If $A$ is a $n\times n$ matrix with complex numbers for elements, and $C$ the $2\times2$ matrix defined by $$\begin{bmatrix} -2&4\\-3&5 \end{bmatrix}.$$ How do you prove that the Kronecker product $C\otimes A$ is diagonalisable if and only if $A$ is diagonalisable?
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2 Answers
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Use the properties of the Kronecker product: $$ (P^{-1}\otimes Q^{-1})(C \otimes A) (P \otimes Q) = (P^{-1}\otimes Q^{-1})( CP\otimes AQ) = (P^{-1}CP)\otimes (Q^{-1}AQ) $$ The given matrix $C$ is diagonalizable (it has two different eigenvalues), thus, there exists $P$ such that $P^{-1}CP$ is diagonal.
If $A$ is diagonalizable, then clearly $C\otimes A$ is diagonalizable.
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This claim is not correct. See exercise 15 in chapter 4.3 in Horn & Johnson's Topics in Matrix Analysis, which states that
$C \otimes A$ is diagonalisable if and only if both $A$ and $C$ is diagonalisable.
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2$\begingroup$ Given the fact that $C \otimes A$ is diagonalisable if and only if both $A$ and $C$ are diagonalisable and the fact that the given $C$ matrix is diagonalisable results in the fact that $C \otimes A$ is diagonalisable if and only if $A$ is diagonalisable. $\endgroup$– EdGCommented Nov 23, 2017 at 13:03