Mathematics - 2013-02-19 (page 1 of 5)

user43758

00:02

I think I know

we will do f of the first colum of the matrix we found

no wait I think I know

Let me just try it on paper

Peter Tamaroff

OK.

Recall how we builld our matrix with the bases.

user43758

Yes it's basically Mat_{B,C}(f)

such that B,C are the basis of V an W respectively

Peter Tamaroff

Yes, yes, but recall how we obtain it.

user43758

ok

Peter Tamaroff

Take the first vector of the basis, and apply $f$.

user43758

00:05

yes let me just do it on the paper. I will tell you what I get

Peter Tamaroff

Then take $f(v)$ and put it in the 2nd basis coords.

What you want is think about "imaginary" basis. I think you should reach an impass, as DonAntonio shows.

Remember you want to model a diagonal matrix.

Something like
a 0 0 0
0 b 0 0
0 0 c 0
0 0 0 d

user43758

The only thing that I see works is the null matrix

Peter Tamaroff

Good. But that is clearly not our matrix.

So, you've reached an impass.

user43758

So what is the conclusion ?

Peter Tamaroff

Well, if you have done things right, that this bases and consequently the matrix cannot exist.

My question is.

Haven't you reached some awkward conclusion about what $B_1$ should be?

We clearly see $B_2$ is not a basis.

Since it is $\{ \vec 0\}$

And bases do not contain the origin.

You assumed $B_2$ was a basis, but what is it, really?

user43758

00:13

Yes. When I expressed the image of each term of the original basis. I obtained all of it 0

Peter Tamaroff

Wait. Did you write an "incognito" basis?

Like $B=\{v_1,v_2,v_3,v_4\}$?

We want to find $v_i$

Similarily $B'=\{w_1,w_2,w_3,w_4\}$ is the other basis.

user43758

This is what I did

Peter Tamaroff

Say $D=\{a_1,a_2,a_3,a_4\}$ are your diagonal values.

user43758

I used the following fact

Peter Tamaroff

You shouild get a big system.

@user43758 Hm?

user43758

00:15

Let E,F be 2 vector spaces, p=dim(E), n=dim(F) (this is from my lesson)

B=(e_1,...,e_n) a base of E, C=(f_1,...,f_n) a base of F

f is a linear application from E to F

Peter Tamaroff

Aha.

user43758

For each j in {1,...,p}, we denote (a1_j,...,an_j) the components of f(e_j) in C

therefore f(e_j)=\sum_{i=1}^n aij fi

Peter Tamaroff

@user43758 OK.

And with the $a_{ij}$ you build up your matrix.

user43758

Yes

Peter Tamaroff

Good.

user43758

00:18

So I started by expressing TD of each of the components of the matrix we got

and we obtain a null matrix

0000
0020
0006
0000

TD(2)=2TD(1)=0

TD(6)=6TD(1)=0

no?

Peter Tamaroff

Darn. You seem to be inputing just numbers. But $(0,0,6,0)$ is really $6x^2$.

You see?

user43758

AH

ok

and 2 is actually 2x

ok let me redo

Peter Tamaroff

Exactly.

But wait.

user43758

we get

0000
0000
00012
0000

Peter Tamaroff

What are you applying TD to?

Did you read what I wrote?

user43758

00:23

Yes

TD(2x)=2TD(x)=2*0=0

Peter Tamaroff

You might also show that your matrix cannot be conjugate to a diagonal matrix.

user43758

I am applying TD to each of the components of the matrix

Peter Tamaroff

That is, if your matrix were diagonalizable, there'd exists a matrix $D$ diagonal, not identically null and another matric $M$ such that, if say $N$ is your matrix, $MNM^{-1}=D$

But your matrix is nilpotent.

Contradiction.

Done

user43758

So I suppose the existence of a matrix of TD such that it is in diagonal form right

?

I am just a little confused with the M and N notations

Peter Tamaroff

Do you know what a matrix to change basis is?

user43758

00:29

as in Pass(B,B') ?

the matrix of passage from B to B' ?

Peter Tamaroff

Yeah

user43758

I know but I can't use it here

because the problem is regarding the material before

Peter Tamaroff

@user43758 Well, I really think this is the best way to go.

Maybe your problem is ill posed.

user43758

Well it is in the Apostoll book

Peter Tamaroff

Else, to diagonalize, I'd just use eigenvectors and eigenvalues.

@user43758 Oh, I have it.

That is useful for you to say!

Which page?

user43758

00:31

It is in the volumee 2

Peter Tamaroff

@user43758 No problem.

user43758

page 51

Peter Tamaroff

@user43758 Add it to the question.

user43758

problem 20 in the 2.12

You have it?

@PeterTamaroff

Peter Tamaroff

@user43758 Found it.

user43758

00:34

@PeterTamaroff It is the last one of the 2.12 section and I had to do the 19

That is why I was confused on what to put or not

anon

@Ethan The quotient of two dirichlet series will also be a dirichlet series. I feel it should be the case that if $q(s)$ is bounded as $s\to1^+$, then $\lim\limits_{s\to1^+}q(s)$ should exist. (Also, I think any dirichlet series will have unconditional convergence past a certain point, but I have nowhere to point to.)

user43758

@PeterTamaroff So to use the nilpotence argument in order to reach a contradiction

?

Peter Tamaroff

@user43758 I must be misreading something. How can Apostol be wrong?

Let me re-read a little.

user43758

ok

math101

Would you classify the average mathematician as egocentric?

user43758

00:39

Yes but it's not a bad thing :)

math101

egocentric has a negative connotation

Peter Tamaroff

@math101 Bleh.

I think not.

To your question, I mean.

@user43758 Note that ${\rm Im}(V)$ is the span of $(x,x^2)$

The polynomials of the form $ax+bx^2$.

I'm thinking what would happen if he really menat $DT$

That is, $p'+xp''$.

Maybe it would work out.

user43758

ok

Peter Tamaroff

Nope.

It is still nilpotent.

user43758

@PeterTamaroff Yes

So I can use the nilpotence argument for this problem right ?

Peter Tamaroff

00:46

I think so, but I feel like I'm missing something.

peoplepower

@JacobBlack I already own it. :) I was looking for a different perspective.

user43758

@PeterTamaroff I was wondering if you knew any good websites to practice problems?

Peter Tamaroff

@user43758 Not really. I prefer books.

You should find a lot of exercises in math.SE though.

user43758

I prefer books too but I assumed you would know more website

Do you recommend any?

I am taking Multivariable Calculus with Theory this term

I am a freshman

Peter Tamaroff

I know none.

user43758

00:52

ok..

anon

books on multivar calc?

user43758

Yes and perhaps Real Analysis

anon

I used Vector Calculus by Susan Colley in high school, I liked it.

user43758

@Ethan removed

:)

Ethan

I didn't read your last comment lol

Now I know

user43758

00:55

@anon

I am not in high school

I am in university freshman

anon

The calc III class I took in HS was twice as rigorous and enlightening than the one I took in university. Your point?

Then again maybe I only felt that way because I had already seen the material, who knows.

user43758

I wasn't making any point lol

Ethan

@anon Where did you take a course like that in hs

anon

my high school was blessed with a mathematical man who had delusions he was really a college professor...

(I think he would appreciate that epithet)

user43758

@anon you are lucky.

Peter Tamaroff

00:59

@anon Lucky you man!

I know you profited from that!

Ethan

@anon I still don't understand, did you take a course some where else?

anon

What do you mean? I took it once in HS, and once in uni last semester.

Ethan

@anon Well I assume your high school didn't offer calc 3 as a course to take

user43758

@PeterTamaroff Maybe we have to find a basis for both V and W ie not consider the given basis?

anon

@Ethan It did. It even offered Advanced Applied for one year, which was differential equations at a graduate level (but going 2-3 times slower). Only two other students took it with me. Both classes due entirely to the teacher I mentioned.

user43758

01:02

@PeterTamaroff I mean maybe we have to find a basis for V and W such that the matrix of TD is in diagonal form

Ethan

@anon How? Was it a special technical school or somthing?

anon

It was a private Jesuit school. It was not a special technical school, but it did emphasize academics. (Unfortunately as I left the direction seemed to be tipping in favor of sports - which I couldn't care less about.)

Three. Why not? My current p-adic class is only me and one other student.

user43758

I was in such a class as well

I was only with another student

Ethan

@anon I am not very familear with academia, also I suppose it depends where you go to school

The Substitute

Howdy Y'all. If I can find an upper bound on the Hausdorff dimension of $A_i$, can anything be said about the Hausdorff dimension of $\cup A_i$ ?

user43758

01:06

It is the sup of the dimension of the A_i

@TheSubstitute

dimH(XxY)\geq dimH(X)+dimH(Y)

@PeterTamaroff Am I misreading something in the book ?

The Substitute

Thanks, this follows from the Def. Thanks @user43758

Ethan

@anon, If I was to substitute a complex number z,with magnitude <1, into $$f(z)=-\ln(1-z)=\sum_{n=1}^\infty \frac{z^n}{n}$$, what exactly does $f(z)$ represent? In what sense is it still a logarithm? Would it satisfy $e^{f(z)}=z$?

user19161

@anon He is probably a banana.

anon

@Ethan no, it would satisfy $e^{f(z)}=\frac{1}{1-z}$

Ethan

@anon my bad, but why? I can add multiples of $2\pi i$ to an exponential with out changing the value, but that would change the value of the logarithm?

Peter Tamaroff

01:14

@Ethan This would happen only on the disk $|z|<1$.

And as anon corrected.

anon

keywords: "monodromy," "branch cut," "riemann surfaces"

user43758

@PeterTamaroff They are asking me to find bases for V and for W

@PeterTamaroff Therefore perhaps we can't use the one defined in the previous problem

anon

We make an arbitrary choice of confining the logarithm function to take complex values with imaginary part in $[0,2\pi)$, similar to what we would do for say $\sin^{-1}$ or $\cos^{-1}$ - a noninjective function does not have an inverse, so we are forced to make a choice of where in the image we want to confine ourselves to in order to define an inverse map that is not multi-valued.

Peter Tamaroff

Any homomorphic image of $G$ is isomorphic to a factor group $G/K$ by a normal subgroup $K$.

Jonas Teuwen

@Ilya Yes.

Peter Tamaroff

01:19

@JonasTeuwen Jonas. How's it chillin, brozzam?

anon

@PeterTamaroff Yes. You want all of those isomorphism theorems to be engraved in your intuition eventually.

Peter Tamaroff

@anon My "intuition"?

saadtaame

What's a manifold?

user43758

@PeterTamaroff Am I reading something wrong?

Peter Tamaroff

@user43758 I don't know

I really gave up on Apostol this time.

user43758

01:21

because they say find one for V

Not use the same one as the one before

Chokladkakan

It is late at night, but I felt I needed to have something in print here: http://math.stackexchange.com/a/307641/12880

Given the time I may very well be off. I would appreciate any and all scrutiny.

With that, I need to bed myself. Sleep well, MSE.

user43758

@PeterTamaroff

Peter Tamaroff

@user43758 What do you want to do?

user43758

Is it possible to proceed by constructing a base for V and for W

?

Peter Tamaroff

No, I don't think so.

@anon I think I have some nice exercise to do

user43758

01:25

Ok well thank you for your help

Peter Tamaroff

@user43758 No problem.

user43758

Have a good evening

Peter Tamaroff

It's night here, hehe.

10:25 pm

But thanks.

user43758

it's 8:25 still

:)

Jonas Teuwen

@PeterTamaroff Good. I moved my blog.

Peter Tamaroff

01:29

@JonasTeuwen Wiiiiiiiii!

Jonas Teuwen

Yes.

Peter Tamaroff

@JonasTeuwen During my vacations, I went to a brewery and tried a bunch of beers.

Some were pretty good.

Jonas Teuwen

That's great!!!

Peter Tamaroff

They served me a tray with their blends.

@JonasTeuwen I signed up for my math career today!

Jonas Teuwen

How so?

Peter Tamaroff

01:33

I went to my uni's "mothership" and made a life-sucking time-consuming mind-killing errand, and now I have a auxiliary credential till I get the official one.

Jonas Teuwen

Excellent.

Good boy.

Peter Tamaroff

And I can sign up for courses in a few days.

Jonas Teuwen

Now treat yourself with a beer.

Peter Tamaroff

@JonasTeuwen Good idea

Orange Harvester

@JonasTeuwen If you install your wordpress instance on your own server, you can have MathJAX.

Peter Tamaroff

01:36

@JonasTeuwen What do you think of Corona beer?

Jonas Teuwen

Like the joke with the cannoo.

@OrangeHarvester I have MathJax.

peoplepower

@JacobBlack Ah, this post is swimming in references.

Orange Harvester

@JonasTeuwen Yes, I saw that later. :-)

Peter Tamaroff

Dawg. THis is my uni's bibliography for Measure and Prob and Real Analisis

J. Cerdà, Análisis Real. Universitat de Barcelona, 1996.
N. Fava y F. Zó, Medida e Integral de Lebesgue. Red Olímpica, 1996.
G. B. Folland, Real Analysis - Modern Techniques And Their Applications. John Wiley & Sons, 1984.
P. R. Halmos, Measure Theory. Van Nostrand, Princeton, 1950.
S. Igari, Real Analysis - With an Introduction to Wavelet Theory. American Mathematical Society, Volume 177, 1998.
H. L. Royden, Real Analysis. Mc Millan, 1968.

What do you think about it?

Jonas Teuwen

Looks good.

Peter Tamaroff

01:44

@JonasTeuwen Thanks.

@anon

anon

?

Peter Tamaroff

I am determining $\text{aut}G$ for $G$ ciclic, infinite and then finite.

I' starting with infinite.

I'm thinking it is "simply" determined by $\eta(a)$

I mean, you take a value $\eta(a)$ and then it is $\eta(a)^k$

For any $k$.

anon

what are $a$ and $\eta$?

Peter Tamaroff

$\eta$ would be the automorphism.

$a$ would be $\langle a \rangle =G$

anon

a cyclic infinite group is just $\bf Z$ up to isomorphism, so might as well work with $\bf Z$

Peter Tamaroff

01:50

Yes, I know.

anon

any homomorphism out of a cyclic group is determined by where the generator is sent, yes

Peter Tamaroff

Oh, OK.

anon

an automorphism is a homomorphism

just like a square is a rectangle

Peter Tamaroff

Sure.

Thus there are $|G|$ automorphisms?

anon

there are |G| endomorphisms. the automorphisms are the invertible ones.

Peter Tamaroff

01:53

Oh, silly me.

Wait.

But if $G$ is infinite, $\eta(1)=1$ only for the identity.

anon

rephrase

Peter Tamaroff

So any endomorphism is injective.

@anon It is cyclic of infinite order.

anon

yes, endomorphisms on Z are injective

not necessarily surjective though

Peter Tamaroff

Well, so there are $|G|$ endomorphisms.

@anon No?

anon

pick some number a and consider the endomorphism $x\mapsto ax$

with $a\ne\pm1$ of course. $a=0$ would be the most extreme example. (indeed, the trivial endomorphism is never an automorphism, unless we are working with the trivial group.)

Peter Tamaroff

01:56

Oh, OK.

So the only homomorphism is the identity, yes?

anon

cough -1 cough

Peter Tamaroff

But that is just the same.

Since we go through negatives.

Oh, well, OK.

anon

no, $x\mapsto x$ and $x\mapsto-x$ are not the same map

Peter Tamaroff

Sure, sure.

I was being an ass to $x\mapsto -x$

anon

the inverse map is an automorphism in any abelian group. it is only identical to the identity map if the group is of exponent 2, which is necessarily abelian, and hence necessarily a direct sum of cyclic groups of order 2 (perhaps infinitely many - no conditions on cardinality are made)

Peter Tamaroff

02:00

Wait, isn't any infinite cyclic group isomorphic to $(\Bbb Z,+)$?

...

SHUSH ME!

anon

Both $x\mapsto x$ (the identity map) and $x\mapsto-x$ are automorphisms of $({\bf Z},+)$.

Peter Tamaroff

Yes, yes,.

Moving on to finite cyclic groups.

Case $|G|=6$.

Any finite cyclic group is isomorphic to a factor group $\Bbb Z_n$, yes?

@anon

anon

yes

Peter Tamaroff

So for this case it suffices that $\mod 6, \eta(1)=\pm 1$.

Jonas Teuwen

02:20

Alright. Good night guys.

Peter Tamaroff

@anon So $x\mapsto 5x$ would be an automorphism, yes?

And so will $x\mapsto x$

anon

in Z6 yes

Peter Tamaroff

OK, and those are them.

Wiat.

Wait.

Yes, I think that is it.

anon

show that if (a,n)>1 then the map x->ax in Zn will have nontrivial kernel. hence you can identify Aut(Cn) with multiplicative by units in the ring Zn.

Peter Tamaroff

My question is: why do we talk about $5\times x$ when our operation is $+$?

@anon Yes, indeed. I was thinking about $\varphi(n)$

anon

02:25

it is $x\mapsto x+x+x+x+x=-x$. IOW any abelian group is a Z-module.

peoplepower

$\eta(n)=\sum^n \eta(1)=n\eta(1)$.

Peter Tamaroff

@peoplepower Yes, I know.

peoplepower

That's where multiplication by a constant factor comes from.

Peter Tamaroff

I mean.

We're being lazy.

But how would you write it in terms of $+$?

peoplepower

@PeterTamaroff Be more specific please. I view multiplication here as repeated addition anyway.

Peter Tamaroff

02:27

@peoplepower Yes, nevermind.

I'm being a little silly.

peoplepower

It really is just a module structure, like anon said, but the scalar operation is canonical automatically being repeated addition.

Peter Tamaroff

Oh, OK.

Then $|\text{Aut}(\Bbb Z/n\Bbb Z)|=\varphi(n)$.

peoplepower

Indeed. Do you just need size?

Peter Tamaroff

My books says "Determine $\operatorname{Aut} G$ for..."

peoplepower

Ok, let's characterize the composition operation on it.

Peter Tamaroff

02:33

$x\mapsto ax\mapsto (ab)x$?

peoplepower

Yes. So we can write the first automorphism as $m_a$ and the second as $m_b$, what is the composite?

Peter Tamaroff

Since $(a,n)=1,(b,n)=1$,we have$(ab,n)=1$.

$m_{ab}=m_a\circ m_b$

peoplepower

Exactly, so it is just multiplication on the congruence classes of integers relatively prime to $n$.

I.e. it is the group of units of the ring $\mathbb Z_n$.

Peter Tamaroff

@peoplepower It is the reduced congruence class?

peoplepower

We don't care, $a$ could have been $a+n,a+2n,\dots$.

Peter Tamaroff

02:36

Yes.

peoplepower

We're just interested in the congruence class.

Peter Tamaroff

OK, I'm done with this. Moving on.

Determine $\operatorname{Aut}S_3$

anon

hint: transpositions are sent to transpositions

Peter Tamaroff

$S_3=\{{\rm id},(23),(12),(12)(23),(23)(12),(13) \}$

MSEoris

hey sorry to interject, but i need a quick sanity check: whats the first component of $1/2 \nabla (v_1^2 + v_2^2 + v_3^2)$

i get $v_1 dv_1 / dx_1 + v_2 dv_2 / dx_1+v_3 dv_3 / dx_1$

anon

02:50

yes

MSEoris

was that yes for me @anon

anon

yes

MSEoris

thanks!

Peter Tamaroff

@anon It suffices to determine values for $(12);(13),(23)$, =)

anon

group homomorphisms always send id to id

Peter Tamaroff

02:54

Yes, yes.

So I'm guessing we get $6$ automorphisms.

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