TITLE: Equivalence between complex and real operator norms
QUESTION [8 upvotes]: Consider a real $m\times n$ matrix $A$ and the $p$-norms in $\mathbb{C}^n$ and $\mathbb{C}^m: \|x\|=\left(\sum|x_i|^p\right)^{1/p}$.
One defines the real $p$-norm of $A$ as $\|A\|=\sup\frac{\|Ax\|}{\|x\|}$, where the $\sup$ ranges over $0\neq x\in\mathbb{R}^n$.
Similarly, one can define the $p$-norm of the complexified operator $A$, that is, $\|A\|_c=\sup\frac{\|Ax\|}{\|x\|}$, where the $\sup$ ranges over $0\neq x\in\mathbb{C}^n$.
I'm looking for a simple proof of the fact that $\|A\|=\|A\|_c$, or, in particular, that $\|A\|\geq|\lambda|$, for any (possibly complex) eigenvalue $\lambda$ of $A$ (when $m=n$). Clearly, it holds that $\|A\|_c\geq|\lambda|$.
I have seen some results on this issue, but in a more abstract context. My question is about a simple proof of this fact for the finite dimensional case. Also, does this result hold for other norms?

REPLY [2 votes]: A discussion of the relation of norms on real and complexified spaces can be found in the paper
O. Holtz, M. Karow. Real and complex operator norms. arxiv-paper
For the finite-dimensional case, where $A$ maps from $(\mathbb{C}^m,\|\cdot\|_p)$  to $(\mathbb{C}^n,\|\cdot\|_q)$ the paper presents an elementary proof for the equality of real and complex operator norm in the case $1 \leq p \leq q \leq \infty$. This inequality is sharp, i.e. for other values of $p,q$ it does not hold in general.
There are some further cases treated in the paper. 
All this probably follows from the more abstract results you have already found, but maybe the presentation will appeal to you in terms of having a simple proof.