TITLE: Matrix norm of Kronecker product
QUESTION [9 upvotes]: Is it true that $ \| A \otimes B \| = \|A\|\|B\| $ for any matrix norm $ \|\cdot \| $? If not, does this identity hold for matrix norms induced by $ \ell_p $ vector norms?

REPLY [4 votes]: Here is a proof for the lazy: Let $A = \sum_i \sigma_i u_i v_i^T$ and $B = \sum_j \lambda_j x_j y_j^T$ be the singular value decomposition (SVD) of the two matrices. Then,
\begin{align}
A \otimes B &= \sum_{i,j}\sigma_i \lambda_j (u_i v_i^T) \otimes (x_jy_j^T) \\
&= \sum_{i,j} \sigma_i \lambda_j (u_i \otimes x_j) (v_i^T \otimes y_j^T)  \\
&= \sum_{i,j} \sigma_i \lambda_j (u_i \otimes x_j) (v_i \otimes y_j)^T
\end{align}
where the first equality is by the bilinearity of $\otimes$, the second by the "mixed product" property of Kronecker product and last one by $(A\otimes B)^T = A^T \otimes B^T$.
It is not hard to see that $\{u_i \otimes x_j, \forall i,j \}$ is an orthonormal collection of vectors and similarly for $\{v_i \otimes y_j, \forall i,j \}$. It follows that the last line above is the SVD of $A \otimes B$, with singular values $\{\sigma_i \lambda_j, \forall i,j\}$. Hence,
\begin{align}
\| A \otimes B\| = \max_{i,j} \sigma_i \lambda_j = (\max_i \sigma_i)(\max_j \lambda_j) = \|A\| \|B\|.
\end{align}
for the $\ell_2$ operator norm.