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The question:

Let $A,B \in M_n(\mathbb R)$ be two matrices in the space of all square matrices, and let $\lVert \cdot \rVert$ be a norm over it.
Show there exist $C>0$ such that $\lVert AB \rVert \le C \lVert A \rVert \lVert B \rVert$.

The only matrix norm that was introduced so far is the operator norm (Sup$Ax$ such that $ \lVert x \rVert = 1$ ). I know this property holds for the operator norm with $C=1$ but the proof we've seen of it utlizes the property of the norm itself and I couldn't come up with a way to generalize that for all norms over square matrices.

Another confusion I have is that in the proof of this property for the operator norm, we've used $\lVert Ax \rVert \le \lVert A \rVert \lVert x \rVert$, but how is this true? $Ax$ is a vector, not a matrix (and in particular not a square matrix), how can we apply the matrix norm to $Ax$?

RatherAmusing
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1 Answers1

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Let $U,V,W$ be normed vector spaces.

  1. If $U,V$ are finite dimensional then every bilinear map $f:U\times V\to W$ is continuous, i.e. $$\exists C\quad\forall(u,v)\in U\times V\quad \|f(u,v)\|\le C\|u\|\|v\|.$$ This applies to $U=V=W=M_n(\Bbb R)$ and $f=$ matrix multiplication. Alternatively, you could argue that $\lVert AB \rVert \le \lVert A \rVert \lVert B \rVert$ for your favorite operator norm, and that on $M_n(\Bbb R),$ every norm is equivalent to that one.
  2. For a continuous linear map $A:U\to V,$ $\|Ax\|\le\|A\|\|x\|$ by definition of the operator norm $\|A\|,$ relatively to the given norms $\|~\|_U$ on $U$ and $\|~\|_V$ on $V.$ More precisely: $\|Ax\|_V\le\|A\|\|x\|_U.$
Anne Bauval
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  • Thank you!
    1. We've actually never seen the definition using a bilinear map. But we did see that every two norms are equivalent over a finite dimensional norm space. I wasn't sure that the map used for this equivalence is the identity, similar to equivalence of norms over $\mathbb R^k$.
     – RatherAmusing Feb 11 '23 at 18:06
  • Not only is it "similar": it is exactly that, when identifying $M_n(\Bbb R)$ to $\Bbb R^k$ with $k=n^2.$ – Anne Bauval Feb 11 '23 at 18:49
  • Oh I see! Thank you so much! – RatherAmusing Feb 11 '23 at 19:22