I'm not able to understand why this equivalences are true for the kronecker product of a matrix and why a the eigenvalues of a kronecker product of two matrixes are the product of their eigenvalues.
The book goes like this:
For $A_{nxn}$ matrix with (possibly nondistinct) eigenvalues ($\lambda_{1},\lambda_{2},...\lambda_{n}$) and $B_{pxp}$ with eigenvalues ($\mu_{1},\mu_{2},...\mu_{p}$) then (np) eigenvalues of $A\otimes B$ are given by $\mu_{j}\lambda_{i}$ for i= 1,2,....n and j=1,2,3....p.
To see this write $A$ and $B$ in their jordan form:
$A=M_{A}J_{A}M^{-1}_{A}$
$B=M_{B}J_{B}M^{-1}_{B}$
Then $M_{A}\otimes M_{B}$ has inverse given by $M^{-1}_{A}\otimes M^{-1}_{A}$ Moreover, we know that the eigenvalues of $A\otimes B$ are the same as the eigenvalues of: $(M_{A}\otimes M_{B})(A\otimes B)(M^{-1}_{A}\otimes M^{-1}_{A})= (M_{A}AM^{-1}_{A})\otimes (M_{B}BM^{-1}_{B})=J_{A}\otimes J_{B}$
Can someone explain to me why: $(M_{A}\otimes M_{B})(A\otimes B)(M^{-1}_{A}\otimes M^{-1}_{A})= (M_{A}AM^{-1}_{A})\otimes (M_{B}BM^{-1}_{B})$ is true?
Thanks in advance.
The point is that for matrices of the appropriate sizes, $(A \otimes B)(C \otimes D) = (AC) \otimes (BD)$. This is easy to see from the action on vectors: $(C \otimes D)(u \otimes v) = (C u) \otimes (D v)$ etc.