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?