User:J D Bowen/Math725 HW13

1) Given an operator $$T:V\to V \ $$, let $$B \ $$ be the Jordan basis of $$V \ $$ with respect to $$T \ $$. Then

$$\mathfrak{M}_B^B(T) = \begin{pmatrix} J_1 & \;    & \; \\ \; & \ddots & \; \\ \; & \;     & J_p\end{pmatrix}$$

where each block Ji is a square matrix of the form


 * $$J_i =

\begin{pmatrix} \lambda_i & 0          & \;     & 0  \\ 1       & \lambda_i    & \ddots & \;  \\ \;       & 1          & \ddots & 0  \\ 0       & \;           & 1     & \lambda_i \end{pmatrix},$$

and $$\text{dim}(J_i) = n_{\lambda_i} \ $$.

Hence $$\text{Tr}(T)= \Sigma \ \text{diagonal} = \Sigma n_{\lambda_i}\lambda_i \ $$.

2) Suppose $$\text{dim}(V)=n \ $$ and the characteristic polynomial of $$T \ $$ is $$c_T(x)=x^n+c_{n-1}x^{n-1}+\dots+c_0 \ $$.

We have $$c_T(x)=x^n+c_{n-1}x^{n-1}+\dots+c_0=\Pi_i (x-\lambda_i)^{n_{\lambda_i}} \ $$, but observe that in this product, terms of the power $$x^{n-1} \ $$ can only come from multiplying every single $$x \ $$ in the product except one, which is multiplied by $$-\lambda_i \ $$. Collecting all the powers of $$x^{n-1} \ $$, we see this is $$-\Sigma n_{\lambda_i}\lambda_i = -\text{Tr}(T) \ $$ by problem 1.

3)