Let $|| = ||_{p,q}$ be an operator norm on $\mathbb R^{n \times m}$. In General, $\|AB\|\le \|A\|\|B\|$. Is there some criterion on $A, B$ (at least for some operator norms) so that $\|AB\| = \lVert A \rVert \lVert B\rVert$?
$\begingroup$
$\endgroup$
4
-
6$\begingroup$ A remark: you cannot have a multiplicative norm on the whole $\mathrm{Mat}(n×m, \, \mathbb{R})$, because there are non-zero matrices whose product is zero. $\endgroup$– Francesco PolizziCommented Feb 12, 2019 at 10:37
-
$\begingroup$ @Francesco Polizzi unless $m = n = 1$... $\endgroup$– KConradCommented Feb 12, 2019 at 12:52
-
$\begingroup$ @KConrad: of course, thank you. $\endgroup$– Francesco PolizziCommented Feb 12, 2019 at 13:44
-
$\begingroup$ There is also the wide class of unitarily-invariant norms (i.e., norms depending only on singular values, like the operator norm, trace norm, Frobenius norm, Ky Fan norms, or Schatten norms). As their name suggests, the desired property holds for them if either of A or B is unitary. $\endgroup$– Nathaniel JohnstonCommented Feb 13, 2019 at 21:34
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
1
At least, there is this important case: in $C^*$-algebras, there is an involution and the norm has the property that $$\|x\|=\|x^*\|=\|xx^*\|^{1/2}.$$ In the special case of ${\bf M}_n({\mathbb C})$, this is $$\|AA^*\|=\|A\|\cdot\|A^*\|$$ where the norm is the usual operator norm $\|\cdot\|_{2,2}$.
Actually this formula extends to the case of rectangular matrices.
answered Feb 12, 2019 at 10:35
-
$\begingroup$ Thanks for the reply. In my case the matrices are usually not square. Do you know of a special family of matrices where $||AB|| = ||A||||B||$ $\endgroup$– N_SegolCommented Feb 12, 2019 at 12:00