Show that normal matrices are diagonalizable.
Consider the Vandermonde matrix: \begin{align*} A \coloneqq \left(\begin{array}{ccc} 1 & \cdots & 1 \\ \lambda_{1} & \cdots & \lambda_{k} \\ \vdots & & \vdots \\ \lambda_{1}^{k-1} & \cdots & \lambda_{k}^{k-1} \end{array}\right) .\end{align*}
Show that \begin{align*} \operatorname{det}A = \prod_{i < j} (\lambda_i - \lambda_j) .\end{align*}
Show that a nonzero nilpotent matrix \(A\) is not diagonalizable over any field. Some useful facts:
- \(\operatorname{Spec}A = \left\{{0}\right\}\), since \(Ax=\lambda x \implies A^n=\lambda^nx\), so \(A^n=0\) forces \(\lambda =0\). This forces \(\operatorname{JCF}(A)\) to be strictly upper-triangular.
- \(\min_A(x) = x^n\).
- If \(A\) were diagonalizable, \(\operatorname{JCF}(A) = 0\).
Prove Cayley-Hamilton in the following way. Let \(V = \mathop{\mathrm{span}}\left\{{\mathbf{v}_1, \cdots, \mathbf{v}_n}\right\}\) and define the \(i\)th flag as \({\operatorname{Fil}}_i V \coloneqq\mathop{\mathrm{span}}\left\{{\mathbf{v}_1, \cdots, \mathbf{v}_i}\right\}\) for all \(1\leq i \leq n\), and set \({\operatorname{Fil}}_0 V \coloneqq\left\{{0}\right\}\). Show that if if \(A\) is upper triangular, then \(A({\operatorname{Fil}}_i V) \subseteq {\operatorname{Fil}}_i V\). Now supposing \(\mathbf{v}_i\) are eigenvectors for \(\lambda_i\), show that \begin{align*} (A-\lambda_n I) {\operatorname{Fil}}_n V &\subseteq {\operatorname{Fil}}_{n-1} V \\ (A-\lambda_{n-1} I) (A-\lambda_n I ) {\operatorname{Fil}}_n V &\subseteq {\operatorname{Fil}}_{n-2} V \\ &\vdots \\ \prod_i (A-\lambda_{n-i} I) {\operatorname{Fil}}_n V &\subseteq {\operatorname{Fil}}_0 V = \left\{{0}\right\} .\end{align*} Conclude that \(\chi_A(A) = 0\).