I apologize if the following question ends up being too elementary for this website; I [asked it](https://math.stackexchange.com/questions/103160/do-real-vectors-attain-matrix-norms) 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](https://math.stackexchange.com/a/104461/9182) 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. 



Do real vectors attain matrix norms?
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.


