Question: $P_{T}$:= characteristic polynomial of T; $minP_T$:= minimal polynomial of T. Consider $P_{T}=minP_T$. If $g$ is a polynomial which divides $minP_T$, then dim(Ker($g(T)$))=deg $g$.
My Attempt: $g(T)$ can be written as a product of powers of irreducible polynomials. Suppose $g(T)=\prod_{i=1}^{k}\Phi_i^{m_i}$ where each $\Phi_i$ is irreducible. Now we know the minimal polynomial of the restriction of $T$ in Ker $\Phi_{i}^{m_i}$ i.e. $minP_{T_{Ker\Phi_{i}^{m_i}}}=\Phi_{i}^{m_i}$. Therefore dim Ker $\Phi_{i}^{m_i}$=deg $\Phi_{i}^{m_i}$. Now since each of the $\Phi_i$'s are relatively prime, their Kernels only have the zero intersection. Hence $dim(Ker(g(T))=\sum_{i=1}^k dim({Ker\Phi_{i}^{m_i}})= \sum_{i=1}^k deg \Phi_{i}^{m_i}=$deg $g$.
Question: I don't see in the question where the condition of the minimal polynomial being same as the characteristic polynomial is required. What I think is that, whenever any polynomial divides the minimal polynomial, it's Kernel has the dimension of the polynomial's degree.
I would like to know if there is any fault in my approach.
Thanks in advance
Best Answer
You had the correct idea to start the proof by induction on the number of distinct irreducibles, but made a serious mistake in not having the base case of your induction.
So let $V$ be a finite-dimensional $k$-vectorspace, and $T\in\operatorname{End}_kV$ be such that (in your notation) $\operatorname{P}_T(x)=\Phi(x)^n$, $\Phi(x)\in k[x]$ irreducible and $\operatorname{minP}_T=\operatorname{P}_T$. We need to prove that $\dim\ker\Phi^m(T)=\deg\Phi^m$ for all $0\leq m\leq n$. The obvious way to do so is to put $T$ into generalized Jordan normal form. Since $\operatorname{minP}_T=\operatorname{P}_T$, there is some $v\in V$ such that $\Phi^{m-1}(T)(v)\neq 0$. Then $$ \begin{matrix} v,&Tv,&\dots,&T^{\deg\Phi-1}v,\\ \Phi(T)v,&\Phi(T)Tv,&\dots,&\Phi(T)T^{\deg\Phi-1}v,\\ \Phi(T)^2v,&\dots,\\ &&\dots,&\Phi(T)^{n-1}T^{\deg\Phi-1}v \end{matrix} $$ is a basis of $V$ and with respect to this basis, $T$ is \begin{align*} T&=\begin{pmatrix}C\\ U & C\\ & U & C\\ &&\ddots & \ddots\\ &&&U&C \end{pmatrix},\\ C&=\begin{pmatrix} &&&\ast\\ 1&&&\ast\\ &\ddots&&\vdots\\ &&1&\ast \end{pmatrix},\quad U=\begin{pmatrix} 0&\dots&0&1\\ &&&0\\ &&&\vdots\\ &&&0 \end{pmatrix} \end{align*} where $C$ is the companion matrix of $\Phi$. The effect of $\Phi(T)$ is to push the blocks down: $$ \Phi(T)= \begin{pmatrix} 0\\ I&0\\ &I&0\\ &&I&0\\ &&&\ddots&\ddots\\ &&&&I&0 \end{pmatrix}, \Phi(T)^2= \begin{pmatrix} 0\\ 0&0\\ I&0&0\\ &I&0&0\\ &&\ddots&\ddots&\ddots\\ &&&I&0&0 \end{pmatrix}, \dots $$ So $\dim\ker\Phi^m(T)=m\deg\Phi=\deg\Phi^m$.