Real and complex operator norms

Question

Let $A\in M_n(\mathbb{R})$ be a matrix and $\|\cdot\|$ be a norm on $\mathbb{C}^n$. When we look at the operator norm of $A$ with respect to $\|\cdot\|$ we can either consider the inclusion of $M_n(\mathbb{R})$ in $M_n(\mathbb{C})$ or the restriction of $\|\cdot\|$ to $\mathbb{R}^n$. Are these points of view equivalent?

In other words do we always have: $$ \sup_{x\in\mathbb{R}^n\setminus\lbrace0\rbrace}\frac{\|Ax\|}{\|x\|}=\sup_{x\in\mathbb{C}^n\setminus\lbrace0\rbrace}\frac{\|Ax\|}{\|x\|} $$ for $A\in M_n(\mathbb{R})$.

I hoped that we could use as an intermediary step this other question however, I was a bit optimistic.

Answer 1

$\newcommand\R{\mathbb R}\newcommand\C{\mathbb C}$No. E.g., let $$A=\begin{bmatrix}1&0\\0&0 \end{bmatrix}$$ and $$\left\|\begin{bmatrix}z_1\\z_2 \end{bmatrix}\right\|=|z_1|+|z_1+iz_2|.$$ Then the real norm of $A$ is $1$ and the complex norm of $A$ is $2$.

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Indeed, if $x=\begin{bmatrix}x_1\\x_2\end{bmatrix}\in\R^2$ with $x_1\ne0$, then $$\frac{\|Ax\|}{\|x\|}=\frac{|x_1|+|x_1+0i|}{|x_1|+|x_1+ix_2|} \le\frac{2|x_1|}{|x_1|+|x_1|}=1,$$ with $\frac{\|Ax\|}{\|x\|}=1$ if $x_2=0$. So, the real norm of $A$ is $1$.

If now $z=\begin{bmatrix}z_1\\z_2\end{bmatrix}\in\C^2$ with $z_1\ne0$, then $$\frac{\|Az\|}{\|z\|}=\frac{2|z_1|}{|z_1|+|z_1+iz_2|} \le2,$$ with $\frac{\|Az\|}{\|z\|}=2$ if $z_2=iz_1$. So, the complex norm of $A$ is $2$.

Answer 2

What you can do is to reverse the question:

Given a norm $\|\cdot\|$ over $\mathbb R^n$, does there exist a norm $N$ over $\mathbb C^n$, whose restriction to $\mathbb R^n$ is $\|\cdot\|$, and such that for every $M\in{\bf M}_n(\mathbb R)$, the subordinated norms coincide : $$N(M)=\|M\|\quad ?$$

Notice that we have obviously $N(M)\ge\|M\|$, a larger domain implying a larger upper bound. Thus only the reverse inequality is at stake.

This question was addressed in the Exercises 5 and 6, Chapter 7 of my book Matrices (Springer-Verlag GTM 216, 2nd edition). A solution is as follows.

Define $N$ by $$\forall z\in\mathbb C^n,\quad N(z)=\inf\left\{\sum_\ell|\alpha_\ell|\|x^\ell\|;z=\sum_\ell\alpha_\ell x^\ell\right\},$$ where the decompositions obey to $\alpha_\ell\in\mathbb C$ and $x^\ell\in\mathbb R^n$. That $N\ge0$ is $\mathbb C$-homogeneous ($N(\mu z)=|\mu| N(z)$) and satisfies the triangle inequality are pretty obvious. If $z=u+iv$ with $u,v\in\mathbb R^n$, then any decomposition yields $$u=\sum_\ell a_\ell x^\ell,\quad v=\sum_\ell b_\ell x^\ell$$ where $\alpha_\ell=a_\ell+ib_\ell$. We have $$\|u\|\le\sum_\ell|a_\ell| \|x^\ell\|\le\sum_\ell|\alpha_\ell| \|x^\ell\|.$$ Taking the infimum over decompositions of $z$, this yields $\|u\|\le N(z)$. Likewise $\|v\|\le N(z)$. Thus $z\ne0$ implies $N(z)>0$. Thus $N$ is a norm over $\mathbb C^n$. If $z\in\mathbb R^n$, then $u=z$ and (see above) $\|z\|\le N(z)$. Since the reverse inequality is obvious, we obtain that the restriction of $N$ to $\mathbb R^n$ is $\|\cdot\|$.

Given $z\in\mathbb C^n$, and $z=\sum_\ell\alpha_\ell x^\ell$ being a decomposition as above, we have the decomposition $Mz=\sum_\ell\alpha_\ell Mx^\ell$. Therefore $$N(Mz)\le\sum_\ell|\alpha_\ell| \|Mx^\ell\|\le\|M\|\sum_\ell|\alpha_\ell| \|x^\ell\|.$$ Taking the infimum over all decompositions of $z$, we obtain $N(Mz)\le\|M\|N(z)$. Taking the supremum over $z$, this gives $N(M)\le\|M\|$, hence the equality.