Any way to inverse a permutated Kronecker product when one participant matrix is known?

Asked May 08 '15 at 12:23

Active May 08 '15 at 12:23

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I know there is a way to inverse a Kronecker product. See this question reverse Kronecker product.

If I have a permutated Kronecker product of $A$ and $B$, that is $P(A\otimes B)Q$ where $P$ and $Q$ are unknown permutation matrices of rows and columns, and $B$ is known, is there any way to obtain $A$?

Thought it would be easy, but it really exhaust me.

edited Apr 13 '17 at 12:20

Community

asked May 08 '15 at 12:23

Jianting Wang

No. In particular, $(e_1e_1^T) \otimes I$ can be permuted to $$(e_ne_n^T) \otimes I$$ with the right choice of $P$ and $Q$. – Ben Grossmann May 08 '15 at 16:10
@Omnomnomnom Not quite sure I've got your idea. What's $e_{1}$ and $e_{n}$? – Jianting Wang May 09 '15 at 01:01
Standard basis vectors – Ben Grossmann May 09 '15 at 13:00

Any way to inverse a permutated Kronecker product when one participant matrix is known?

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