Question: Let $T:V\rightarrow V$ be a linear transformation defined on the finite dimensional vector space $V$. Let $\lambda$ be an eigenvalue of $T$, and set $W_i=\{v\in V|(T-\lambda I)^i(v)=0\}$. If $m$ is the multiplicity of $\lambda$ as a root of the characteristic polynomial of $T$, prove that $W_m=W_{m+1}$.
My thoughts: Since $m$ is the multiplicity of $\lambda$ as a root of $\Delta(v)$, we have that $\Delta(v)=(v-\lambda)^m$. So, the minimal polynomial must divide $(v-\lambda)^m$ (not sure if that helps here). Now, for $W^m$ we have that $(T-\lambda I)v=0\implies Tv=\lambda v\implies T^mv=\lambda^mv$... but now I am stuck. Any help is greatly appreciated! Thank you.
Obviously $W_m \subseteq W_{m + 1}$. So it remains to show $W_{m + 1} \subseteq W_m$. Otherwise there exists $\alpha \in W_{m + 1}$ but $\alpha \notin W_m$, i.e., $\beta := (T - \lambda I)^m(\alpha) \neq 0$. To arrive contradiction, let $\lambda_1, \ldots, \lambda_t$ be all the remaining distinct eigenvalues of $T$, with multiplicities $m_1, \ldots, m_t$ respectively. Therefore the characteristic polynomial of $T$ can be written as $p(x) = (x - \lambda)^m(x - \lambda_1)^{m_1}\cdots(x - \lambda_t)^{m_t}$. By Bezout's identity, it can be shown that $$W_m \cap \text{Ker}((T - \lambda_1 I)^{m_1}\cdots(T - \lambda_t I)^{m_t}) = 0. \tag{$*$}$$
This and $\beta \in W_m$ (as $(T - \lambda I)^m(\beta) = (T - \lambda I)^{2m}(\alpha) = (T - \lambda I)^{m - 1}(T - \lambda I)^{m + 1}(\alpha) = 0$) together imply $\beta \notin \text{Ker}((T - \lambda_1 I)^{m_1}\cdots(T - \lambda_t I)^{m_t})$, whence $$p(T)(\alpha) = (T - \lambda I)^m(T - \lambda_1 I)^{m_1}\cdots(T - \lambda_t I)^{m_t}(\alpha) = (T - \lambda_1 I)^{m_1}\cdots(T - \lambda_t I)^{m_t}(\beta) \neq 0,$$ which contradicts with $p(x)$ is an annihilating polynomial of $T$. This completes the proof.
Notice that the conclusion of this exercise can be generalized to that $m$ is the geometric multiplicity of $\lambda$. And the proof is completely the same as above.
Proof of $(*)$: Since $\lambda, \lambda_1, \ldots, \lambda_t$ are distinct, the polynomial $(x - \lambda)^m$ and the polynomial $(x - \lambda_1)^{m_1}\cdots(x - \lambda_t)^{m_t}$ are coprime, hence by Bezout's identity, there exist $f(x), g(x) \in F[x]$ such that \begin{align*} 1 = f(x)(x - \lambda)^m + g(x)(x - \lambda_1)^{m_1}\cdots(x - \lambda_t)^{m_t}. \end{align*} Therefore, if $v \in W_m \cap \text{Ker}((T - \lambda_1 I)^{m_1}\cdots(T - \lambda_t I)^{m_t})$, then \begin{align*} v = I(v) = f(T)(T - \lambda I)^m(v) + g(T)(T - \lambda_1 I)^{m_1}\cdots(T - \lambda_t I)^{m_t}(v) = 0 + 0 = 0. \end{align*} This shows $(*)$.