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In matrix analysis, a matrix norm can be induced by a vector norm by the following.

$$|||A|||_\alpha = \max _{||x||_\alpha =1} ||Ax||_{\alpha}$$ where $||\cdot||_\alpha$ is a vector norm and $A$ is a square matrix.

One can verify that the $|||\cdot|||_\alpha$ is a matrix norm.

However suppose there is another vector norm $|||\cdot||_\beta$ and define

$$|||A|||_{\alpha, \beta} = \max _{||x||_\alpha =1} ||Ax||_\beta.$$

Is the above still a matrix norm? What I cannot establish is the submultiplicative property, i.e.

$$|||AB|||_{\alpha, \beta} \leq |||A|||_{\alpha, \beta}|||B|||_{\alpha, \beta}$$

where $A$ and $B$ are square matrices of the same size.

If not, can someone give me a counterexample that makes the above false?

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  • $\begingroup$ How would you define $\Vert \cdot \Vert_{\alpha}$ ? Is it $$ \forall x \in \mathbb{R}^n, \; \Vert x \Vert_{\alpha} = \Big( \sum_{i=1}^{n} \vert x_i \vert^{\alpha} \Big)^{1/\alpha} $$ ? $\endgroup$ – jibounet Oct 9 '17 at 14:46
  • $\begingroup$ you can assume that i guess, then just add in the l-infinity norm $\endgroup$ – user489434 Oct 9 '17 at 14:55
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Matrix norms defined like this are known as consistent norms. Consistent norms are not necessarily submultiplicative. Here is an easy way to construct a non-submultiplicative consistent norm: just define $\|\cdot\|_\beta=\epsilon\|\cdot\|_\alpha$ for some sufficiently small $\epsilon>0$. Then $|||\cdot|||_{\alpha,\beta}=\epsilon|||\cdot|||_\alpha$ and hence $$ |||I^2|||_{\alpha,\beta}=\epsilon|||I|||_\alpha >\epsilon^2|||I|||_\alpha=|||I|||_{\alpha,\beta}^2 $$ when $\epsilon$ is small enough.

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  • $\begingroup$ How does this answer my question? I'm confused, sorry. $\endgroup$ – user489434 Oct 9 '17 at 19:11
  • $\begingroup$ You said we can construct such norm with no submultiplicative property. Okay, however, I defined a specific norm above. $\endgroup$ – user489434 Oct 9 '17 at 19:13
  • $\begingroup$ user1551 said that a matrix norm is not necessariily submultiplicative. That is, your defintion of "matrix norm" is wrong. They also gave you an example of two norms in which the corresponding matrix norm is not submultiplicative. $\endgroup$ – Friedrich Philipp Oct 9 '17 at 19:31
  • $\begingroup$ @SumRay If the definition you quoted comes from a textbook, I think you have some misunderstandings. In the definition of consistent norm, the symbols $\|\cdot\|_\alpha$ and $\|\cdot\|_\beta$ simply refer to two vector norms rather than the $\ell_\alpha$- and $\ell_\beta$-norms. Some consistent norms are submultiplicative, but some of them are not. What I have demonstrated here is an example of the latter case. $\endgroup$ – user1551 Oct 9 '17 at 19:46
  • $\begingroup$ If you are asking specifically whether the matrix norm consistent with the $\ell_\alpha$ and $\ell_\beta$ norms is submultiplicative or not, please make it clear in your question and I'll delete my answer. $\endgroup$ – user1551 Oct 9 '17 at 19:49

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