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.

Now observe that in the product $$\Pi_i (x-\lambda_i)^{n_{\lambda_i}} \ $$, the only way we can form a term without any powers of $$x \ $$ is to multiply all the $$-\lambda_i \ $$ terms. Thus, we have $$c_0 = \Pi (-\lambda_i)^{n_{\lambda_i}} = (-1)^n \Pi \lambda_i^{n_{\lambda_i}} \ $$.

3) Given $$\left\{{0}\right\} = V_0 \subset \dots \subset V_i \subset \dots \subset V_t = V \ $$, observe that the projection map $$\pi_t:V_t\to V_t/V_{t-1} \ $$ has kernel $$V_{t-1} \ $$ and image V_t/V_{t-1} \.

Therefore, $$\text{dim}(V)=\text{dim}(V_{t-1})+\text{dim}(V_t/V_{t-1}) \ $$.

Now suppose that there exists an $$i, \ 1\leq i \leq t \ $$, such that $$\text{dim}(V)= \text{dim}(V_i)+\Sigma \text{dim}(V_j/V_{j-1}) \ $$. Then since the map $$\pi_i \ $$ has kernel $$V_i \ $$ and image $$V_i/V_{i-1} \ $$, and so $$\text{dim}(V)= \text{dim}(V_{i-1})+\Sigma \text{dim}(V_j/V_{j-1}) \ $$.

Notably, this means $$\text{dim}(V)=\text{dim}(V_1)+\Sigma \text{dim}(V_j/V_{j-1}) \ $$.

Since $$V_0 = \left\{{0}\right\}, \ V_1=V_1/V_0 \ $$ and so $$\text{dim}(V)=\Sigma \text{dim}(V_j/V_{j-1}) \ $$.

4) Suppose $$T:V\to V \ $$ is linear with eigenvalue $$\lambda \ $$ and let $$p(x)\in\mathbb{C}[x] \ $$. We aim to show that $$p(\lambda) \ $$ is an eigenvalue of $$p(T) \ $$.

Let us observe that if we have $$Tv=\lambda v \ $$, then $$T^n v = T^{n-1}(Tv)=T^{n-1}(\lambda v)= \lambda T^{n-1}v = \lambda T^{n-2}(Tv)= \lambda T^{n-2}(\lambda v) = \lambda^2 T^{n-2}v=\dots=\lambda^{n-1}Tv=\lambda^n v \ $$.

Let $$a_i \ $$ be such that $$p(x)=a_nx^n+\dots+a_0 \ $$. Let $$v \ $$ be an eigenvector of $$T \ $$ associated with $$\lambda \ $$, so that $$Tv=\lambda v \ $$.

Then $$p(T)v = a_nT^n v +\dots +a_0v = a_n\lambda^n v+\dots+a_0v= p(\lambda)v \ $$ and so

Suppose $$p(T) = 0 \ $$ and $$\lambda \ $$ is an eigenvalue of $$T \ $$. Then $$p(\lambda) \ $$ is an eigenvalue of the zero operator, and so $$p(\lambda)=0 \ $$. Hence, the eigenvalues of $$T \ $$ are among the roots of $$p \ $$.

5) Let $$V, W \ $$ be vector spaces, with $$U\subset V \ $$ a subspace and $$T:V\to W \ $$ linear, $$U\subset \text{ker}(T) \ $$. Define $$\pi:V\to V/U \ $$.

Given $$x\in V/U \ $$, let $$x^* \in V \ $$ be such that $$\pi(x^*)=x \ $$. Define $$\overline{T}(x)=T(x^*) \ $$.

Observe that if $$x^*_1, x^*_2 \ $$ both satisfy $$\pi(x^*_1)=\pi(x^*_2)=x \ $$, then they must differ by an element $$u\in U \ $$. Then $$T(x^*_1)=T(x^*_2+u)=T(x^*_2)+0 $$ and so the function $$\overline{T} \ $$ is well defined.

Observe that $$(\overline{T}\circ \pi)(x^*)=\overline{T}(x)=T(x^*) \ $$.

Now suppose that $$S:V/U \to W \ $$ satisfies $$S\circ \pi = T \ $$. Then $$T(x^*)=S(\pi(x^*))=S(x)=\overline{T}(x) \ $$, and so $$\overline{T} \ $$ is unique.

6) Suppose we have $$V_1\subset V_2 \subset V_3 \ $$ and $$T:V_3\to V_3 \ $$ such that $$T(V_3)\subset V_2, \ T(V_2)\subset V_1 \ $$.

Given $$x\in V_3/V_2 \ $$, let $$x^* \in V_3 \ $$ be such that $$\pi_3(x^*)=x \ $$. Define $$\overline{T}(x)=\pi_2(T(x^*)) \ $$. Since $$\text{ker}(\pi_2)=V_2 \ $$, by the previous problem this is well-defined and unique.