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.

  • $\begingroup$ 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
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    $\begingroup$ 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
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    $\begingroup$ 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


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.


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