Exercises

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 $\chi_A(A) = 0$.