Mathematics - 2017-07-30 (page 1 of 4)

Akiva Weinberger

00:00

Oh OK hm

Daminark

If you just choose an eigenvector for each eigenvalue they have to be linearly independent so you can choose a basis in which the matrix of $T$ is diagonal

gian

Gtg. Talk to you all later.

Ted Shifrin

bye

Daminark

See you!

Ted Shifrin

DogAteMy needs to read my linear algebra chapter.

Akiva Weinberger

00:05

Pretty sure I did, it was just a while ago @TedShifrin

Ted Shifrin

uh huh

Daminark

Well, I'll give you that and the last problem

So let's say you have a matrix $A$, and let $f$ and $g$ be coprime polynomials whose product is the minimal polynomial of $A$

Show that $\ker(f(A)) \oplus \ker(g(A)) = \mathbb{F}^n$

Where $A\in \mathbb{F}^{n\times n}$

Ted Shifrin

Interesting ... that's the key to proving that a matrix whose minimal polynomial has distinct roots is always diagonalizable.

Daminark

Actually I think this is a problem that we were given after we already had that theorem

I don't remember how we did this with Laci, I think HK used T-annihilators for that purpose

("this" = minimal polynomial has distinct roots => diagonalizable)

Lucas Henrique

@Ted, since you're the master of geometric approaches (:P), can you help me?

I want to know if this is a valid argument.

Ted Shifrin

00:15

Maybe, maybe not.

Lucas Henrique

Ted Shifrin

Oh, similar triangles?

I've never seen that. That's cute.

What's your question, @Lucas?

Daminark

Okay so we proved it via this lemma

Lucas Henrique

If the argument is valid.

Ted Shifrin

Well, without worrying about why the infinite series converges to a finite number ...

Lucas Henrique

00:20

I mean... this r power makes me uncomfortable

exactly

Daminark

So let's say $V$ is a finite-dimensional vector space, $T$ is a linear operator whose minimal polynomial factors completely, and let $W$ be a proper $T$-invariant subspace. Then you can find a vector $\alpha$ outside $W$ such that $(T-cI)\alpha \in W$ where $c$ is some eigenvalue of $T$

Ted Shifrin

I guess you could start with the first bit, draw the line (through $(1,r)$) and then deduce that the rest of the picture is correct.

Quantifiers don't seem right, Demonark.

But I don't want to think about it.

Daminark

I think it does work, there exists $\alpha\in V\setminus W$ and $c\in Spec(T)$ such that blah blah

Ted Shifrin

As I said, I don't wanna think.

Daminark

Fair

But yeah that lemma is nifty because it both gives you that matrices are triangulable iff the minimal polynomial factors completely, and diagonalizable iff minimal polynomial has distinct roots

Ted Shifrin

00:25

I can do a more elementary argument. I wrote a problem for the linear algebra book doing this with partial fraction decomposition.

Daminark

Well, it gives you that knowing the facts about the minimal polynomial gives you the facts about matrices

Ted Shifrin

@Lucas: So you understand it?

That's not surprising if you think of vector spaces as $k[T]$-modules, Demonark.

But you haven't got there yet.

Lucas Henrique

@TedShifrin What part, exactly?

Ted Shifrin

The whole thing, Lucas.

You never really asked me a question.

Daminark

Wait, what's $T$ in that context?

Ted Shifrin

00:28

the linear map, Demonark

Daminark

Is that not circular?

Ted Shifrin

but I should have said $k[t]$ ... and $t$ acts by $T$.

Daminark

Oh I mean... I guess in afterthought that works

Wait hmm

Actually I gotta think about this more, gimme a second

Lucas Henrique

I asked you if it's a valid argument.

Geometrically speaking.

Ted Shifrin

I think so, if you do the construction as I said it.

You start with the first square, then draw the line through $(0,1)$ and $(1,r)$, and then construct all the rest. It's cool.

Lucas Henrique

00:35

What about the $r^n$ down there?

Ted Shifrin

Do it step by step.

Lucas Henrique

oh, so I think there's no problem at all

Ted Shifrin

No, I like it.

I mean, the fact that the lines aren't parallel will ultimately tell you the series converges.

Lucas Henrique

it's a valid construction as long as the slope is such that the $(0,1)$, $(1,r)$ line touches the base of the square

Ted Shifrin

That sentence made no sense (in English, anyhow).

Lucas Henrique

00:37

Didn't it? :(

Daminark

Okay wait I just noticed your thing about partial fractions... it might be more elementary but somehow... I think I prefer this way of doing it

Ted Shifrin

well, that's ok ... both have their advantages

PVAL-inactive

really the bottom thing isn't a triangle.

Ted Shifrin

you can prefer super fancy stuff, but it's not always pedagogically appropriate. We've had this argument before.

PVAL-inactive

or wait

Ted Shifrin

00:38

It limits to one, PVAL.

Akiva Weinberger

@TedShifrin He edited the sentence (though I don't know what was wrong with the first version?)

PVAL-inactive

I guess you have to assume the sum is finite for it to work then.

Ted Shifrin

Oh ...

Well, I thought that at first, PVAL, but if you construct the figure recursively, you prove it because the line you drew isn't parallel to the $x$-axis.

Akiva Weinberger

I mean if you really want to you can go the measure theory route

Ted Shifrin

WTF?

Akiva Weinberger

00:39

That closed line segment is the disjoint countable union of lots of half-open intervals

and a point

PVAL-inactive

What line are you drawing

Akiva Weinberger

Use countable additivity

Ted Shifrin

oh good grief

the line through $(0,1)$ and $(1,r)$.

Akiva Weinberger

@PVAL-inactive The one made of infinitely many smaller ones

Ted Shifrin

NO.

Akiva Weinberger

00:40

@TedShifrin No I mean the flat one

Ted Shifrin

I know you meant that.

Akiva Weinberger

from $(0,0)$ to $(0,\frac1{1-r})$

Ted Shifrin

The $x$-axis is just sitting there. Good grief.

Akiva Weinberger

What?

Lucas Henrique

how is the 0 < r < 1 restriction obvious?

Ted Shifrin

00:41

You mark off distances along it.

Akiva Weinberger

@TedShifrin What did I do

Ted Shifrin

Because if $r\ge 1$ the line (ray) and the $x$-axis do not intersect and you get no figure.

Akiva Weinberger

The picture shows that the interval $[0,\frac1{1-r}]=\{\frac1{1-r}\}\cup\bigcup_k[\sum_{n=0}^k r^n,\sum_{n=0}^{k+1} r^n)$, and that's a disjoint union

is all I was saying

Wait no

Fixed I think

Ted Shifrin

I'm not being that level of pedantic. I don't know what the issue is.

Akiva Weinberger

I don't know why you yelled "NO" at me

Ted Shifrin

00:46

I don't either. I was talking about the other line. I don't get the issue.

anyhow, I'm out of here.

Daminark

See you!

Lucas Henrique

01:20

cya

Damn, I always fail to understand $\exists a, b \in \Bbb N: p = a^2 + b^2 \iff p \equiv 1 \, \mathrm{mod} \, 4$, $p$ prime

Akiva Weinberger

01:34

Well, one direction is easy, at least

but yeah

IIRC, you first prove that $x^2\equiv-1\pmod p$ has a solution, then you factor $x+i$ in $\Bbb Z[i]$

Daminark

Isn't that putting the cart before the horse somehow?

Oh well maybe not

Akiva Weinberger

Like, take $p=13$. We know that $5^2\equiv-1\pmod p$, so $5+i$ has norm $26$

The prime factorization of $5+i$ is $(3-2i)(1+i)$ I think

Daminark

So, say that $p = a^2 + b^2$, then one of those is odd and the other is even, so that $(2k+1)^2 + (2n)^2 = 4k^2 + 4k + 1 + 4n^2$, so mod 4 that's just 1

Akiva Weinberger

@Daminark That's the easy direction

Lucas Henrique

yeah.

Akiva Weinberger

01:37

(Cont'd) and the norm of $3-2i$ is $3^2+2^2=13$

Lucas Henrique

the hard part is to prove the such $a, b$ exist

Akiva Weinberger

Proving that $x^2\equiv-1\pmod p$ always has a solution when $p$ is $1$ mod $4$ is genuinely hard, and proving the relevant lemmas about prime factorization in $\Bbb Z[i]$ (the Gaussian integers) is interesting

but those are the main ingredients

And the $x^2\equiv-1\pmod p$ thing has a very quick and clever proof.

To be clear, $\Bbb Z[i]$ (aka the set of "Gaussian integers") is the set of numbers of the form $a+bi$ where $a$ and $b$ are integers

Lucas Henrique

@AkivaWeinberger yeah, ikr

even tho idk what "prime factorization" in Z[i] means

(i'm too lazy to type this in LaTeX)

Akiva Weinberger

@LucasHenrique Well. You know how $5$ is "prime" in the integers, even though $5=(5)(1)=(-5)(-1)$

You can't write it as the product of other things… unless you use $1$ or $-1$

Daminark

Is there not a way to do it using Fermat's little theorem?

Lucas Henrique

01:43

I think that congruences won't help that much

Akiva Weinberger

@LucasHenrique Now, $5$ isn't prime in the Gaussian integers, because it equals $(2+i)(2-i)$

Lucas Henrique

Since we're looking for """concrete""" numbers

@AkivaWeinberger wow.

Akiva Weinberger

$2+i$ is prime in the Gaussian integers, though.

Daminark

Something like, $(-1)^{\frac{p-1}{2}} = 1 \mod p$

Lucas Henrique

what is a norm, again? in Z[i], I mean

Akiva Weinberger

01:45

But there's still a subtlety because you can write it as $(2+i)(1)=(-2-i)(-1)=(1-2i)(i)=(-1+2i)(-i)$

So the way primes work in the Gaussian integers, is that they're considered prime if you can't write them as the product of other stuff, without using $1$, $-1$, $i$ or $-i$

They're like $1$ and $-1$ for the integers

The reason they cause problems is 'cause you can divide by them

so everything is something times $-1$, and everything is something times $i$, etc

Daminark

Oh wait hold on a second

Akiva Weinberger

@LucasHenrique The norm of $a+bi$ is $a^2+b^2$.

Daminark

So assume $x^2 \equiv -1 \mod p$

Then $x$ has order 4

So by Lagrange's theorem $4\mid p-1$

Akiva Weinberger

This has the really nice property that the norm of the product of things is the product of the norms

Daminark

That's a nice way of thinking about that direction

Lucas Henrique

01:47

so as usual? $N(a) = a \overline{a}$?

Akiva Weinberger

@Daminark That's not the direction we want though

@LucasHenrique Yeah

Daminark

Oh I know, the other direction can be done with Wilson's theorem

Akiva Weinberger

So like the norm of $(1+2i)(3+4i)$ is equal to the norm of $1+2i$ times the norm of $3+4i$

Daminark

It's just that I only realized now that there's a way of doing this as such

Akiva Weinberger

and you prove it by writing it like $a\bar a$

Daminark

01:48

$(p-1)! \equiv -1 \mod p$

Akiva Weinberger

@Daminark There's a much cleverer way

@LucasHenrique So, the norm of $5$ is $25$, and the norm of $2+i$ is $5$

There are two important things to know here: (1) Every Gaussian integer can be written as the product of Gaussian primes in a unique way (up to factors of $-1$ and $i$ and $-i$), and

(2) the norm of a Gaussian prime (assuming the Gaussian prime isn't real) is a prime in the integers

So like $2+i$ is a Gaussian prime, and its norm, $5$, is a prime in the integers

Lucas Henrique

@AkivaWeinberger OOOOH.

This is so clear, now.

Akiva Weinberger

So once you've solved $x^2\equiv-1\pmod p$, then you know that $x^2+1$ is a multiple of $p$

so the norm of $x+i$ is a multiple of $p$

so you can factor $x+i$ into the product of Gaussian primes

and one of those Gaussian primes will have to have norm $p$

Daminark

Oh there's a fucky way to do this if you use Fermat's little theorem actually

Okay so

Akiva Weinberger

and so if that Gaussian prime is $a+bi$, then $a$ and $b$ solve your thing

Daminark

01:52

$x^{p-1} - 1 = (x^{\frac{p-1}{2}} - 1)(x^{\frac{p-1}{2}} + 1)$

Akiva Weinberger

We still need to show that $x^2\equiv-1\pmod p$ always has a solution when $p$ is $1$ mod $4$

Lucas Henrique

Conjecture: Let $z \in \Bbb Z[i]$. If $\mathrm{N}(z)$ is composite, $\exists z_1, z_2 \in \Bbb Z[i] : z = z_1 z_2$

Akiva Weinberger

@LucasHenrique Yeah that follows from my statement labeled (2) above, which I wrote without proof

Daminark

Now, we know that $x^{p-1} - 1$ has $p-1$ roots

So in particular, $x^{\frac{p-1}{2}} + 1$ has a root

Akiva Weinberger

Wait @Daminark

Oh I see you're using that a polynomial in $\Bbb Z/p\Bbb Z$ can't have more roots than its degree

Daminark

01:55

waiting

Yeah exactly

So, take the root of that equation, $r$

And look at $r^{\frac{p-1}{4}}$

Square that, you get $-1$

Akiva Weinberger

Yeah that works

Should I present the clever solution of the $x^2\equiv-1$ thing that I was thinking of?

Daminark

Sure

Akiva Weinberger

So the idea is that we can take the integers mod $p$, so there's $p$ of those,

and partition the nonzero ones into sets of the form $\{x,x^{-1},-x,-x^{-1}\}$

Essentially you're writing an equivalence where $x\sim y$ iff one of $x$, $x^{-1}$, $-x$, and $-x^{-1}$ is $\equiv y$

One of these sets, the equivalence classes, is $\{1,-1\}$

Lucas Henrique

@AkivaWeinberger are a professor or something?

Akiva Weinberger

For all other $x$, all of the elements of the equivalence class containing $x$ are distinct… unless $x=-x^{-1}$

@LucasHenrique Nah

Lucas Henrique

02:00

you guys, in general, are so clever.

Akiva Weinberger

I'm quoting a proof I didn't come up with

Daminark

^He knows the great Lobachevsky

Akiva Weinberger

Now if $p$ is $1$ mod $4$ then the number of nonzero things is a multiple of $4$

^^That's a Tom Lehrer reference

Look up the song "Lobachevsky"

Daminark

Also while you're at it, look up "Numberwang" if you want to know what math is all about

Akiva Weinberger

So if most of those equivalence classes have four elements, and one of them has two elements, we need another one to have two elements as well for it to work

(otherwise we'd have that the number of nonzero things is 2 mod 4)

So, we need an element such that $x\equiv-x^{-1}$

Multiply both sides by $x$, you get $x^2\equiv -1$

QED

(I implicitly assumed that the only solutions to $x\equiv x^{-1}$ were $1$ and $-1$.

But that has to be true, because that's equivalent to $x^2\equiv1$,

and so any solution satisfies $(x+1)(x-1)=x^2-1\equiv0$.

When working modulo a prime number, the product of nonzero things is nonzero, so either $x+1\equiv0$ or $x-1\equiv0$,

Daminark

02:07

Nifty

Akiva Weinberger

and so $x$ is either $1$ or $-1$.)

Daminark

Actually wait I don't remember how the proof that $p = a^2 + b^2 \iff x^2 \equiv -1 \mod p$ is solvable goes

Gimme a sec

Akiva Weinberger

@Daminark I did it above

$x+i$ has norm divisible by $p$

Factor it into Gaussian primes

Nonreal Gaussian primes have prime norm

Zophikel

For Integrals in the form $\frac{1}{2 \pi i} \oint_{\gamma} \frac{f(\zeta)}{(\zeta - z)} d \zeta;$ as seen in complex analysis is their a way to apporximate them

Akiva Weinberger

so one of them must have norm equal to $p$

user2457324

02:09

Hi -- can someone please help me with this question: https://math.stackexchange.com/questions/2376054/satisfying-the-following-determinant-inequality
thank you!

Daminark

Well that's $f(z)$

Akiva Weinberger

@Zophikel I think it's always less than the circumference of the path times the maximum value of the integrand's absolute value on the path

Zophikel

@Daminark I know but how would one approimate an integral in that form "

Daminark

Well, I know there's the length of $\gamma$ times a bound on the integrand, beyond that I dunno

Akiva Weinberger

To approximate it, reparametrize it into a real integral and use stuff like the trapezoid rule and stuff, I guess

Daminark

02:10

@Akiva somehow I think there might be another way to do this

Zophikel

@AkivaWeinberger you have an example of how to do that

Akiva Weinberger

I'm forgetting why nonreal Gaussian primes have prime norm

Zophikel

@AkivaWeinberger would the ML Inequality work ?

Akiva Weinberger

I forget what that is

Zophikel

Estimation lemma

In mathematics the estimation lemma, also known as the ML inequality, gives an upper bound for a contour integral. If f is a complex-valued, continuous function on the contour Γ and if its absolute value | f (z) | is bounded by a constant M for all z on Γ, then | ∫ Γ f ( z ) d z | ≤ M l ( Γ ) , {\displaystyle \left|\int...

Daminark

02:14

@Akiva think about this

Zophikel

@AkivaWeinberger note for the integral I gave $\gamma$ would be the closed disk

Daminark

If $a^2 + b^2 = p$, then $a^2 \equiv -b^2 \mod p$

Akiva Weinberger

Oh yeah so isn't that what both Dami and I already mentioned? @Zophikel

Zophikel

@Akiva true I just didn't think the ML-Inequality would work for the case of a the integral being definted on a closed disk

Since I usually see it being employed on Contours such as the semi=circular contour

Daminark

@Zophikel you can't integrate over the disk in this case, you've got a singularity

Zophikel

02:22

@Daminark oh dang :(

Daminark

Also what you wrote was a 1-form

Akiva Weinberger

I broke Reddit?

Daminark

:O

Zophikel

@Daminark then how would you go about evaluating the integral ?

Akiva Weinberger

You want to integrate over the circle, not the disk

"Disk" means the 2D thing that includes its interior

Zophikel

02:28

ahhhh ok Cauchy's INtegral Formula

bassically

Akiva Weinberger

The "circle" is the boundary of the disk

Zophikel

ahhh ok

Akiva Weinberger

Also I love this error picture Reddit also has

BAYMAX

:)

that's a cat and ?

Daminark

That looks so sad

Zophikel

02:35

so @AkivaWeinberger evaulating the integral: $\frac{1}{2 \pi i} \oint_{\gamma} \frac{f(\zeta)}{(\zeta - z)} d \zeta; = f(z)$

I feel like I did something wrong

Akiva Weinberger

Isn't that literally the statement of the Cauchy integral formula

@BAYMAX The Reddit alien

Zophikel

yeah pretty much

BAYMAX

ohh...I see..looks like little Baymax with red eyes :)

Zophikel

lol

Leaky Nun

03:12

@Abcd I just haven't appeared in this chat.

Pythonista

03:35

Can anyone point out what's wrong with my expression for this integral (mathematica): s11.postimg.org/wm5tr85z7/Capture.png I have the partial (wrt x) of an unknown function (in x and y) = function * constants. Not sure how to enter this abstractly in mathematica

Akiva Weinberger

04:30

Is "derpy" in Merriam-Webster yet

No but Oxford has it

Lucas Henrique

@AkivaWeinberger I actually didn't understand your argument here.

BAYMAX

if $\sum_{n=1}^{\infty}a_{n}$ converges then $\sum_{n=1}^{\infty} \frac{a_{n}}{1 + a_{n}}$ diverges I think?

as $\sum_{n=1}^{\infty} 1 - \frac{1}{1 + a_{n}}$

Lucas Henrique

well, writing that as a partial fraction does not help. :p

BAYMAX

the first term itself diverges

Lucas Henrique

yes. does the second one also diverge? idk, i suck at series

if the first diverges and the second converges, I think that the series should diverge

BAYMAX

04:40

If any one diverges then the whole will diverge

its like a single bad grape will spoil the whole bunch of good grapes :)

I also think it doesnot even matter whether $\sum_{n=1}^{\infty} a_{n}$ converges or not

Lucas Henrique

@BAYMAX not, maybe?

I mean

BAYMAX

any how the summing of $1'$s will make the series $\sum_{n=1}^{\infty}\frac{a_{n}}{1 + a_{n}}$

divergent!

Lucas Henrique

$\Sigma n - n$ converges

BAYMAX

I mean $\frac{a_{n}}{1 + a_{n}}$

Lucas Henrique

you have two divergent series

BAYMAX

04:43

form only

Lucas Henrique

oh, ok.

BAYMAX

but still the question spirals

may be $\frac{1}{1 +a_{n}} $ is divergent

and the difference of two divergent series may or may not be convergent!

so it matters that does $\frac{1}{1 + a_{n}}$ is cvgt or dvgt

also how that will affect

when $a_{n}$ are positive real numbers

for all n

Lucas Henrique

@AkivaWeinberger, @Daminark: we could have applied Euler's criterion directly, so we could jump to the abstract algebra part.

Akiva Weinberger

@LucasHenrique Most of those sets have 4 elements. At least one of them has 2 elements. They're all disjoint. Thus, the total amount of elements is going to be 4+4+…+4+4+2

I mean, that's what's gonna happen if there's only one with 2 elements

If there's two with 2 elements, the total is gonna be 4+4+…+4+4+2+2

Note that the total should be p-1, which is a multiple of 4 (we're looking at the nonzero numbers mod p)

4+4+…+4+4+2 is not a multiple of 4, so that can't happen

To take an example, let's say $p=13$.

The sets are then {1,12}, {2,11,7,6}, {3,10,9,4}, {5,8}

(Remember, each set is of the form $\{x,-x,x^{-1},-x^{-1}\}$)

Because the total has to be 13-1=12, a multiple of four, there has to be a set like {5,8} in there

which only has two elements because $-5^{-1}=5$

And, indeed, $5^2=25=-1$

(mod 13)

user84215

05:05

After your astonishing participation in the first event, The Ways That New Groups Can Be Constructed From Given Groups, I have decided to create the second one. Please inform me of your suggestions by pinging me or posting in Discussing Specific Topics.

user84215

Thanks.

Lucas Henrique

why there can't be a set with only one element? and why "most" of the sets have 4 elements? why do at least one have 2 elements?

Semiclassical

To have a set of one element, it sounds like you'd need $x=-x=x^{-1}=-x^{-1}$.

But $x=-x\implies 2x=0$ mod $p$. Which means $p$ would have to be even so that $x=p/2$.

So $p$ would have to be even.

Lucas Henrique

OOOOOH.

Equivalence classes.

That's simply brilliant

@Akiva tbh, Euler's method is easier, but your approach is very cool

yeah. now it's easy to see why the sets can only be of sizes 4 or two.

BAYMAX

I see this $\sum_{n=1}^{\infty} \frac{a_{n}}{1 + a_{n}} < \sum_{n=1}^{\infty}a_{n}$

so if $\sum_{n=1}^{\infty}a_{n}$ converges

then so does $\sum_{n=1}^{\infty}\frac{a_{n}}{1 + a_{n}}$

user2457324

05:23

Hi -- can someone please help me with this question: https://math.stackexchange.com/questions/2376054/satisfying-the-following-determinant-inequality
thank you!

user462339

05:37

What is the smartest way to determine the values of the killing form for the adjoint representation (Lie algebras)?

Say for example I want to calculate the killing form for $\mathfrak{sl}(2,\Bbb C)$. Then I can write out the adjoint matrices, and then due to the trace satisfying $\text{tr}(AB)=\text{tr}(BA)$ I only have to calculate $6$ products $k(e,e),k(e,f),k(e,h),k(f,f),k(f,h),k(h,h)$. Is there a smarter way?

pbeentje

Hi all. I came across this post and am trying to understand the step that u/quantumelixir posted. He eliminates t from two equations and ends up with a clean equation, but I haven't yet been able to reproduce his result. Can someone give me any pointers?

Lucas Henrique

can someone provide a proof for this?

Let $\mathrm{N}(\alpha)$ denote the norm of $\alpha \in \Bbb Z [i]$. Show that $\mathrm{N}(\alpha)$ is composite $\implies \alpha$ is composite.

user462339

05:54

What have you tried?

Lucas Henrique

Nothing. I mean, I don't even know where to start.

user462339

Tried writing out $N(\alpha)$ for an arbitary $\alpha=a+bi$?

Lucas Henrique

That's why I'll start @Ted's book on abstract algebra

user462339

Where did you come across this?

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