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I wonder if the $2$-norm or spectral norm is also submultiplicative for non-square matrices, i.e.,

$$\| A B \|_2 \leq \| A \|_2 \cdot \| B \|_2$$

if the number of columns of $A$ coincides with the number of rows of $B$. In the literature I can only find a statement about square matrices. Thanks a lot for any remarks.

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I take it that if $A$ is $m\times n$ then $||A||_2$ is the norm of $A$ as a map from $\mathbb R^m$ to $\mathbb R^n$, where both spaces have the Euclidean norm? If so then this is obvious; it's trivial that operator norms are submultiplicative: $$||STx||\le||S||\,||Tx||\le||S||\,||T||\,||x||,$$so $||ST||\le||S||\,||T||$.

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    $\begingroup$ Thanks a lot, The usual definition: $\lVert A \rVert_2=\sqrt{\rho(A^{\top}A)}$ applies to non-square matrices too in my opinion, doesnt it? $\endgroup$ Jan 12, 2018 at 10:20
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    $\begingroup$ $\rho$ is the largest absolute eigenvalue $\endgroup$ Jan 14, 2018 at 14:45
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    $\begingroup$ @DavidC.Ullrich Why is it trivial that operator norms are submultiplicative? Based on the following, we can not get $\|ST\|\le \|S\|\|T\|$. $$\|S(Tx)\|\le \|S\|\|Tx\|\le\|S\|\|T\|\|x\|,$$ $$\|(ST)x\|\le \|ST\|\|x\|$$ $\endgroup$
    – suineg
    Oct 21, 2021 at 2:30
  • $\begingroup$ This question has been answered here. $\endgroup$
    – suineg
    Oct 21, 2021 at 2:53
  • $\begingroup$ @suineg What??? If $T$ is linear and $c>0$ then it's clear from the definition that $||Tx||\le c||x||$ isequivalent to $||T||\le c$. $\endgroup$ Oct 21, 2021 at 10:35

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