91Ó°ÊÓ

The Kronecker product (tensor product) of two matrices \(A \in \mathbb{F}^{n\times n}\) and \(B \in \mathbb{F}^{m \times m}\) is a block matrix defined as follows: \[ (A \otimes B)_{(i - 1)m + x, (j - 1)m + y} = a_{ij} b_{xy} \] where \(a_{ij}\) are the entries of matrix A, and \(b_{xy}\) are the entries of matrix B.

Given that \(A\) has eigenvalues \(\lambda_1, \ldots, \lambda_n\) and \(B\) has eigenvalues \(\mu_1, \ldots, \mu_m\), there exist eigenvectors \(v_1, \ldots, v_n\) of A and \(w_1, \ldots, w_m\) of B such that: \[ A v_i = \lambda_i v_i , \quad B w_j = \mu_j w_j, \] for \(i \leq n, j \leq m\).

Consider the tensor product of the eigenvectors of \(A\) and \(B\), defined by \[ v_i \otimes w_j = \begin{pmatrix} v_{i1} w_j \\ \vdots \\ v_{in} w_j \end{pmatrix}, \] where \(v_{ik}\) is the k-th entry of eigenvector \(v_i\). Now, compute the product \((A \otimes B)(v_i \otimes w_j)\) as follows: \[ \begin{aligned} (A \otimes B)(v_i \otimes w_j) &= (A \otimes B)\begin{pmatrix} v_{i1} w_j \\ \vdots \\ v_{in} w_j \end{pmatrix} \\ &= \begin{pmatrix} a_{11}B(v_{i1}w_j) & \cdots & a_{1n}B(v_{in}w_j) \\ \vdots & \ddots & \vdots \\ a_{n1}B(v_{i1}w_j) & \cdots & a_{nn}B(v_{in}w_j) \end{pmatrix} \\ &= \begin{pmatrix} a_{11}(\lambda_iv_{i1})w_j & \cdots & a_{1n}(\lambda_iv_{in})w_j \\ \vdots & \ddots & \vdots \\ a_{n1}(\lambda_iv_{i1})w_j & \cdots & a_{nn}(\lambda_iv_{in})w_j \end{pmatrix} \\ &= \begin{pmatrix} \lambda_i v_{i1} a_{11}w_j & \cdots & \lambda_i v_{in} a_{1n}w_j \\ \vdots & \ddots & \vdots \\ \lambda_i v_{i1} a_{n1}w_j & \cdots & \lambda_i v_{in} a_{nn}w_j \end{pmatrix} \\ &= \lambda_i \begin{pmatrix} v_{i1} a_{11}w_j & \cdots & v_{in} a_{1n}w_j \\ \vdots & \ddots & \vdots \\ v_{i1} a_{n1}w_j & \cdots & v_{in} a_{nn}w_j \end{pmatrix} \\ &= \lambda_i (v_i \otimes a_1w_j) \\ &= \lambda_i (\lambda_1 v_i \otimes \mu_j w_j) \\ &= (\lambda_i \mu_j) (v_i \otimes w_j) \\ \end{aligned} \] We have shown that \((A \otimes B)(v_i \otimes w_j) = (\lambda_i \mu_j) (v_i \otimes w_j)\), which means that the eigenvalues of \(A \otimes B\) are indeed the products of the eigenvalues of A and B, as stated in the exercise, with their eigenvectors being the tensor product of the eigenvectors of \(A\) and \(B\). This completes the proof.