infinite schminfinite

Heya @Sanath

I have to leave in, say, 15 minutes or so.

But we can chat till then.

@SanathDevalapurkar What have you been upto?

Any fun stuff?

Oh, I just saw your blogpost in here, @Sanath. Glad to know you've been reading some number theory.

If I have function $f:\mathbb{R}^3\rightarrow\mathbb{R}$ where $f(x,y,z)=xy-z^3$ then is the ith component function is just $f_1(x,y,z)=xy-z^3$?

@MatsGranvik it means a pair (A,B) of matrices, where we distinguish (A,B) from (B,A) if A,B are distinct

OK, I gotta go now.

Hi all! How could I compute the square matrix of an $n \times n$ diagonal matrix?

@SanathDevalapurkar are you a prodigy?

@nullgeppetto what is square matrix ? :o

@r9m, sorry, I mean square root of a diagonal matrix...

:)

I am working for many hours now.. sorry

I've made this typo even in my notes (handwritten), so thank!

*thanks!

@r9m, by the way, do you have an answer?

@nullgeppetto take a square root of each entry

(there are many square roots of a matrix)

@nullgeppetto en.wikipedia.org/wiki/Square_root_of_a_matrix .. :D I didn't know there were so many methods :o

@blue if a diagonal entry is negative ? :o

@r9m oh, so now you want a square root that's real?

you gotta say things like that from the get-go

@blue yes ! :D ^_^

there may or may not be a real square root

the number of negative diagonal entries must be even

are you working with a specific matrix, or do you really want to know a general method?

@blue general :)

@blue herro

@r9m, thanks! (I need some sleep...), @blue, :D

@r9m What with?

Oh.

I read "help" not "sleep."

I'm off to study algebra.

@Chris'ssis "Who the hell cares?"

@PedroTamaroff I care ... ;)

@Chris'ssis It is just some random sequence.

@PedroTamaroff Not really. Its solution would amaze you.

Anyway.

@Chris'ssis I have been staring at it since I woke up half an hour ago ... lol .. its crazy :P

brb

Honestly, if an interviewer thinks solving that reflects intelligence...

@PedroTamaroff That's true. Most of the interviewers are really stupid (at least the ones I met).

@r9m Indeed. :-)

I don't know why interviewers here try to show off during the interviews, and they simply think they are smarter if they ask impossible questions.

Every time they ask me these questions, I immediately ask them a question of mine.

@Chris'ssis lol :P ... you are aggressive ?! :P

@r9m I just present them the reality. Some are shocked to realize they aren't able to solve a problem for kids. I'm not aggressive, but I hate the way they try to show off.

Some are decent, and simply ask good questions you'd love to answer.

@Chris'ssis So you squeal interviewers during interviews ? :P .. thought its the task of interviewers to squeal the candidates .. not the other way round :P ..

@r9m Would you like to work for someone that does not respect you and that would probably continue to show off when you occupy that position?

It's really bad to work for someone that is not able to correctly appreciate the quality of your work.

@Chris'ssis does not respect me => nope, won't work there ... shows off => I don't care as long as I can learn something :P

@r9m If I'm right, I cannot say my boss is right. I simply cannot. I better die.

@Chris'ssis oh .. I'm possibly misinterpreting that .. :|

I love to work with open-minded people ...

@r9m I mean that we should be correct in everything we do. If what I say, do is correct, then it is correct, no matter if some like this or not.

Hi @DanielFischer! May I havea minute of your time? If so, that's my question: Am I possible to define a function that given an $n\times n$ matrix (let it be symmetric, positive-definite) it returns the orthogonal matrix after applying SVD to it? I mean, I would like to present it in a rigorous way, so, could I find a function, or operator, or something? Thanks a lot!

@Chris'ssis hmm .... we can't behave like perfect machines that makes no mistake :P

@r9m not perfect, but correct. :-)

@Chris'ssis They only ask questions to which they have been told the answer :-)

@robjohn I know it! True! :-)))))))

:16527700 Yeah, but I think you didn't get my point. I wanted to say that people should accept you're right when you're right and when you can prove this. You cannot say you're wrong just because you have a discussion with your boss.

@Chris'ssis Ya .. they definitely should do that :D

@Chris'ssis I don't get it :|

hint?

:16527700 I kid my wife all the time that I am omniscient. She just laughs.

@robjohn :-)

@nullgeppetto I'm not sure which orthogonal matrix you mean. The orthogonal matrices in an SVD as well as in an orthogonal diagonalization of a symmetric matrix are not unique. So you'd have to make a somewhat arbitrary choice. But in principle, you can make that choice and define such a function. You can even describe such a function explicitly for invertible symmetric matrices $A$. Then $A^TA$ has a unique positive definite square root, and $\sqrt{A^TA}^{-1}A$ is orthogonal.

However, computing $\sqrt{A^TA}$ is not entirely trivial.

@DavidSpeyer FUUUUUUUUUUUUUUUUUU

you wrote your answer before me

@DanielFischer, thanks! I mean that, if $A$ is a symmetric positive definite $n\times n$ matrix, then its SVD is $A=VD^TV$, where $D=\text{diag}(\lambda_1,\ldots,\lambda_n)$ is the diagonal matrix containing the positive eigenvalues of $A$, and $V$ is an orthogonal matrix. Isn't $V$ unique?

@blue Hey.

That's a lot of rage.

@nullgeppetto That is $VD({}^TV)$, is it? Then the columns of $V$ are eigenvectors of $A$. If all eigenvalues of $A$ are distinct, then you only can multiply any subset of the columns by $-1$. But if $A$ has eigenspaces of dimension greater than $1$, there is more freedom in the choice of $V$.

Another sick downvote here math.stackexchange.com/questions/215352/…

@blue $(3,1+\sqrt 10)$ is not a principal ideal in $\Bbb Z[\sqrt 10]$; amirite?

@DanielFischer, and thus it cannot be defined as a function? Sorry for my unawareness! Thanks a lot!

@Chris'ssis I think the other half of this answer would have been better to use.

@PedroTamaroff $\sqrt10$? ;-)

@PedroTamaroff hint: 3 is not a quadratic residue mod 10. also, rage rage rage rage rage

@Balarka iirc there is some kind of a method of 'infinite' induction, from set theory..

transfinite

@nullgeppetto You can define a function that returns one possible $V$. But that involves an arbitrary choice (that's not necessarily a bad thing). Mathematically, you can also define a function returning the set of all possible $V$.

Thanks a lot @DanielFischer! You helped once more! I think I need some sleep now! (By the way, where are you living? Germany?)

@nullgeppetto Bremen. Kali Nichta.

@DanielFischer! Ok then! Angenehme Nacht!

Danke

:)

@blue I was trying to show 3 is irreducible but not prime in such ring

so if 3 is a product of thingies

3=aa'+bb'10+(ab'+a'b)sqrt 10

so ab'+a'b=0.

and 3 = aa' mod 10

or N(3)=9 so a nonassociate divisor must have norm 3

Oh. =P

@blue N(a+b sqtr 10 ) = a^2-10b^2, yes?

I hesitated for a second.

yes, +/- sqrt10 are conjugates, so the conjugate of a+bsqrt10 is a-bsqrt10, and their product is a^2-10b^2

So, now I have shown 3 is irred.

To show it is not prime, I want to observe that it divides 9 = (sqrt 10-1)(sqrt 10 +1 ) but doesn't divide any of those factors.

if 3x= sqrt 10 -1, taking norms gives 9 Nx = - 99 , so Nx=-11

don't bother with norms there

OK.

Right, I am being stupid.

ever element of $\Bbb Q(\sqrt{10})$ has a unique representation as $a+b\sqrt{10}$ for $a,b\in\Bbb Q$

Yes.

so $\frac{1}{3}+\frac{1}{3}\sqrt{10}$ cannot be written with integer coefficients there

Well, that does it.

@blue I have to do the same with $\Bbb Z[\sqrt{-10}]$: show it is not a DFU, and exhibit a nonprincipal ideal.

is dfu the espanol ordering?

Yes, sorry. =)

"dominio de factorización única"

It is funny it is exactly backwards.

It take it you have a better perspective of this from field theory (Since you mentioned $\Bbb Q(\alpha)$s, norms, conjugates...)

galois theory and algebraic number theory

@PedroTamaroff: does my answer intrude on yours. Would you like me to delete mine?

@PedroTamaroff hint: the norm can only take certain values. so, as last time, if N(x) is the product of two rationally prime values N() can't take, then x must be irreducible.

@robjohn i was kidding

np

@PedroTamaroff Okay, just checking. I don't want to step on any toes.

@blue "rationally prime" means?

(though I get your idea)

@robjohn Is it okay(by community rules) If I put sth like 'I'll reward a +100 bounty for a nice answer' in a question that I asked (without actually putting the bounty .. coz it expires in 7 days and I loose the points for nothing :P ) ? :D

@PedroTamaroff it means prime in $\mathbb{Z}$.

@r9m There was a meta question about this, and I think the consensus was that circumventing the bounty system was not good. Let me look for the thread.

@robjohn is it possible to put a user on 'ignore' on main (e.g., so you don't see comments they make on your stuff)?

@MikeMiller I don't think that is possible. Chat has that, but not main.

@r9m I can't find the thread. I wonder if the question was deleted.

@robjohn Okay thanks ^_^

@DanielFischer Oh, the "rational integers".

>not the integers

@r9m take a look at this proposal

@PedroTamaroff which could include algebraic integers.

@blue OK. I obtained $2+\sqrt{-10}$ that has norm $2\cdot 7$ and $2$ and $7$ are not QR mod 10.

So this is irreducible.

mmhmm

and fails to be prime because (same reason as last time)

rationally prime means prime in Z

@blue that times conjugate is 14, but it doesn't divide either.

mmhmm

can you show $(2+\sqrt{-10},2)$ is not free as a $\Bbb Z[\sqrt{-10}]$ module? :)