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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?

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  • $\begingroup$ 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. $\endgroup$ – xavierm02 Jun 30 '17 at 17:40
  • $\begingroup$ Yes, it's true for the spectral norm. That's the only case I know for certain. $\endgroup$ – Rob Jun 30 '17 at 18:08
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    $\begingroup$ 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. $\endgroup$ – xavierm02 Jun 30 '17 at 18:12
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On page 149 exercise 6 in book: Matrix analysis for scientists and engineers, this is true for operator norm. You can see chapter 13 of the book by the link: http://www.siam.org/books/textbooks/OT91sample.pdf

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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.

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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.

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