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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): $\displaystyle det(A\otimes BB\otimes A)=0$ Do you know a proof of this fact? (I have tried to use this identity to proove CayleyHamilton 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 
Global Moderator Joined: May 2007 Posts: 6,768 Thanks: 698  
September 18th, 2016, 02:34 PM  #5  
Member Joined: Jun 2009 Posts: 83 Thanks: 1  Quote:
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(XY) even for det(X)=det(Y). Last edited by honzik; September 18th, 2016 at 02:47 PM. Reason: better explaining  

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determinant, identity, kronecker, kronecker product, product 
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