Is every norm for real matrices extensible to a norm for complex matrices?

Question

I was browsing this website haphardly and hit upon this question: Do we have for all $M \in SL_n(\Bbb K)$, $\lVert M \rVert \geq 1$ when $\lVert \cdot \rVert$ is a matrix norm? . It suddenly dawned on me that all submultiplicative matrix norms I have ever seen are actually defined for complex matrices. None of them is defined specifically for real matrices. Is it because every submultiplicative norm for real matrices is extensible to the complex case? More specifically:

Suppose $\|\cdot\|_r$ is a submultiplicative matrix norm on $M_n(\mathbb R)$. (The subscript $r$ is just a notation; it is not a number and it doesn't signify any specific matrix norm.)

Does there always exist a submultiplicative norm $\|\cdot\|_c$ on $M_n(\mathbb C)$ such that $\|A\|_r=\|A\|_c$ for every $A\in M_n(\mathbb R)$?

If homogenity and submultiplicativity are not required, we can pick a vector norm on $\mathbb R^2$ and define $\|X+iY\|_c=\|(\|X\|_r,\|Y\|_r)\|$. However, I don't see any way to implement the homogenity condition $\|(a+ib)(X+iY)\|_c=|a+ib|\|X+iY\|_c$, not to mention submultiplicativity.

I have also considered the case where there exists a submultiplicative norm $\|\cdot\|_R$ on $M_{2n}(\mathbb R)$ such that $$\left\|\pmatrix{A&0\\ 0&A}\right\|_R=\|A\|_r$$ for every $A\in M_n(\mathbb R)$. My first thought was to define $$\|X+iY\|_c=\left\|\pmatrix{X&-Y\\ Y&X}\right\|_R.$$ If such a norm $\|\cdot\|_R$ does exist, then $\|\cdot\|_c$ is automatically submultiplicative. However, I don't see any reason why $\|\cdot\|_R$ should exist and why the homogenity condition $$\left\|\pmatrix{aI&-bI\\ bI&aI}\pmatrix{X&-Y\\ Y&X}\right\|_R=|a+ib|\left\|\pmatrix{X&-Y\\ Y&X}\right\|_R.$$ is satisfied.

Answer 1

Here is a partial answer: every submultiplicative norm $\|\cdot\|_r$ on $M_n(\mathbb R)$ $\underline{\text{with $\|I\|_r=1$}}$ can be extended to a submultiplicative norm $\|\cdot\|_c$ on $M_n(\mathbb C)$.

First, define a vector norm on $M_n(\mathbb C)$ by $$ \|A\|_v=\max_{\omega\in\mathbb C,\,|\omega|=1}\|\Re(\omega A)\|_r. $$ Then define a matrix norm on $M_n(\mathbb C)$ by $$ \|A\|_c=\max_{Z\in M_n(\mathbb C)\setminus0}\frac{\|AZ\|_v}{\|Z\|_v}. $$ This, by definition, is an induced norm (although it is induced by a vector norm on $M_n(\mathbb C)$ rather than by a vector norm on $\mathbb C^n$). Hence it is submultiplicative.

Now suppose $A$ is real. We want to show that $\|A\|_c=\|A\|_r$. First, note that $\|A\|_v=\|A\|_r$ because $\Re(\omega A)=\Re(\omega)A$ in this case. It follows that $$ \|A\|_c=\max_{X\ne0}\frac{\|AZ\|_v}{\|Z\|_v} \ge\frac{\|AI\|_v}{\|I\|_v} =\frac{\|AI\|_r}{\|I\|_r} =\|A\|_r. $$ Next, if $A=0$, surely $\|A\|_c=\|A\|_r$. Suppose $A$ is a nonzero real matrix. Let $Z$ be a nonzero complex matrix such that $\|A\|_c=\frac{\|AZ\|_v}{\|Z\|_v}$. Since $A\ne0$, we must have $\|A\|_c>0$ and hence $AZ\ne0$. Let $\omega$ be any complex number on the unit circle such that $\|AZ\|_v=\|\Re(\omega AZ)\|_r$. Let $X=\Re(\omega Z)$. Then $\|AZ\|_v=\|\Re(\omega AZ)\|_r=\|AX\|_r$ and $\|Z\|_v\ge\|\Re(\omega Z)\|_r=\|X\|_r$. Since $AZ\ne0$, $X$ is necessarily nonzero. Therefore $$ \|A\|_c=\frac{\|AZ\|_v}{\|Z\|_v} \le\frac{\|AX\|_r}{\|X\|_r} \le\frac{\|A\|_r\|X\|_r}{\|X\|_r} =\|A\|_r. $$ Consequently, $\|A\|_c=\|A\|_r$ when $A$ is real.