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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.

Rodrigo de Azevedo
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Tim vor der Brück
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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 $||ST||\le||S||\,||T||$.

David C. Ullrich
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    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? – Tim vor der Brück Jan 12 '18 at 10:20
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    $\rho$ is the largest absolute eigenvalue – Tim vor der Brück Jan 14 '18 at 14:45
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    @DavidC.Ullrich Why is it trivial that operator norms are submultiplicative? Based on the following, we can not get $|ST|\le |S||T|$. $$|S(Tx)|\le |S||Tx|\le|S||T||x|,$$ $$|(ST)x|\le |ST||x|$$ – suineg Oct 21 '21 at 02:30
  • This question has been answered here. – suineg Oct 21 '21 at 02:53
  • @suineg What??? If $T$ is linear and $c>0$ then it's clear from the definition that $||Tx||\le c||x||$ isequivalent to $||T||\le c$. – David C. Ullrich Oct 21 '21 at 10:35
  • @DavidC.Ullrich I still can not get $|ST|\le |S||T|$ from your argument. From the words ''clear''''obvious'', I think you take everything granted. Can you make your argument convincing with mathematical words, not the other way around? – suineg Oct 21 '21 at 10:59
  • There's nothing I could be taking for granted here - everything is immediate from the definitions. Tell me what version of the definition of $||T||$ you want to use. And tell me which half you're having trouble with:(i)Lemma: $||Tx||\le c||x|$ for all $x$ if and only if $||t||\le c$, (ii) it follows from the lemma that $||ST||\le||S||,||T||$. – David C. Ullrich Oct 21 '21 at 11:13