7
$\begingroup$

I apologize if the following question ends up being too elementary for this website; I asked it on math.SE a week ago and it remains unanswered.

Let $A$ be an $n \times n$ matrix with real entries and let $p \geq 1$. I'm wondering if $$ \max_{ x \in \mathbb{C}^N, \|x\|_p = 1} \|Ax\|_p$$ is the same as

$$ \max_{ x \in \mathbb{R}^N, \|x\|_p = 1} \|Ax\|_p.$$ The only difference is the replacement of $\mathbb{C}^n$ by $\mathbb{R}^n$. Certainly, the answer is yes if $p=1,2,\infty$; on the other, it is pointed out in this answer that the answer is no for mixed $(p,q)$-norms.

Edit, Will: I can't get this stupid thing to work. On MSE, Robert Israel introduced a $p,q$ mixed matrix norm with suitable notation, where the vectors $x$ are measured with the $p$-norm, but the $Ax$ are measured with the $q$-norm.

$\endgroup$
9
  • $\begingroup$ I don't think it's true for reals. Otherwise, Perron-Frobenius would be a stupidly weak assertion. $\endgroup$ Feb 5, 2012 at 22:25
  • $\begingroup$ Sorry if I'm being thick, but is there a connection between Perron-Frobenius and matrix norms? I thought Perron-Frobenius tells us that the largest eigenvalue of a nonnegative matrix is real, which is only a lower bounded for its $p$-norms. $\endgroup$
    – user21162
    Feb 5, 2012 at 22:29
  • $\begingroup$ Ah, I thought you were speaking of the "standard" matrix norm. What is the $p$-norm of a matrix? Hilbert-Schmidt with $p$ instead of $2$? $\endgroup$ Feb 5, 2012 at 22:35
  • 1
    $\begingroup$ As currently constituted, there are no matrix norms in the question; only $p$-norms of vectors - these, presumably, are unambiguous $\endgroup$ Feb 5, 2012 at 22:51
  • 1
    $\begingroup$ Doesn't this paper: arxiv.org/pdf/math/0512608v1.pdf answer the question (Theorem 3.1 specifically)? $\endgroup$ Feb 6, 2012 at 0:14

1 Answer 1

7
$\begingroup$

Okay, so basically the answer can be found in here: http://arxiv.org/pdf/math/0512608v1.pdf

Here's how the argument works (a simplified version of what is done in the paper with finite dimensions and $p=q$):

(Note: we define "$\Re$" of a vector by taking the real part componentswise)

Lemma 3.4 says (applied to the finite dimensional situation) $$ \int_0^{2\pi} \| \Re(e^{i\varphi} x) \|_p^p d\varphi = \int_0^{2\pi} |\cos(\varphi)|^p d\varphi $$ for any $x\in \mathbb C^n$ with $\|x\|_p=1$. This is fairly elementary to verify.

Therefore, whenever $x,y\in \mathbb C^n$ both have norm $1$, we will find a $\varphi\in[0,2\pi]$ such that $$\|\Re(e^{i\varphi}x)\|_p \leq \|\Re(e^{i\varphi}y)\|_p$$ since the integral $$ \int_0^{2\pi}\|\Re(e^{i\varphi}y)\|_p^p - \|\Re(e^{i\varphi}x)\|_p^p d\varphi = 0 $$ is zero und thus the integrand has to be non-negative somewhere.

Then Lemma 3.2 of the paper yields the result. What the authors do here is to take a vector $0\neq x\in \mathbb C^n$ such that $\|Ax\|_p/\|x\|_p$ is maximal (assume also that $A\neq 0$; then $Ax\neq 0$ follows automatically and we can divide by its norm below). Then they take a $\varphi$ such that $$ \left\|\Re(e^{i\varphi} \frac{x}{\|x\|_p})\right\|_p \leq \left\| \Re(e^{i\varphi} \frac{Ax}{\|Ax\|_p})\right\|_p $$ which is possible by the above. If $\Re(e^{i\varphi}x)\neq 0$, this can then be rewritten as $$ \frac{\|Ax\|_p}{\|x\|_p} \leq \frac{\|A \Re(e^{i\varphi}x)\|_p}{\|\Re(e^{i\varphi}x)\|_p} $$ which shows that the maximum is also attained at the real vector $\Re(e^{i\varphi}x)\in \mathbb R^n$. If $\Re(e^{i\varphi}x)=0$, then $i\cdot e^{i\varphi}x$ is a real vector at which the maximum is attained.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.