$||A\times B||\le ||A||\cdot ||B||$ is not always correct. But which kind of matrix norm satisifies this formula for square matrix $A$ and arbitrary matrix $B$?
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$\begingroup$ Your use of the singular form "norm" (rather than "norms") is confusing. Since $B$ is not necessarily square, $AB,\,A$ and $B$ in general live in different matrix spaces. Therefore, in the inequality $\|AB\|\le\|A\|\|B\|$, three norms are actually involved. $\endgroup$– user1551May 3, 2015 at 13:25
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$\begingroup$ Yes, but I just want a norm defined for matrix of arbitrary shape to meet the inequality. $\endgroup$– CatDogMay 4, 2015 at 14:59
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Operator norms (induced by any norms on vectors of the appropriate dimensions) satisfy $\|A B\| \le \|A \| \|B \|$.
answered Apr 30, 2015 at 4:28
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$\begingroup$ I do not think so, if A and B is [3,1] and [4;1], respectively. and define the norm as the max absolute value of matrix elements. ||A*B|| = 14 > ||A||*||B|| = 12 $\endgroup$– CatDogMay 3, 2015 at 9:31
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1$\begingroup$ That is not an operator norm. Given norms $\| \cdot\|_m$ and $\|\cdot \|_n$ on $\mathbb R^m$ and $\mathbb R^n$, the corresponding operator norm on $m \times n$ real matrices is $\|A\| = \max(\|A x\|_m : \|x\|_n = 1)$. $\endgroup$ May 3, 2015 at 18:07
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$\begingroup$ I read the wiki page. I see the matrix is treated as a linear operator mapping vector in $R^m$ to $R^n$. You mean for operator norm, given a unit vector, mapping of $A\times B$ produce a vector with lesser length than the product of separate action of $A$ and $B$? $\endgroup$– CatDogMay 4, 2015 at 14:56
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1$\begingroup$ $\|ABx\| \le \|A\| \|Bx\| \le \|A\| \|B\| \|x\|$ $\endgroup$ May 4, 2015 at 15:13
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