# Is the spectral norm submultiplicative? [duplicate]

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.

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 $$||ST||\le||S||\,||T||$$.
• 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? Jan 12, 2018 at 10:20
• $\rho$ is the largest absolute eigenvalue Jan 14, 2018 at 14:45
• @DavidC.Ullrich Why is it trivial that operator norms are submultiplicative? Based on the following, we can not get $\|ST\|\le \|S\|\|T\|$. $$\|S(Tx)\|\le \|S\|\|Tx\|\le\|S\|\|T\|\|x\|,$$ $$\|(ST)x\|\le \|ST\|\|x\|$$ Oct 21, 2021 at 2:30
• @suineg What??? If $T$ is linear and $c>0$ then it's clear from the definition that $||Tx||\le c||x||$ isequivalent to $||T||\le c$. Oct 21, 2021 at 10:35