Let $A$ be a square matrix and $I$ the $k \times k$ identity matrix. Then the identity $$ \det(A \otimes I) = \det(A)^k,$$ holds as can be seen from a general result on the determinant of block matrices.
Since the structure of $A \otimes I$ is quite beautiful I am wondering
Can someone find a short proof of the above equality?
Consider the canonique basis $e_1,\ldots,e_k$ of $\Bbb R^k$. Furthermore let $\lambda_1,\ldots,\lambda_n$ be the eigenvalues of $A \in \Bbb R^{n\times n}$ and $v_1,\ldots,v_n$ their associated eigenvectors. Then $v_i \otimes e_j$ is an eigenvector of $A \otimes I$ associated to the eigenvalue $\lambda_i$ for every $j = 1,\ldots,k$. Since $v_i \otimes e_l$ and $v_i \otimes e_j$ and linearly independent for every $l \neq j$. It is clear that the eigenvalues of $A \otimes I$ are exactly $\underbrace{\lambda_1,\ldots,\lambda_1}_{k \text{ times}},\ldots,\underbrace{\lambda_n,\ldots,\lambda_n}_{k \text{ times}}$ it follows that $\det(A\otimes I) = \det(A)^k$.