Wikipedia:Reference desk/Archives/Mathematics/2009 August 8

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August 8

Alternative proof that a square matrix is similar to its transpose

A square matrix (over C) is similar to its transpose. This can be shown by a Jordan form. I know it's actually true for any field. For this, I suppose you use Smith normal form. Question: Is there any other way to prove this? (that doesn't use the structure results) Maybe more elementary or conceptually simpler way? More importantly, is there any similar fact or formulation that leads to a generalization of this to a (linear) operator acting on an infinite-dimensional space (e.g., Hilbert space)? -- Taku (talk) 02:03, 8 August 2009 (UTC)

Not quite true in a Hilbert space (without further assumptions). Consider the right-shift operator on l₂(N) taking e_n in e_n+1: its transpose is the left-shift operator, and they are obviously not similar, for the former is injective and the latter is not. --pma (talk) 10:02, 8 August 2009 (UTC)

Definitely not in the fullest generality, because the index of a Fredholm operator (between the same space) doesn't vanish in general. But this actually gives a clue: somehow the result above holds because of the simplicity of geometry, and a Jordan form doesn't capture geometry. I'm actually quite positive that the answer is affirmative. It follows from that A is similar to A^T, that, for example, $\sigma (A)=\sigma (A^{T})$. But the proof of this relies on a Jordan form? which doesn't seem right. (Yes, you can use determinant, but that's basically a weaker result.) -- Taku (talk) 12:06, 8 August 2009 (UTC)

Absolute Values with Factorials

As you know, absolute values are identified conventionally with the formula $|x|={\sqrt {x^{2}}}$. But the problem is that although conventional square roots can only be positive, the true square root can be negative too. Therefore, I set of to find another formula of absolute values. I found this formula $|x|=x^{2}-x\Pi ^{-1}({\frac {\Pi (x)}{x}})$ . Can anyone help me confirm it? Thanks!
The Successor of Physics 04:32, 8 August 2009 (UTC)

The "square root of the square" formula is pretty silly - it only works because the square root symbol is defined to give the positive square root. Why do you want a formula for it? What is wrong with $|x|={\begin{cases}x\,\mathrm {if} \,x\geq 0\\-x\,\mathrm {if} \,x<0\end{cases}}$? --Tango (talk) 04:46, 8 August 2009 (UTC)

The "square root of the square" formula does reference how the Euclidean norm generalize to larger spaces. But I do agree that Tango's formula is more straight forward if we're just dealing with the reals. I also have no idea what Π is in that equation up there. Rckrone (talk) 06:31, 8 August 2009 (UTC)

I'll make it clearer! Thanks!

The Successor of Physics 13:19, 8 August 2009 (UTC)

Oh I see, the Pi function as related to the Gamma function. Pretty sure that formula does not work. First of all Π isn't defined everywhere on the real line, second of all it's not injective on the real line, so Π^-1 is not well defined. Even ignoring those problems the formula would just comes out to x, even for negative x. You can show by Euler's reflection formula that zΓ(z) = Γ(z+1) everywhere that Γ is defined, not just for for z with Re(z) > 0. Rckrone (talk) 07:55, 8 August 2009 (UTC)

But I tested it for a few values and it worked!

The Successor of Physics 13:19, 8 August 2009 (UTC)

1) For x where Π is defined, Π(x)/x = Π(x-1). If we just pretend that Π is injective then Π^-1(Π(x-1)) = x-1 unambiguously. I assume that's the value you want. So the expression on the right is just x² - x(x-1) = x. There could be other values of Π^-1, but you would need Π(x-1) = Π(x+1) for all x < 0 and that is clearly not true in general since Π(x+1)/Π(x-1) = x(x+1).

2) Assume there is a function f on the reals such that |x| = x² - xf^-1(f(x)/x) for all x. Then f^-1(f(x)/x) = x-1 for x > 0 and f^-1(f(x)/x) = x+1 for x < 0. So for all 0 < x < 1, f(x)/x = f(x-1) and f(x-1)/(x-1) = f(x). This implies f(x) = f(x-1) = 0. Then f^-1(f(x)/x) = f^-1(0) = x-1 for all 0 < x < 1 which is a contradiction. No such function f can exist. Rckrone (talk) 17:13, 8 August 2009 (UTC)

Unless you know exactly what you're doing, "I tested it for a few values and it worked" doesn't usually cut it in mathematics. Try applying this approach to the claim "$x^{2}+x+41$ is prime for any integer x" and see what you get.

You should read the introduction of Square root to clear up your confusion on the matter (at least with regards to terminology). Absolute value is obviously also relevant. -- Meni Rosenfeld (talk) 18:12, 8 August 2009 (UTC)

The formula $|x|={\sqrt {x^{2}}}$ (where x is real) is correct, concise, and compatible with some generalizations of real numbers (including complex numbers, where the radicand involves the real and imaginary parts). There is no need to replace it with a different formula, and even if there was, your suggestion is wrong on many levels. -- Meni Rosenfeld (talk) 18:42, 8 August 2009 (UTC)

May I ask what you're doing? In the complex numbers, $|x|={\sqrt {x^{*}x}}$ is what you want. In other systems, maybe something different. Some structures don't really have "absolute values" defined per se. Perhaps an understanding of your needs would help us in gauging how to respond. --Leon (talk) 18:53, 8 August 2009 (UTC)

I always thought $|x|=x^{+}-x^{-}$, where $x^{+}=x\vee 0$ and $x^{-}=x\wedge 0$, was a bit more conventional than $|x|={\sqrt {x^{2}}}$. Nice and concise...Nm420 (talk) 18:29, 10 August 2009 (UTC)

If you're going to use the R² → R max function (which I assume is what you mean by $\vee$ here), then why not just $|x|=x\vee -x$? —JAO • T • C 19:14, 10 August 2009 (UTC)

The formula $|x|={\sqrt {x^{2}}}$ is quite a nice formula because it allows multiple differentiation. And not just of $|x|$. Consider the following example, assuming that f(x) ≠ 0 then:

${\frac {d|f|}{dx}}={\frac {d}{dx}}{\sqrt {f^{2}}}={\frac {ff'}{\sqrt {f^{2}}}}=f'|f|=|f|{\frac {df}{dx}}\ .$

ANd you could differentiate until you're dizzy. I know it's a bit naughty but the expressions that you get hold true. It reminds me of the tricks one plays to solve ODEs and PDEs. There's no need to worry the square root symbol when it comes to real numbers. The convention is that ${\sqrt {x}}$ is positive: that's why we write $\pm {\sqrt {b^{2}-4ac}}$ in the quadratic formula. Besides |x| ≥ 0 so the only choice of sign for ${\sqrt {x^{2}}}$ which makes sense is the positive one. ~~ Dr Dec (Talk) ~~ 22:46, 14 August 2009 (UTC)

Correction:

${\frac {d|f|}{dx}}={\frac {d}{dx}}{\sqrt {f^{2}}}={\frac {ff'}{\sqrt {f^{2}}}}={\frac {f'f}{|f|}}={\frac {|f|f'}{f}}={\frac {|f|}{f}}{\frac {df}{dx}}\ .$

--69.91.95.139 (talk) 23:16, 14 August 2009 (UTC)

Quite right, thanks! ~~ Dr Dec (Talk) ~~ 10:56, 15 August 2009 (UTC)

Set of points with density 1

For a Lebesgue measurable set of real numbers A, let B={X:X is a real number & d(X)=1}, where d(X) is the density of the set A at X. Note that B may not be a subset of A. What property characterizes B? My intuition tells me it should always be an open set, and that B could be any open set of real numbers by choosing the right example of A (eg. B itself); am I right or wrong?

And I have the same question for higher-dimensional Lebesgue measures. --COVIZAPIBETEFOKY (talk) 22:31, 8 August 2009 (UTC)

How about the density at 0 of [-1, 1] with the sets [1/n, 1/n + 1/n³] removed? Rckrone (talk) 23:14, 8 August 2009 (UTC)

That would be a counter example, wouldn't it? Ok, so it doesn't seem like there is a very succinct characterization of B, but now that I think about it, that wasn't really what I needed for my purposes, so it doesn't matter. Just one more question, on which I'm pretty sure I'm correct, but I would like confirmation:

Am I correct in believing that every Lebesgue measurable set is a null set different from exactly one open set of real numbers? A is a null set different from B iff (A-B)U(B-A) is a null set. --COVIZAPIBETEFOKY (talk) 15:25, 9 August 2009 (UTC)

No, you are not. E.g. R itself is a null set different from R\F for any finite F. Algebraist 15:56, 9 August 2009 (UTC)

Right.

Okay, there has to be a name for what I'm looking for, as you have pointed out that merely "open" is not sufficient. I want the set to be open, and no isolated points not in the set.

Let me redefine my B as {X:X is a real number & there is a (sufficiently small) positive real number r such that μ([X-r, X+r] ∩ A) = 2r}.

Now, is there a simple name for the property that characterizes B, completely? --COVIZAPIBETEFOKY (talk) 16:14, 9 August 2009 (UTC)

I believe this is the difference between the two:

openness: X is in the set implies that it has a neighborhood in the set.

my property: X is in the set iff it has a neighborhood in the set. --COVIZAPIBETEFOKY (talk) 16:42, 9 August 2009 (UTC)

No, never mind, that doesn't quite work. Would "... iff it has a neighborhood in the set, excluding itself" be sufficient? —Preceding unsigned comment added by COVIZAPIBETEFOKY (talk • contribs) 16:45, 9 August 2009 (UTC)

You can say that every measurable set differs from exactly one set like B by measure zero. In other words, you can set up equivalence classes of sets that have a symmetric difference of measure zero, and then every equivalence class has a unique element B such that B is exactly the set of points that have Lebesgue density 1 with respect to B. It seems like there should be some unifying property of the sets like B, but if there is I have no idea what it is. The property you mention doesn't work, since that would imply that every set like B is open, which is not the case. It's definitely an interesting question. Rckrone (talk) 17:36, 9 August 2009 (UTC)

You seem to forget that I redefined B. It's now defined so that it will always be open. I made the redefinition bold above. --COVIZAPIBETEFOKY (talk) 00:07, 10 August 2009 (UTC)

There are measurable sets that cannot be written at all as the symmetric difference of an open set and a null set. For example, take the Smith–Volterra–Cantor set. — Emil J. 16:57, 10 August 2009 (UTC)

Ok, clearly I'm completely lost in my understanding of the Lebesgue measure. So let me cut to the chase:

Is there a general collection of relatively simple "representative" sets for which every Lebesgue measurable set can be expressed as a symmetric difference between one of the sets in said collection and a null set? --COVIZAPIBETEFOKY (talk) 19:42, 10 August 2009 (UTC)

Ah. That's a different question entirely. And the answer is "yes." Every Lebesgue measureable set differs from a $G_{\delta }$ or $F_{\sigma }$ set by a set of measure 0. Ray^Talk 20:28, 10 August 2009 (UTC)

An application of the Lebesgue differentiation theorem to the characteristic function of A tells us that almost every point of A is a point of density of A. Also, it says that almost every point in the complement of A is not a point of density of A.

I think it's pretty clear that any point in the interior of the closure of A is a density point of A, and similarly that. Any point on the boundary of A is not going to be a density point.

However, no variety of "open" or "closed" is going to characterize density points. We know that this set cannot be closed, since you need only consider a point on the boundary to have issues. Another example tells us that the set of density points need not be open: if you let A be a fat Cantor set (the set of remaining points on the unit interval after following the Cantor process, but taking care so that the sum total of the lengths of all intervals removed is less than 1), then A is a set of positive measure, which is closed, completely disconnected, etc. By the Lebesgue differentiation theorem, there *must* be density points of A. However, the density points of A cannot be open, since any open interval would contain a subinterval which is in the complement of A.

That's the best I could come up with, hope it helps some. Ray^Talk 20:14, 10 August 2009 (UTC)

I think it's pretty clear that any point in the interior of the closure of A is a density point of A: not at all. The rationals have measure zero, hence no density point, nevertheless the interior of their closure is the whole real line. — Emil J. 11:00, 11 August 2009 (UTC)

Oops. Thanks for the correction. Ray^Talk 11:39, 11 August 2009 (UTC)

Well, if you are going to correct that, then the other half of the sentence is false too, for much the same reason (the irrationals). — Emil J. 12:22, 11 August 2009 (UTC)

So humiliating :) Note to self: in the future, do not make statements one cannot prove on the refdesk page. Ray^Talk 14:42, 11 August 2009 (UTC)

Thanks, everyone, for your help. --COVIZAPIBETEFOKY (talk) 17:52, 11 August 2009 (UTC)