# Matrix norm of Kronecker product

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?

• According to wikipedia, you can relate the eigenvalues of the Kronecker product to that of the operands. This should give you something for the spectral norm. – xavierm02 Jun 30 '17 at 17:40
• Yes, it's true for the spectral norm. That's the only case I know for certain. – Rob Jun 30 '17 at 18:08
• Compute everything for two arbitrary $2\times 2$ matrices (i.e., get both sides as an expression of $a_{11}, a_{12},\dots ,b_{22}$). I'd expect counter examples to be easy to find once you have done that. – xavierm02 Jun 30 '17 at 18:12

Theorem 8 here provides the answer: http://www.ams.org/journals/mcom/1972-26-118/S0025-5718-1972-0305099-X/S0025-5718-1972-0305099-X.pdf. As discussed on page 413, the identity holds for all matrix norms induced by $\ell_p$ vector norms. In fact, it seems to hold for any induced vector norm, or any submultiplicative norm.
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.