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$\det(\mathbf{A}\otimes \mathbf{B})=(\det(\mathbf{A})^m)(\det(\mathbf{B})^n)$. $\mathbf{A}$ is $n\times n$ matrix. $\mathbf{B}$ is $m\times m$ matrix. $\otimes$ is Kronecker product. How can i prove that?

Michael Hardy
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user137231
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1 Answers1

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If $A$ and $B$ are diagonal matrices, it is easy to image that the equation is satisfied.

If $A$ and $B$ are not diagonal matrices, we can do some transformation like this:

$A^{\prime} = P A P^{-1}$

where $A^{\prime}$ is a diagonal matrix. $P$ is transformation matrix. we can do similar:

$B^{\prime} = Q B Q^{-1}$

Note: ${\rm det}(A^\prime) = {\rm det}(P) {\rm det}(A) {\rm det}(P^{-1}) = {\rm det}(A)$

Using the relation $(AC)\otimes(BD)=(A \otimes B)(C \otimes D)$

$A \otimes B = (P^{-1} A^\prime P) \otimes (Q^{-1} B^\prime Q) = [(P^{-1}A^\prime)\otimes(Q^{-1}B^\prime)][P \otimes Q]=[P^{-1} \otimes Q^{-1}][A^\prime \otimes B^\prime][P\otimes Q]=[P\otimes Q]^{-1}[A^\prime \otimes B^\prime][P \otimes Q] $

So

$\det(A \otimes B) = \det([P\otimes Q]^{-1}[A^\prime \otimes B^\prime][P \otimes Q]) = \det [A^\prime \otimes B^\prime]$

Since $A^\prime$ and $B^\prime$ are diagonal matrices, easy to image $\det [A^\prime \otimes B^\prime] = \det(A^\prime)^m \det(B^\prime)^n = \det(A)^m \det(B)^n$

complete.

Fei Zhang
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  • If A,B is not diagonalizable? – user137231 Nov 11 '14 at 06:33
  • using similar strategy, you may try LU decomposition way, which is more general than diaganolization, and might also be easier to prove – Fei Zhang Nov 11 '14 at 08:10
  • For the non-diagonalizable case, use the Schur decomposition. The unitary matrices do not contribute anything to the determinant since $\det(U)\det(U^*)=1$. The determinant of a triangular matrix is simply the product of the diagonal elements. In the Kronecker product of triangular matrices, only the diagonal elements of the factors affect the diagonal elements of the product. – greg Dec 06 '15 at 21:46