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?



Equivalence between complex and real operator norms
Since you seem really interested in the inequality $\rho(A)\le\|A\|$ ($\rho$ the spectral radius), here is a simple and elegant proof. In the 2nd edition of my book *Matrices* (Springer Verlag GTM**216**), it is Proposition 7.6.


Take any operator norm $N$ on ${\bf M}\_n({\mathbb C})$ ; you may take $N=\|\cdot\|\_c$, but this is not necessary. Because of finite dimension, the restriction of $N$ to ${\bf M}\_n({\mathbb R})$ is equivalent to $\|\cdot\|$ : there exists a finite constant $C$ such that $N(M)\le C\|M\|$ for every $M\in {\bf M}\_n({\mathbb R})$. On the other hand, $N$ satisfies $\rho(M)\le N(M)$ for every $M\in {\bf M}\_n({\mathbb R})$. Apply these to a power of $A$ :
$$\rho(A)^m=\rho(A^m)\le N(A^m)\le C\|A^m\|\le C\|A\|^m.$$
Take the $m$-root in the inequality above
$$\rho(A)\le C^{1/m}\|A\|,$$
and let $m\rightarrow+\infty$.


Remark that this proof is valid for every norm over ${\bf M}\_n({\mathbb R})$ satisfying the inequality $\|A^m\|\le\|A\|^m$. Such a norm is called *super-stable*. Super-stable norms include all the operator norms. The *numerical radius* 
$$r(A)=\sup\_{x\in{\mathbb C}^n\,;\,\|x\|\_2=1}|x^\*Ax|$$
is a super-stable norm, though not an operator norm. Actually, there exist two matrices such $r(A)r(B)<r(AB)$.


