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](https://mathoverflow.net/questions/440771/existence-of-weird-complex-norms) however, I was a bit optimistic.



Real and complex operator norms
$\newcommand\R{\mathbb R}\newcommand\C{\mathbb C}$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 $2$.


