My Math Forum Kronecker product identity

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 September 15th, 2016, 12:42 PM #1 Member   Joined: Jun 2009 Posts: 83 Thanks: 1 Kronecker product identity Hi, during my exploration of some matrix operations I came to a suspicion that following itentity holds (A,B are square matrices both with n rows and n columns and $\displaystyle \otimes$ is so called Kronecker product): $\displaystyle det(A\otimes B-B\otimes A)=0$ Do you know a proof of this fact? (I have tried to use this identity to proove Cayley-Hamilton theorem but I have also failed.) Thank you very much for any answers.
 September 15th, 2016, 01:49 PM #2 Global Moderator   Joined: May 2007 Posts: 6,768 Thanks: 698 I'm pleading ignorance, but wouldn't the determinate of a K. product be the same irrespective of the order of the product?
 September 18th, 2016, 08:11 AM #3 Member   Joined: Jun 2009 Posts: 83 Thanks: 1 No, Kronecker product is not commutative as can be seen on some "random" matrices.
September 18th, 2016, 01:21 PM   #4
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 Originally Posted by honzik No, Kronecker product is not commutative as can be seen on some "random" matrices.
I didn't say it was commutative, only the determinant is the same.

September 18th, 2016, 02:34 PM   #5
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 Originally Posted by mathman I didn't say it was commutative, only the determinant is the same.
Ok, but $\displaystyle det(-X)=(-1)^n.det(X)$ for all nxn matrices X so when both determinants are equal - and therefore 0 - is valid reasoning only for n odd. For even n determinants are equal always.

Maybe you wold like to say that $\displaystyle det(A\otimes B)=det(B\otimes A)$. It really seems so. But can it help with the proof? because generally it doesn't hold true thar det(X)-det(Y)=det(X-Y) even for det(X)=det(Y).

Last edited by honzik; September 18th, 2016 at 02:47 PM. Reason: better explaining

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