Mathematics - 2013-10-20 (page 1 of 4)

anon

12:17 AM

@KevinDriscoll what is the ISC?

Pedro Tamaroff

12:46 AM

@anon Inverse Symbolic...

@KevinDriscoll Square root!

@anilorap With what?

user60887

So I want to show that if G is abelian then every subgroup of G must be abelian. Suppose G is abelian and there is subgroup H that is not abelian. Let $a,b\in H$ it follows that $ab\in H$ Since H is not abelian then $ab\neq ba$. But since $H\subseteq G$ it follows that $ab\in G$. But G is abelian so $ab=ba$ a contradiction.

Pedro Tamaroff

@user60887 Wait wait wait.

You need no contradiction whatsoever.

Pick two $a,b\in H$.

user60887

oh is it just by definition of subgroups?

Pedro Tamaroff

@user60887 Well, yes.

You're restricting the operation to $H$.

The restriction preserves commutativity.

You have $\eta:G\times G\to G$.

user60887

oh thats what ithought i should in my proof

Kevin Driscoll

12:52 AM

@Pedro Holy shit..... you're right

sr(2) is a REALLY SHITTY notation for $\sqrt{2}$

3

Pedro Tamaroff

@KevinDriscoll There are two options.

Either you buy a spaceship, and propel yourself straight into the sun.

Or you listen to PUDDI PUDDI for 15 minutes straight.

skullpatrol

@KevinDriscoll I agree!

Kevin Driscoll

Start time 8:55 god damnit

Pedro Tamaroff

@KevinDriscoll What was the problem you were looking at?

@KevinDriscoll I already did 2:30.

Kevin Driscoll

@Pedro suppose for $a>0$, $b>0$, $\int_{-a}^{b} f(\alpha,x) dx$ exists in the usual Riemann sense for some values of $\alpha$

Pedro Tamaroff

12:58 AM

@KevinDriscoll "...some values"?

Kevin Driscoll

I am trying to determine under what conditions we can analytically continue the resulting function $g(\alpha)$

and when the resulting analytic continuation equals the finite part of those integrals for values of $\alpha$ where the integral diverges

Pedro Tamaroff

@KevinDriscoll Heh, good luck with that! =D

I will keep reading FDVS.

Kevin Driscoll

(wow.... its only been 6 minutes and I'm already about to die. This was fun for the first 2 minutes or so and then VERY QUICKLY deteriorated)

@Pedro In particular I was numerically computing the value of $F.P. \ \int_{-1}^{2} \frac{\log{\left|y\right|}}{y} \ dy$

Pedro Tamaroff

@KevinDriscoll You want $\log |y|$?

Kevin Driscoll

@pedro Maybe. I'm not sure yet.

Pedro Tamaroff

1:03 AM

@KevinDriscoll Well, how would you define $\log y$ for $y<0$?

Kevin Driscoll

@Pedro I think you can view Log as a single-valued function on the Riemann surface, and then you basically get to pick. I think the definition Mathematica uses is $\log{-x} = \pi i + \log{x}, \ \ \ x>0$

I'm worried about this log business cuz its not analytic

User10001

How can I find 'i' if I have p, m, and n in the recurrence equation p[n] = p[n−1]*(1 + i) − m or its closed form p = (m/i)(1-(1+i)^-n)

Is there a way to turn the equation in respect to i? I cannot figure it out.

Kevin Driscoll

@Pedro 15 minutes. DONE. That was actually pretty annoying.

Pedro Tamaroff

@KevinDriscoll Your soul is saved.

Take joy in that.

Kevin Driscoll

I do. I just hope next time the contrition isn't worse.

Pedro Tamaroff

1:14 AM

Mother's Day tomorrow. Went to buy something, and I hear a man talk with his wife, telling her about what he picked, asking for opinion. "What?! Hats have sizes too?!" Made my day.

Charlie

@skull

@PedroTamaroff funny, mothers day here is in may :)

@kevin what is your avatar?

skullpatrol

@Charlie hi/bye

Pedro Tamaroff

@Charlie Dem looks like 3D tilings.

Charlie

@skullpatrol bye

Kevin Driscoll

@Charlie Its a quasicrystal formed solely by putting polyhedral dice in a cubic container and shaking it for a long time

Charlie

1:25 AM

@KevinDriscoll are you sure?

Karl Kronenfeld

@PedroTamaroff Tries to read while listening to PUDDI PUDDI. "The PUDDI is PUDDI if and only if PUDDI PUDDI."

3

Pedro Tamaroff

@KarlKronenfeld It will deplete yer brainz.

Kevin Driscoll

@Charlie Yup. I was at a talk given by the person who made it

Charlie

@KarlKronenfeld HAHAHAHAHA

Kevin Driscoll

@Pedro @Karl However, I find endless Smooth jazz Nyan Cat to be quite relaxing

Charlie

1:34 AM

@KevinDriscoll fascinating

Karl Kronenfeld

@KevinDriscoll Oh, nyah, I see what you mean.

Charlie

Oh no @karl I can't stop.laughing :')

agent154

I'm wondering if I did something wrong here, or if I'm actually on the right track and don't see it... Can somebody give me any advice?

"If $x$ and $y$ are odd, prove that $x^2+y^2$ cannot be a perfect square."

If $x$ and $y$ are odd, then $x^2$ and $y^2$ are also odd; An odd number plus an odd number is an even number. Assume that $x^2+y^2=z^2$ for some integer $z$. Therefore, $(2k+1)^2+(2m+1)^2=(2l)^2$ for some $k,m,l\in\mathbb{Z}$.

$$(2k+1)^2+(2m+1)^2=(2l)^2$$
$$4k^2+4k+4m^2+4m+2=4l^2$$
$$4(k^2+k+m^2+m)+2=4l^2$$

Pedro Tamaroff

@agent154 Oh, good.

But note this.

agent154

@PedroTamaroff I suppose I could have said it explicitly, but if x and y are odd, then x+y=z is even

Pedro Tamaroff

1:45 AM

There is an important detail, that it seems you didn't leave explicit.

agent154

and if z is even then $z^2$ is also even

Pedro Tamaroff

@agent154 Yes, so $2\mid z^2\implies 2\mid z$.

That is important.

Note that $\mod 4$ the odds are $\pm 1$.

But $2\not\equiv 0\mod 4$.

agent154

I was wondering how I can look at this with congruences, as this question is part of the congruence chapter in my book. I just couldn't think of how to apply that concept to this.

Pedro Tamaroff

@agent154 What I said above =)

If $x,y$ are odd, then $x,y\equiv \pm 1\mod 4$.

Thus $x^2+y^2\equiv 2\mod 4$ and $\equiv 0\mod 2$.

Thus we have $2\mid z^2$, this gives $z^2=4\ell^2$, so $z^2\equiv 0\mod 4$.

But $2\not\equiv 0\mod 4$, thus the above is impossible.

You get that in this line $4(k^2+k+m^2+m)+2=4l^2$, @agent154

Modulo $4$ that gives $2=0$.

agent154

Is it important that I do this modulo 4? I'm looking at this on paper trying to think from the start using modulo 2... $x\equiv 1\mod{2}$ and $y\equiv 1\mod{2}\Rightarrow x^2\equiv 1\mod{2}$ and $y^2\equiv 1\mod{2}\Rightarrow x^2+y^2\equiv 1+1\mod{2}\equiv 0\mod{2}$. But where can I go from here to show that $x^2+y^2$ is not a perfect square?

Pedro Tamaroff

1:54 AM

@agent154 Well, yes, we need $\mod 4$. $\mod 2$ it is true that $2=0$.

Kevin Driscoll

Anyone know if $P.V. \int_{-1}^{2} \frac{\log{y}}{y} \ dy$ exists?

agent154

@PedroTamaroff Could just scale the equation up by two since $cx\equiv cy\mod{cm}$...

Both should be equivalent

Pedro Tamaroff

@agent154 You want $c\neq 0$.

Here $2=0$, @agent154

agent154

I think it's just as simple for me to stick with my first explanation and make sure I don't leave out anything important. It's not wrong, just not complete.

Pedro Tamaroff

@agent154 Your explanation is fine.

Karl Kronenfeld

2:00 AM

@agent154 A generalization: Let $x_1,x_2,\dots x_n$ be odd integers. What we have is $\sum_{i=0}^n x_i^2\equiv n\pmod 4$. Thus, the sum cannot possibly be a square when $n$ is not a square modulo $4$. In general, let $m$ be any integer such that all odd squares are $1$ mod $m$. Then $\sum_{i=0}^nx_i^2\equiv n\pmod m$. You get that the sum is definitely not a square when $n$ is not a square modulo $m$.

Pedro Tamaroff

@KarlKronenfeld One more dollar.

Karl Kronenfeld

@PedroTamaroff Thanks, I was proofreading the whole thing--I left out a word in my second sentence.

@PedroTamaroff

Sorry

Pedro Tamaroff

@KarlKronenfeld What?

Karl Kronenfeld

@PedroTamaroff (I accidentally hit enter last time.) Is there a characterization of those $m$ for which $x^2\equiv 1\pmod m$ when $x$ is odd?

agent154

Hmm.. Sometimes I regret taking Number Theory. I like the material, but it isn't easily absorbed. I guess I know the theorems but I just don't know how to apply them.

Pedro Tamaroff

2:04 AM

@agent154 You'll get used to it. =)

@KarlKronenfeld Other than reciprocity?

Let me think.

agent154

I spent hours on one question the other day because I didn't realize that $a^{p-1}\equiv 1\mod{p}$ was Fermat's Little Theorem... and how to apply it to my question "Prove $42\mid n^7-n$".

Pedro Tamaroff

@agent154 If you could solve it, the hours weren't wasted.

agent154

I had to look it up.

Found the exact question asked on mse by somebody else

Pedro Tamaroff

@agent154 Did you try to factor $42$, say?

Note it is $6\times 7$.

agent154

I do know how it works... I studied it enough that I understand Fermat's Theorem quite well now

at least given a question like that

Pedro Tamaroff

2:09 AM

@KarlKronenfeld If $m=2,4$, the any $x$ works.

Now, suppose $m>4$.

agent154

But what I didn't realize was that $42\mid n$ if and only if $3\mid n$, $2\mid n$ and $7\mid n$.

I originally set out to prove it by induction... :|

Didn't quite work as I had hoped... though maybe I just did something wrong.

Pedro Tamaroff

@agent154 Prime numbers.

If $(4,m)=1$, we may just look at $y(y+1)=0\mod m$.

@KarlKronenfeld I'm just brainstorming here.

If we decompose $m$ into prime factors $m_1,\ldots,m_r$, with powers, then we need that $y(y+1)=0\mod m_i$.

agent154

What is $y$?

Pedro Tamaroff

@agent154 $x=2y+1$.

agent154

OK, so we're talking again about my first question.

Pedro Tamaroff

2:12 AM

@agent154 Not really, Karl wants to know if we can characterize those $m$ for which $x^2=1\mod m$ where $x$ is odd.

Karl Kronenfeld

@PedroTamaroff If $m$ is not a power of $2$ then it has an odd divisor $\ell$ such that $\ell^2$ is either $0$ or some even number mod $m$.

Or maybe not.

Pedro Tamaroff

@KarlKronenfeld What about...

Karl Kronenfeld

But this would work if we decompose into $m_i$.

Pedro Tamaroff

@KarlKronenfeld Oh, well.

Note this.

If $m=2^{k}$.

Then $(2^{k-1}+1)^2=2^{2k-2}+2^k+1$

So if $2k-2\geqslant k\iff k\geqslant 2$ that works.

agent154

Here's another question...

Pedro Tamaroff

2:17 AM

@KarlKronenfeld So off go powers of $2$.

@agent154 OK.

Karl Kronenfeld

@PedroTamaroff Derp. Isn't this wrong?

Pedro Tamaroff

@KarlKronenfeld Whutz?

agent154

"Let $f(x)=375x^5-131x^4+15x^2-435x-2$. Find the remainder when $f(97)$ is divided by $11$."

I know that $97\equiv 9\mod{11}$, so therefore $f(97)\equiv f(9)\mod {11}$. So should I just let $x=9$ for that polynomial and then solve from there? Or is there an easier way?

Karl Kronenfeld

@PedroTamaroff That doesn't rule out powers of $2$, since $(2^{k-1}+1)^2$ is actually $1$ mod $2^k$ when $k\geq 2$.

Pedro Tamaroff

@KarlKronenfeld Oh, maybe I completely misunderstood yer questiunn.

agent154

2:19 AM

Or am I misunderstanding a theorem?

Karl Kronenfeld

@PedroTamaroff This is the correct statement. (for all $x$ at the end)

Pedro Tamaroff

@agent154 First, reduce the polynomial $\mod 11$.

Bill

Hello

Pedro Tamaroff

I would rather use $-2$ instead of $9$.

@Bill HOLIS.

agent154

I'm not sure what you mean by reduce...

Bill

2:21 AM

I'm reading a bit about acyclic models.

Pedro Tamaroff

@agent154 I mean obtain the least positive remainder.

Karl Kronenfeld

@PedroTamaroff Note that $2^{k-1}+1$ is not odd when $k=1$.

Pedro Tamaroff

@KarlKronenfeld I said $k\geqslant 2$.

@KarlKronenfeld But dully noted.

=D

agent154

Well, yes... that's my goal. But to do that don't I just find $f(9)\mod{11}$?

Pedro Tamaroff

@agent154 Yeah.

@KarlKronenfeld Could you restate the question?

@Bill Let me google that. =)

Karl Kronenfeld

2:23 AM

@PedroTamaroff Determine all integers $m$ for which the following statement holds. For all odd $x$, $x^2\equiv 1\pmod m$.

Pedro Tamaroff

@KarlKronenfeld Oh...

@Bill NOPE NOPE NOPE.

@KarlKronenfeld Cannot you state than in group theoretic terms?

agent154

I seem to not be seeing some messages...

Bill

The proof of the acyclic model theorem has a suspicious resemblance to a lemma about projective complexes. Let $C$,$\vareps$ be a projective complex over $M$, $C'$,$\vareps '$ a resolution of $M'$. If $\mu:M\rightarrow M'$ then, there exist a chain morphism $\alpha$ from $C$ to $C'$ such that $\mu\vareps=\vareps '\alpha_0$. Any such two morphisms are homotopic.

agent154

I see "NOPE NOPE NOPE" directed at @Bill, but I don't see what he wrote first...

Karl Kronenfeld

Yeah, if $m$ is a power of $2$ we can note that the odds should all be in the kernel of the homomorphism $\mathbb Z/m\mathbb Z^\times\to\mathbb Z/m\mathbb Z^\times:x\mapsto x^2$.

Bill

2:27 AM

sorry

Karl Kronenfeld

@PedroTamaroff In fact, the odds should comprise that kernel by counting.

Bill

The proof of the acyclic model theorem has a suspicious resemblance to a lemma about projective complexes. Let $C$,$\varepsilon$ be a projective complex over $M$, $C'$,$\varepsilon '$ a resolution of $M'$. If $\mu:M\rightarrow M'$ then, there exist a chain morphism $\alpha$ from $C$ to $C'$ such that $\mu\varepsilon=\varepsilon '\alpha_0$. Any such two morphisms are homotopic.

Is there more than just a resemblance?

Pedro Tamaroff

@Bill Don't ask me. =)

@KarlKronenfeld Not really.

$m=16$, $x=5$ then $25=9\neq 1$?

Karl Kronenfeld

@PedroTamaroff I am saying should not is.

I.e. I am restating the condition.

Pedro Tamaroff

@KarlKronenfeld Oh, my bad.

Karl Kronenfeld

2:34 AM

But the multiplicative group is just the set of (equivalence classes) of all odds, so we are requiring that squaring is a trivial map. (All when $m=2^k$)

Pedro Tamaroff

@KarlKronenfeld Yeah, the homomorphism is trivial.

Karl Kronenfeld

Oh damn, this is stupid of me.

Just use $3$ when the exponent is large enough.

So $2,4,8$ are the only powers of $2$ which satisfy the given property.

Pedro Tamaroff

@KarlKronenfeld CAAAAAAAAAAAWL.

Karl Kronenfeld

@PedroTamaroff For odd $m$ the problem statement doesn't make much sense, because odds are congruent to evens.

Pedro Tamaroff

@KarlKronenfeld Yeah.

So we're done.

Let's celebrate.

Karl Kronenfeld

2:38 AM

@PedroTamaroff Yes. :)

agent154

OK, so I'm allowed (and it would seem strongly encouraged) to use a calculator for my upcoming test in this class, but I don't know if I should be using a calculator for something like this. Is there an easy way without using a calculator to find $-6293\equiv x\mod{11}$?

Pedro Tamaroff

►

agent154

I suppose the division algorithm would do it

Pedro Tamaroff

@agent154 Yes, sure.

agent154

But this is probably where the calculator comes in handy, because not many people know a multiple of 11 close to -6293

Karl Kronenfeld

2:40 AM

@agent154 Alternately add and subtract the digits. 6293 is congruent to 3-9+2-6 mod 11

negate that.

mod 9 is similar except you add all digits.

agent154

I'm familiar with the divisiblity rules mod 2, 3, 5, 6, 8, 9, 10, and now 11... I didn't now that one

Pedro Tamaroff

@agent154 Well, now you do!

agent154

But that doesn't help much if I have to find mod some slightly bigger number

Pedro Tamaroff

confetti

agent154

So I guess the division algorithm plus a calculator will do the job

That's enough proof that I know the theory behind the question

I'm just wondering if my math is right here: $$375(9)^5-131(9)^4+15(9)^2-435(9)-2\equiv 9^5-10(9)^4+4(9)^2-6(9)-2\mod{11}$$... I figure this because $375\equiv 1\mod{11}, 131\equiv 10\mod{11}, 15\equiv 4\mod{11}, 435\equiv 6\mod{11}$.

Pedro Tamaroff

2:51 AM

@agent154 Yes.

Now $9=-2$.

That is easier to raise to powers.

agent154

That's another trick I guess.. hadn't thought of that

Pedro Tamaroff

@agent154 Yep. You want to work with small numbers.

Twink

3:03 AM

hi

@PedroTamaroff in te book that you gave me they don't talk about the spectral descomposition, right?

Pedro Tamaroff

@Twink What is that, precisely?

Twink

:(

what does it mean for you to find the spectral descomposition of an operator?

Pedro Tamaroff

What is "spectral decomposition"?

Twink

I've been looking for examples

and I can't find any

Pedro Tamaroff

I don't know. I never encountered that term. =)

Twink

3:06 AM

in Hoffman they say it's an operator

it's a linear combination of the projections of the vector space into each eigenspace

with the eigenvalues as the coefficients

but in the problems I don't know how to do that

Pedro Tamaroff

@Twink Well, I am reading that the projections are along the eigenspaces.

So it is doable.

You find the eigenspaces first.

And the eigenvalues.

Then it is a matter of defining the projections properly.

Twink

yes that's the problem

to define the projections

Pedro Tamaroff

@Twink Why?

Twink

remember the problem of symmetric and antisymmetric matrices?

Pedro Tamaroff

@Twink Aha.

Twink

3:10 AM

the operator was $T(A)=A^t$

the eigenvalues where 1 and -1

the eigenspace of 1 was the symmetric matrices

and the eigenspace of -1 was the antisymmetric matrics

Pedro Tamaroff

@Twink Aha.

Twink

but the projection of a matrix $A$ into the space of symmetric matrices

is the sum

Pedro Tamaroff

What?

Twink

of the terms $<A, E_{ij}>E_{ij}$

where the $E_{ij}$ are the elements of the basis of the space of symmetric matrices

agent154

OK, I'm stuck with this one... "Given that $a\equiv b\mod{p}$ for some prime $p$, show that $a^p\equiv b^p\mod{p^2}$.

I know that $a^p\equiv b^p\mod{p}$, but how do I show that they're congruent mod $p^2$?

Twink

3:13 AM

and the dot product is defined as $<A,B>=tr(AB^t)$

Pedro Tamaroff

@agent154 Binomial theorem.

I'm not following

Let me try first.

So it's to tedious

@Twink Computations usually are =D

Twink

3:15 AM

but there must be an easier way since the subspaces are invariant

Pedro Tamaroff

@agent154 Yes.

Twink

everything is so perfect

there must be a theorem or something

but I'll never know

Pedro Tamaroff

Write $a=b+kp$. Then raise to the power of $p$ on both sides. Then clear stuff $\mod p^2$. @agent154

The general claim is that $a=b\mod p^k\implies a^p=b^p\mod p^{k+1}$.

And the proof is the same.

@Twink Read on. Maybe you'll find something. =)

Twink

Pedro and also I need to find the spectral descomposition of this operator math.stackexchange.com/questions/529260/…

agent154

Hmm, ok... I'll give that a try thanks

Twink

3:17 AM

but I don't know which are the characterisic values

I didn't understand the answers :(

Pedro Tamaroff

@Twink I don't know about that, sorry.

Twink

ok but do you know which are the eigenvalues?

of that operator?

Pedro Tamaroff

Ah, OK.

Twink

no

in this case we don't have a formula for the scalar product

agent154

So that's $$\sum_{i=0}^{p}\left.p\choose i\right.b^ikp^{p-i}$$ right?

Twink

3:19 AM

but they gave me hints to find the characteristic and the minimal polynomials

agent154

I just learned the binomial theorem last month... Still trying to remember it

Pedro Tamaroff

@agent154 You need some parentheses.

$(kp)^{p-i}$

agent154

Oh, ok

Pedro Tamaroff

Now, you should know that $p\mid {p\choose i}$ if $1\leqslant i<p$.

So you get a lot of stuff that is $0\mod p^2$.

@anon

agent154

So then I just need to worry about the first and last term? @PedroTamaroff

anon

3:23 AM

yeah?

Pedro Tamaroff

@agent154 Just the first. The last is also $0$.

@anon I wonder if you ever encountered this.

Let $p$ be a projection on $V$, a vector space.

Then define a new action $\lambda\times v=\lambda p(v)$.

Then all previous axioms hold, save $1\times v=v$.

agent154

@PedroTamaroff How is the last term congruent to 0? The last term looks like $b^p$... unless I'm not following.

Pedro Tamaroff

@agent154 Oh, you made it backwards w.r.t. me! =)

I usually start $b^p+\cdots$.

Yeah, that is possibly nonzero.

So keep it.

agent154

OK then, the first one for me is $(kp)^p$

so yeah, I see... $p$ divides that

Pedro Tamaroff

@anon Conversely, is the above the only procedure to obtain a vector space satisfying all axioms save $1x=x$? (Halmos)

@agent154 And all the what comes after, except possibly $b^p$.

agent154

3:28 AM

OK, so from here I have $a^p=p(\dots)+b^p$ where $\dots$ is some integer... I don't see how I can get that $p^2$ divides $a^p$

Pedro Tamaroff

@agent154 $p^2$ should divide $a^p-b^p$.

Which it does.

agent154

OK, I see - looking at the expanded polynomial, I see there's an extra factor of $p$ that I can pull out.

Pedro Tamaroff

@agent154 Because $p\mid {p\choose k}$ if $k=1,\ldots,p-1$.

agent154

Yeah. I had that part already, but I forgot that each term of $(b+pk)^p$ has some power of $(pk)$.

anon

@PedroTamaroff let V be an abelian group and K a field. if you impose all of the vector space axioms on (V,K) except 1v=v then you basically have a nonunital homomorphism K->End(V). show that 1 must act idempotently, hence it acts as a projection. the image of this projection gets a true K vector space structure.

masfenix

3:34 AM

Hello guys, was wondering if anyone can give me an intuition on the following line: "Open sets are arbitrary union of finite intersections of sets open with respect to any $||\cdot||_N$

Pedro Tamaroff

@masfenix Well, you have a few results packed in there.

anon

@masfenix in an open set, every point has wiggle room: there exists a cozy ball around it contained in said set. put such a ball around every single point in the open set A and take the union: now you have a subset of A which admits every point of A as a member. thus, A is a union of open balls. conversely, every union of open balls is open.

Twink

@anon do you know what does it mean to find the spectral descomposition of a linear operator?

Pedro Tamaroff

@anon And every norm on a finite dimensional space is equivalent.

@anon I wonder why they want intersection, i.e. looking at a prebasis instead of basis.

anon

@Twink write $V=V_1\oplus\cdots\oplus V_m$ in such a way that $A(v_1+\cdots+v_m)=\lambda_1v_1+\cdots+\lambda_m v_m$ holds generically

Twink

3:41 AM

yes

thoes $V_i$ are the eigenspaces right?

Pedro Tamaroff

Game devs should take over Hollywood. I say goddamn...

Twink

but I thought it was an function

Pedro Tamaroff

@anon Man, t.b.'s contributions to MSE are awesome.

Twink

$\lambda_1 E_1 + \cdots + \lambda_m E_m$

anon

if $E_i:V\to V_i$ are the projection operators then sure, my equation reads $A=\lambda_1E_1+\cdots+\lambda_mE_m$

Twink

3:43 AM

where $E_i$ is the projection of $V$ into $E_i$

anon

decomposing the space and decomposing the operator are more or less equivalent tasks: do one and you've effectively done the other

Twink

not really

because I already have the descomposition

in one problem

but it's difficult to find the projections

anon

are you talking about that H_v operator?

it's basically a reflection

Twink

no about another one

but that one too

in that one I don't even know which are the eigenvalues

in the H_v projection

anon

its not a projection, its a reflection

I will write an answer to that one illuminating the geometry

Twink

3:47 AM

ok sorry

ok thanks :)

skullpatrol

Hi @HenningMakholm how are you pal?

Twink

I was thinking that maybe there's an easier way to find the spectral descomposition without calculating each projection

or an easier way to find the projection

depending on the problem, of course

hi Pal :D

how are you?

you have another pal?

I see I'm not the only one

skullpatrol

Hi pal; I'm fine thanks, how are you?

Twink

@anon the other problem is this one math.stackexchange.com/questions/529253/…

in this one I already have the eigenvalues 1 and -1

anon

don't accept answers until you are satisfied with them

Twink

3:52 AM

and the subspaces $V_1$ and $V_2$ are the symmetric and antisymmetic matrices

no but in that question I didn't ask anything about the spectral descomposition

so, in this problem, the scalar product is $<A,B>=tr(AB^t)$

and the formula for the projection of $T$ into $V_i$ is

first I need an orthonormal basis for $V_i$

say $e_i$

and then the projection is $\sum <A,e_i>e_i$

Pedro Tamaroff

@Twink Use \langle and \rangle

Twink

but in this problem I need to say which is the basis and then calculate the dot product

and it's a horrible and tedious calculation I think

but I was thinking that maybe there's some trick

$\langle \rangle$

thanks Pedro

I didn't know that :)

I have an exam on monday :(

and I can't compute spectral descompositions

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$Mathematics$ Mathematics