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?

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    $\begingroup$ I do not know, whether is it true, but we should be careful: it is false for operators from $\ell^p_2$ to $\ell^q_2$ with different $p,q$, for example, for the matrix $\pmatrix{1&1\\-1&1}$ with $p=1$, $q=\infty$. $\endgroup$ – Fedor Petrov Jun 16 '16 at 8:15
  • $\begingroup$ or, better to say, for $p=\infty,q=1$ $\endgroup$ – Fedor Petrov Jun 16 '16 at 10:54
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    $\begingroup$ Thanks, but I think it is true if $p\leq q$. $\endgroup$ – Shake Baby Jun 16 '16 at 20:24

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 GTM216), 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 $r(A)r(B)<r(AB)$.

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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.

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