Mathematics - 2019-10-26 (page 2 of 2)

Poline Sandra

19:12

@anakhro I don't understand what you say

anakhro

@PolineSandra what about it do you not understand?

schn

Consider the quadratic form $2xy-3xz-yz$. What is its signature?

Poline Sandra

@anakhro what is G?

Rithaniel

Well, you can also throw your arms up in exasperation, as if to say "I'm done with this."

anakhro

@PolineSandra The geometric sum $\sum_{k=0}^n (1/2)^k$.

Starting at 0, right?

Poline Sandra

19:14

yes

anakhro

@Rithaniel I imagine many things can be expressed with throwing your arms in the air.

schn

Since the trace of the matrix corresponding to $2xy-3xz-yz$ is $0$, counting the sign changes won't work.

Poline Sandra

AAA so S=G-1?

anakhro

@PolineSandra does it make sense that that would be the case?

Rithaniel

Hmmm, how about indifference?

anakhro

19:16

@Rithaniel ¯\_(ツ)_/¯

Rithaniel

I shall now start throwing my arms up in hunger

Poline Sandra

@anakhro yes I think that it make a sens

anakhro

@PolineSandra that's enough to at least try writing down an answer. Make sure to check it after wars with a few examples.

user2154420

If the initial ideal with respect to a monomial ordering in $k[x,y,z]$ is $xyz$, then is it singular?

Woops, I meant, "is the ideal singular"

Blergh. I mean, let $I$ be a $k[x,y,z]$-ideal. Suppose $in_<(I)=xyz$. Then is $I$ singular?

$in_<(I)=(xyz)

genescuba

19:33

1

Q: Strategy for game where larger number wins. Drawn from standard uniform distribution with one redraw allowed

Two players are playing a game where they each draw a secret random number uniformly between 0 and 1. If they are not satisfied with their draw they may redraw. The players do not know whether or not the other has chosen to re-draw. The players then compare their numbers and he/she who holds larg...

anakhro

@genescuba does the redrawing actually matter here?

genescuba

I believe so, apparently the answer is 1- golden ratio. I have no clue how to get there tho

anakhro

Why would the re-draw matter?

I feel like I may be missing something.

genescuba

news.ycombinator.com/item?id=12940675

There is some discussion about this problem on this page.

anakhro

19:48

He gives his solution for it right there.

genescuba

There is another way to do it, without integrals and convolutions. Say player A has threshold a and player B has threshold b, a <= b. Write 3x3 table of Player A win probabilities for ranges 0 to a, a to b, and b to 1. It will have 1/2 on the diagonal, 1 below, 0 above.
After some simplification, total Player A win probabilty is (-ba^2 + (b^2-b+1)a +(b^2-b+1))/2. Find a that maximizes this probability for any given b by taking the derivative and setting it to 0, which would be a = (b^2-b+1)/2b.

I don't understand how he goes from the top to "after some simplification" and gets the player A win probability

anakhro

The guy who that guy is replying to has a more full solution.

Poline Sandra

20:12

@anakhro I don't understand

anakhro

20:22

@PolineSandra what do you not understand?

Stan Shunpike

I am studying Shur's complement and having trouble understanding some basic matrix algebra stuff. I have a matrix as follows

$$\begin{pmatrix}
\epsilon A & -B \\
-B^T & \epsilon^{-1} C\\
\end{pmatrix}
$$

This is positive semidefinite

and, using this info

I want to show that

$$\begin{pmatrix}
\epsilon A & -B \\
-B^T & c\mathbf{I}- C\\
\end{pmatrix}
\succeq
0
$$

but i don't get how to do it

some tools i thought might be relevant would be, for the part $c\mathbf{I}-C$, i might need eigenvalues, or some kind of decomposition, or something

I forgot to note $A,B,C$ are matrices

and $\succeq$ denotes positive semi-definite ordering

@TedShifrin hey ted!

Ted Shifrin

hi @Stan

So we need to know something about $c$ here. And do we know the first statement for one particular $\epsilon>0$ or for all $\epsilon>0$ or ... ?

Stan Shunpike

right sorry $\epsilon > 0$ and I'm supposed to choose the $c$ large enough to make this true

Ted Shifrin

Aha. Wait. One particular $\epsilon>0$? Or all?

Stan Shunpike

and it's for any $\epsilon$

Ted Shifrin

20:37

Do we know $A,C$ symmetric?

To me, one only talks about positive/negative definite for symmetric guys.

Stan Shunpike

No. The goal of this is to show

$$
\begin{pmatrix}
A & B\\
B^T & C\\
\end{pmatrix}^{-1}
\succeq
\begin{pmatrix}
A^{-1} & 0\\
0 & 0\\
\end{pmatrix}$$

and i'm required to use a limit approach

Ted Shifrin

That doesn't address my point.

Well, that's clearly not true as stated. You have a habit of leaving out lots of hypotheses.

Oh, I missed the inverse.

Stan Shunpike

it doesn't say whether they are symmetric in the context we are working in

Ted Shifrin

Really?

Stan Shunpike

i'll send it to you

Ted Shifrin

20:40

So your definition of positive definite has nothing to do with eigenvalues, then.

Only in the symmetric case can you state it in terms of eigenvalues.

Stan Shunpike

ok i sent it to you by email so that we can avoid me mistating the question.

Ted Shifrin

LOL

Stan Shunpike

i will try to improve at this lmao

i accept it cuz ur right

i get that feedback on stack exchange sometimes

its question 4 part d

Ted Shifrin

Regardless, I can't see any connection to the thing you wrote down with $cI$. ...

Stan Shunpike

i proved d.i using determinants

well, I think the first step is to show that if i stick this epsilon in here

Ted Shifrin

20:45

If this is coming out of covariance stuff, these should be symmetric matrices.

Stan Shunpike

it is although the proof of the lemma is separate

but i think they omitted that

and they should have stated it

it was like....stat stats stats....now do real math

Ted Shifrin

But most of the world assumes symmetric matrices when talking about definiteness, as I said.

Stan Shunpike

ok i will add that to my notes

is that in the context of block matrices?

or just generally even for one without block matrices?

Ted Shifrin

Generally.

That's why you have $B$ and $B^\top$ appearing in the non-diagonal blocks.

schn

Consider the quadratic form $2xy-3xz-yz$. How would one go about finding its signature? Counting the number of sign changes in the determinant sequence $D_k$ for the k by k matrix representing the q.f. wouldn't work to find the negative index of inertia, since the second determinant would be $0$.

Ted Shifrin

20:48

So @Stan, you should be writing down $Mx\cdot x\ge 0$ for your matrices, but likewise breaking $x$ into two pieces, according to the blocks, say $x=(y,z)$. Have you done this?

@schn What's your definition of signature?

Stan Shunpike

I wrote down

$$x^T ((cI-C)-(-B)^T(\epsilon A)^{-1}(-B))x $$

Ted Shifrin

Do you diagonalize the quadratic form and look at $Q=PDP^\top$? Just look at the entries of $D$.

Stan Shunpike

and was going to expand it out

Ted Shifrin

@Stan: But break $x$ into pieces, as I said. Wait — why is there an inverse in that one? Which part are you doing?

Stan Shunpike

Uh well for the earlier part I used the determinant of M

Ted Shifrin

20:50

Don't use determinants.

Stan Shunpike

ok

lol i'll have to redo the earlier part

Ted Shifrin

I don't understand the formula you just wrote down. Do it carefully.

Determinants won't tell you anything unless you compute all possible principal determinants.

Stan Shunpike

ok i better try again give me a couple minutes to think about it brb

schn

@TedShifrin It's the pair; (positive index of inertia, negative index of inertia). I considered the change of coordinates $x=x'+y', y=x'-y', z=z'$, but it didn't yield any simple calculations (wolframalpha.com/input/…)

Ted Shifrin

Do you know how to diagonalize the quadratic form (as opposed to looking at eigenvalues)? It is like completing the square in high school algebra. (You can find this, too, in my videos :P)

schn

20:53

Are there any squares to complete in $2xy-3xz-yz$?

Ted Shifrin

Yeah, having no $x^2$ term makes it harder.

So we do need to make a rotational change of variable as you were suggesting. E.g., $2xy = \frac12\big((x+y)^2-(x-y)^2\big)$.

schn

Yes. The computation of finding the eigenvalues of the matrix corresponding to $2xy-3xz-yz$ would be pretty messy too.

Would the change of variables $2xy = \frac12\big((x+y)^2-(x-y)^2\big)$ differ from the one I mentioned previously?

Ted Shifrin

No, you were doing that, so we need to do something similar with $x,z$, then see what happens to the $yz$ term. Your teacher likes yucky problems.

schn

Indeed.

Ted Shifrin

It seems to actually impede learning, rather than improve it.

So I would do $x=x'+y'/2$, $y=x'-y'/2$ and then rewrite the $xz$ and $yz$ terms in terms of $x', y', z$ and then work on it again.

schn

21:01

Let me try.

Stan Shunpike

ok i multiplied it out and got

$$\begin{align*}
x M x^T
&= (y,z) M(y,z)^T\\
&= (yA + zB^T,yB + zC)(y,z)^T\\
&= yAy^T + zB^Tz^T + yBy^T + zCz^T\\
\end{align*}
$$

Ted Shifrin

I think you can deduce it from the determinant sequence, though, @schn. You get $0$, $-1$, and some positive number, right? This tells you that the eigenvalues must be one two positive and one negative.

@Stan: OK, this is good. Now you are free to let $z=0$ if you want, or $z$ very big ...

You have a mistake, I think, @Stan. The $B$ terms should have $y$ and $z$ both, not just one of them.

Stan Shunpike

ok give me a sec

$$\begin{align*}
x M x^T
&= (y,z) M(y,z)^T\\
&= (yA + zB^T,yB + zC)(y,z)^T\\
&= yAy^T + zB^Ty^T + yBz^T + zCz^T\\
\end{align*}
$$

there

Ted Shifrin

OK, I'm happier now.

Now write out the corresponding thing for the right-hand side and see if you can make some deductions. First, I want you to deduce that $A$ and $C$ must each be positive semidefinite.

Stan Shunpike

ok brb

schn

21:09

@TedShifrin Okay. Doing $x=x'+y', y=x'-y', z=z'$ yields a non-squared term $y'z'$, which makes another change of coordinates tricky. I think $x=x'+y'/2, y=x'-y'/2$ would turn out similar. How could one conclude from the determinant sequence that the eigenvalue are positive, positive and negative respectively?

Ted Shifrin

Right, but it's not bad to work with a new problem that's like $x^2 - y^2 + axz + byz$, @schn. Then it really is easy with completing the square.

Think about what you know about the eigenvalues of the first $2\times 2$ matrix to get $0$ and negative determinant.

Stan Shunpike

@TedShifrin for the right hand side, are you referring to

$$
\begin{pmatrix}
\epsilon A & -B \\
-B^T & \epsilon^{-1} C\\
\end{pmatrix}
$$

or the one with $cI-C$?

Ted Shifrin

Yeah, sorry for misspeaking.

Stan Shunpike

cool

Ted Shifrin

I was just following his i.

schn

21:13

Do the eigenvalues of the $2\times 2$ matrix also belong to the $3\times 3$ matrix?

Stan Shunpike

Let
$$
N=
\begin{pmatrix}
\epsilon A & -B \\
-B^T & \epsilon^{-1} C\\
\end{pmatrix}
$$

Then
$$
\begin{align*}
x N x^T
&= (y,z) N(y,z)^T\\
&= (y\epsilon A - zB^T,-yB + z\epsilon^{-1}C)(y,z)^T\\
&= \epsilon yAy^T - zB^Ty^T - yBz^T + \epsilon^{-1} zCz^T\\
\end{align*}
$$

where $\epsilon > 0$

So i guess i can subtract the first one

from the second one

and then show its still positive semidefinite

Ted Shifrin

@schn, No. But it's a matter of understanding the proof of the criterion.

@Stan: Did you first deduce what I said, that each of $A$ and $C$ is positive semidefinite? That might be helpful. Then think about what you do to $y,z$ to get the negatives on the $B$ terms.

Stan Shunpike

ok right ur right brb

Ted Shifrin

@schn: I lied before. It should be 2 negatives and 1 positive. I mistyped.

schn

@TedShifrin Why?

Ted Shifrin

21:26

The final product has to be positive, so what I said was just a mistake. But it works out beautifully the way I said to change variables, following you. You're right that the halves don't help. If you take $x=x'+y'$, $y=x'-y'$, and leave $z$ alone, then you'll get $2xy-3xz-yz = 2x'^2-2y'^2 - 3(x'+y')z -(x'-y')z = 2x'^2-2y'^2 -4x'z+4y'z$?

schn

$2 (x^2 - 2 xz - y^2 - y z)$

Ted Shifrin

But now we can easily complete squares.

schn

But completing the square one is left with an $y'z'$-term, no?

Ted Shifrin

No, I was in too much hurry to finish editing. It should end with $-2y'z$.

schn

Yes

$2((x'-z')^2-(y'+z)^2+y'z')$

Ted Shifrin

21:32

So now we complete the square. $2(x'-z)^2 - 2(y'+z/2)^2 - \frac32 z^2$.

There are our two negatives and one positive :)

Stan Shunpike

So if I combine the two sides, then I get
$$
\begin{align*}
(yAy^T + zB^Ty^T + yBz^T + zCz^T) - (\epsilon yAy^T - zB^Ty^T - yBz^T + \epsilon^{-1} zCz^T) &\geq 0 \\
(1-\epsilon) yAy^T + 2zB^Ty^T + 2yBz^T+(1-\epsilon^{-1})zCz^T &\geq 0
\end{align*}
$$

Ok so suppose $y = e_i$ and $z = e_j$

$$(1-\epsilon) A_{ii} + 2B_{ij} + 2B_{ij} +(1-\epsilon^{-1})C_{jj}$$

Would this help me?

Ted Shifrin

I don't see how combining helps you prove that if (1) is positive, then (2) is positive.

Stan Shunpike

what does $M - N \succeq 0$ tell me then?

Ted Shifrin

You want to deduce that (2) holds for all $y,z$ by choosing them appropriately in terms of $\epsilon$ and the $y,z$ you'll plug into (1).

Nothing about them individually :P You could subtract a negative-definite guy from a positive-definite guy and stay positive-definite, right?

Stan Shunpike

yeah ofc ur right.

Ted Shifrin

21:42

If I tell you $ay^2>0$ for all nonzero $y$, how do you deduce that $b\epsilon y^2>0$ for all nonzero $y$?

Stan Shunpike

oh i see

ok brb let me think

schn

@TedShifrin You're right, we get three squared terms. The last term being $- \frac34 z^2$. Thanks!

Stan Shunpike

Ok, so I can say then

$$
\begin{align*}
x M x^T
&= (\epsilon^{1/2} y, \epsilon^{-1/2}z) M(\epsilon^{-1/2}y,\epsilon^{-1/2}z)^T\\
&= (\epsilon^{1/2}yA + \epsilon^{-1/2}zB^T,\epsilon^{1/2}yB + \epsilon^{-1/2}zC)(y,z)^T\\
&= \epsilon yAy^T + zB^Ty^T + yBz^T + \epsilon^{-1}zCz^T\\
\end{align*}
$$

which shows they are equivalent since i picked $y,z$ arbitrarily

which equals the RHS

i think im too imprecise to talk to mathematicians

hahahahha

@TedShifrin

Ted Shifrin

You're welcome, @schn.

@Stan: I think you've got the idea, now.

Stan Shunpike

and in particular, determinants were a bad idea

Ted Shifrin

21:55

Yup.

Stan Shunpike

when are they useful then?

i read a stack exchange post yesterday that they are often expensive to compute

schn

Say the transformation $x=x'+y',y=x'-y',z=z'$ wouldn't be invertible (which it isn't...). Would that be a problem?

Ted Shifrin

Yes, they're a pain to compute numerically. They are important in telling you how volumes transform. They're important in finding eigenvalues. But I typically would not show a matrix is nonsingular by computing its determinant to see it's nonzero.

@schn: I got lost in your negations. Yes, it's a problem. You have to be using an invertible change of variables to go backwards.

schn

Okay.

Ted Shifrin

Remember that the point of the completing the square argument is to write the quadratic form in terms of the diagonal sum of squares: $A = PDP^\top$, where $P$ is invertible.

schn

21:59

@TedShifrin Yes. Consider the quadratic form $(2x+y+z)^2+(x+2y+2z)^2+(x-y-z)^2$. Since a change of variables here would not be invertible, what does that conclude about the quadratic form?

Stan Shunpike

@TedShifrin My high school physics teacher retired and he had all these colleagues come into visit and it made me remember how much i love being around physicists

schn

The transition matrix would not be invertible.

Stan Shunpike

@TedShifrin was really incredible. he did his phd with Higgs and it was wild hearing stories about his career and seeing how many lives he had changed

Ted Shifrin

Oh, I remember you told me about him.

@schn: You can conclude that the quadratic form is degenerate. It's the sum of two squares, not three.

schn

Does it say anything about the signature? (as I defined before)

Stan Shunpike

22:03

@TedShifrin yeah I have been browsing careers and, back when I started college, I decided I couldn't make an impact on fundamental theories of physics. So I should drop it. But I'm doing my masters now and thinking about careers and I've been wondering if something like quantum computing would be a good match for me. I had lunch with him and we discussed particle physics for like an hour and I hadn't had that much fun in ages.

it's just so abstract lol

but people have said doing something u love is really important

Ted Shifrin

@schn: It tells you that there's a 0. You still have to look at the actual terms to figure out how many + terms and now many - terms.

@Stan: If you can get a job doing quantum computing, go for it.

schn

@TedShifrin Right.

With degenerate you mean it reduces to 2d-space, correct?

Ted Shifrin

In appropriate coordinates, yes.

OK, I'm disappearing. I have neck exercises and cooking to do.

schn

Good luck.

Stan Shunpike

@TedShifrin bye ted! thanks! always fun chatting

Darius

22:55

@TedShifrin, I've just asked a question (math.stackexchange.com/questions/3409363/…) here from your book Multivariable Mathematics. If you have some time, it'll be awesome if you could weigh in!

Basically, the question is if $A$ is a matrix, then what can we conjecture about matrices $A$ satisfying $A^{n-1} \neq \bf{0}$ but $A^n = \bf{0}$.

Semiclassical

23:14

For googling purposes, such a matrix is said to be nilpotent

Leaky Nun

this can be read off from the Jordan normal form

schn

Consider the quadratic form $(2x+y+z)^2+(x+2y+2z)^2+(x-y-z)^2$. A change of variables here would not be invertible. Why does that imply that the quadratic form has an eigenvalue that is 0?

Ted Shifrin

23:37

@Darius: I answered. You're doing just fine.

@schn: I already explained that to you. You can write it as the sum of two or fewer independent squares.

schn

How can you see that?

In the expression, that is.

Ted Shifrin

What is $(2x+y+z) - (x+2y+2z)$?

schn

$x-y-z$

Since every quadratic form can be diagonalized, how would one go about doing it in this specific case? Since any transition matrix is not invertible, how does one find the diagonal matrix?

Ted Shifrin

Well, if $u=2x+y+z$ and $v=x+2y+2z$, you have $u^2+v^2+(u-v)^2$. So diagonalize that quadratic form and then introduce an independent third variable and you'll have 0 times its square.

schn

Okay.

Stan Shunpike

23:46

Does a matrix $M$ have to satisfy $x^T M x \ne 0$ to be invertible? or is definiteness not related to invertibility?

Leaky Nun

@StanShunpike "$x^T M x \ne 0$" is ambiguous because you don't have enough quantifiers

Ted Shifrin

You have an implication but not the right one.

If you have a positive-definite matrix, then of course it is invertible (prove that!). But the converse is most definitely false.

Shaddup, @Leaky.

Leaky Nun

oops

Ted Shifrin

We know you know elementary linear algebra, show-off.

Leaky Nun

sorry I didn't mean to

Ted Shifrin

23:50

Yeah, the keyboard just typed by itself.

Leaky Nun

I thought I somehow proved the opposite direction

then I realized it's the original direction all along so I deleted it

Rithaniel

Hey, my keyboard types by itself all the time. It's honestly quite a frustrating prob-cxnjexdfapsofnapsdfn

Ted Shifrin

@Rithaniel: Usually, most people blame their cat.

Leaky Nun

or columbus

Ted Shifrin

You referring to DogAteMy?

Leaky Nun

23:53

lol

Semiclassical

regarding quadratic forms, the easiest way to see that there's a zero eigenvalue is to, y'know, look at the matrix itself

in this case, it's \begin{pmatrix} 6 & 3 & 3 \\ 3 & 6 & 6 \\ 3 & 6 & 6\end{pmatrix}

Ted Shifrin

But this should be conceptual, @Semiclassic.

Leaky Nun

and here comes the computer

:P

Semiclassical

eh, it doesn't take a computer to write out the matrix

Leaky Nun

as in, one who computes

Ted Shifrin

23:58

I'm fond of doing computations, but not when there's a good reason they're unnecessary.

Semiclassical

I mean, it's great being able to spot that it's of the form $u^2+v^2+(u-v)^2$

but if one didn't spot that, then it's easy enough to write out the matrix and spot that it's obviously singular

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$Mathematics$ Mathematics