Consider two block diagonal matrices $$\mathbf A = \begin{bmatrix} A_{1} & O\\ 0 & A_{2} \end{bmatrix} \mathbf B = \begin{bmatrix} B_{1} & O\\ 0 & B_{2} \end{bmatrix} $$ $A_1, A_2, B_1, B_2$ have same dimension(may not necessary). Could we show after some row and column permutations of $\mathbf A\otimes \mathbf B$, it will be same as $diag\{A_1\otimes B_1, A_2\otimes B_1, A_1\otimes B_2, A_2\otimes B_2\}$?
1 Answer
Hint
Note that
$$A\otimes B=\begin{bmatrix}A_1\otimes B&O\\O&A_2\otimes B\end{bmatrix}$$and $$A_i\otimes B=\begin{bmatrix}A_i\otimes B_1&O\\O&A_i\otimes B_2\end{bmatrix}$$which are direct consequences of the Kronecker product.
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$\begingroup$ Maybe the second equation is not "equal" but "permutation equivalent" $\endgroup$ Sep 7, 2019 at 2:28