I have a question related to reverting kron tensor product. As We can see the in below example:

>> X

X =

     0     1
     1     0

>> Z

Z =

     1     0
     0    -1

>> D=kron(X,Z)

D =

     0     0     1     0
     0     0     0    -1
     1     0     0     0
     0    -1     0     0

So the problem is: -If We know D, also know Z, How can We calculate X? Are there any cmd to do it in MATLAB?

Simply, As I do for some specific case, I investigate X(i,j)*Y --> So I got X(i,j). But I need to solve in general case.

I am looking forward to hearing from you.

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  • $\begingroup$ Are you looking for just the $2\times 2$ case (i.e. with $X,Z$ being $2\times 2$ matrices) or the general $n\times n$ case? $\endgroup$ – Winther Jul 1 '15 at 15:31
  • $\begingroup$ In general the Kronecker Product is not invertible so it's hard to come up with a useful numerical algorithm that works in general. When it is uniquely invertible, see this answer for how you can invert it. $\endgroup$ – Winther Jul 1 '15 at 15:36
  • $\begingroup$ Thanks for your response. I need to solve in general case. As We know: If X(a×b) and Y(c×d) so kron(X,Y) will have size (ac)×(bd) $\endgroup$ – Hai Duc Jul 1 '15 at 15:43

Given $C=A\otimes B$,
Calculate: $D=BA^\dagger$. See Neil de Beaudrap's answer for how to get $D$ without knowing $B$ or $A$
Use $D$ to get $kB$ and $A/k$ where $k$ is a scalar.

Remember that $kA \otimes (1/k)B = A\otimes B$ for all $k$,
so given $A\otimes B$, you cannot possibly know exactly what $A$ and $B$ were, you can only know them up to the factors involving $k$.

This is the best resource I know on the Kronecker quotient, and pages 3 and 4 have some properties that would be fabulous for your application.


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