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.