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.

  • $\begingroup$ I suspect that the condition $\|(a+bi)A\|_c = |a+bi|\cdot\|A\|_r$ defines a unique matrix norm over $M_n(\Bbb C)$, but I don't see an obvious proof. $\endgroup$ – Omnomnomnom Jun 28 '19 at 16:15
  • $\begingroup$ Following Omnomnomnom's idea, if submultiplicativity is not required, I think $\|A\|_c=\sup_{|\omega|=1}\|\Re(\omega A)\|_r$ should work. $\endgroup$ – user1551 Jun 28 '19 at 22:49

(Too long for a comment.) If submultiplicativity is not required, we may simply define $$ \|A\|_c=\sup_{|\omega|=1}\|\Re(\omega A)\|_r. $$ This idea can be applied to the decomplexification trick mentioned in the question, but we need to impose a further condition. Suppose we are so lucky to have a norm $\|\cdot\|_R$ on $M_{2n}(\mathbb R)$ such that

  1. $\|\cdot\|_R$ is a submultiplicative,
  2. $\left\|\pmatrix{X&0\\ 0&X}\right\|_R=\|X\|_r$ for every $X$,
  3. $\sup\limits_{|a+ib|=1}\left\|\pmatrix{aI&-bI\\ bI&aI}\right\|_R=1$.

Then $$ \|X+iY\|_c=\sup_{|a+ib|=1}\left\|\pmatrix{aI&-bI\\ bI&aI}\pmatrix{X&-Y\\ Y&X}\right\|_R $$ will be a submultiplicative norm on $M_n(\mathbb C)$ that preserves the values of $\|\cdot\|_r$ on $M_n(\mathbb R)$. Below are the sanity checks.

  • Preservation of $\|\cdot\|_r$. The following shows that $\|X\|_r\le\|X\|_c\le\|X\|_r$ when $X$ is a real matrix: \begin{aligned} \left\|\pmatrix{X&0\\ 0&X}\right\|_R &=\left\|\pmatrix{I&0\\ 0&I}\pmatrix{X&0\\ 0&X}\right\|_R\\ &\le\sup_{|a+ib|=1}\left\|\pmatrix{aI&-bI\\ bI&aI}\pmatrix{X&0\\ 0&X}\right\|_R\\ &\le\sup_{|a+ib|=1}\left\|\pmatrix{aI&-bI\\ bI&aI}\right\|_R \left\|\pmatrix{X&0\\ 0&X}\right\|_R=\left\|\pmatrix{X&0\\ 0&X}\right\|_R. \end{aligned}
  • Nonnegativity. Clearly $\|X+iY\|_c=0$ iff $X=Y=0$.
  • Triangle inequality. This should be obvious.
  • Homogenity. By homogeneity of $\|\cdot\|_R$, it suffices to prove that $\|(p+iq)(X+iY)\|_c=\|X+iY\|_c$ when $|p+iq|=1$. \begin{aligned} &\|(p+iq)(X+iY)\|_c\\ &=\sup_{|a+ib|=1}\left\|\pmatrix{aI&-bI\\ bI&aI}\pmatrix{pI&-qI\\ qI&pI}\pmatrix{X&-Y\\ Y&X}\right\|_R\\ &=\sup_{|(a+ib)(p+iq)|=1}\left\|\pmatrix{aI&-bI\\ bI&aI}\pmatrix{pI&-qI\\ qI&pI}\pmatrix{X&-Y\\ Y&X}\right\|_R\\ &=\sup_{|a+ib|=1}\left\|\pmatrix{aI&-bI\\ bI&aI}\pmatrix{X&-Y\\ Y&X}\right\|_R\\ &=\|X+iY\|_c. \end{aligned}
  • Submultiplicativity: \begin{aligned} &\|(X+iY)(Z+iW)\|_c\\ &=\sup_{|a+ib|=1}\left\|\pmatrix{aI&-bI\\ bI&aI}\pmatrix{X&-Y\\ Y&X}\pmatrix{Z&-W\\ W&Z}\right\|_R\\ &\le\sup_{|a+ib|=1}\left\|\pmatrix{aI&-bI\\ bI&aI}\pmatrix{X&-Y\\ Y&X}\right\|_R \left\|\pmatrix{Z&-W\\ W&Z}\right\|_R\\ &=\left\|X+iY\right\|_c \left\|\pmatrix{Z&-W\\ W&Z}\right\|_R\\ &\le\left\|X+iY\right\|_c\sup_{|a+ib|=1}\left\|\pmatrix{aI&-bI\\ bI&aI}\pmatrix{Z&-W\\ W&Z}\right\|_R\\ &=\left\|X+iY\right\|_c\left\|Z+iW\right\|_c. \end{aligned}

Unfortunately, the magic norm $\|\cdot\|_R$ required in the construction above cannot possibly exist if $\|I\|_r>1$. And I have absolutely no idea whether $\|\cdot\|_R$ always exists when $\|I\|_r=1$. So, the usefulness of the idea here is somewhat questionable.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.