symmetry $f(x) \to f(1/x)$ called scale invariance?

maybe it's $f(\lambda x)=\lambda^{k} f(x)$

Ted Shifrin

@Clarinet are you sure this isn’t the transpose of the matrix you intend?

Clarinetist

@TedShifrin My understanding is that some people use the transpose; I am more familiar with the convention of the rows summing to 1

Ted Shifrin

Then your vectors are rows, not columns, and the matrix acts on the right.

Clarinetist

@TedShifrin Correct.

Ted Shifrin

Ugh. OK.

Clarinetist

23:50

I'll edit the post to include the transposed version.

Ted Shifrin

The linear algebra is good for steady state distributions, i see a straightforward yucky approach.

It’s straightforward that there’s a bunch of self-similar routes, so you just have a geometric series to solve, since at each non-success you just return to state 1 from some state.

Transcript for

$Mathematics$ Mathematics