# $N$ is nilpotent and $T$ is diagonalizable transformation s.t. $N \circ T = T \circ N$. Show $\ker q_{\lambda} ( T + N ) = V_{\lambda}$

linear algebranilpotence

Problem: Let $$V$$ be finitely generated vector space over $$\mathbb{F}$$, let $$T: V \to V$$ be a diagonalizable linear transformation and $$N : V \to V$$ a nilpotent linear transformation and suppose that $$N \circ T = T \circ N$$.
For every eigenvalue $$\lambda$$ of $$T$$ we'll denote $$V_{\lambda}$$ as the eigenspace of $$T$$ that corresponds to $$\lambda$$. Let $$r$$ be the nilpotency index of $$N$$. Show that the polynomial $$q_\lambda = ( x – \lambda )^r \in \mathbb{F[x]}$$ satisfies
$$\ker q_{\lambda} ( T + N ) = V_{\lambda}$$ .

Attempt: ( I proved the other direction, I have left to prove the following one, $$\subseteq$$ )
Let $$\tau \in \ker q_{\lambda} ( T + N )$$ be arbitrary.
Meaning $$q_{\lambda} ( T + N ) \tau = ( T + N – \lambda \cdot I_d )^r \cdot \tau = 0$$ We'll show that $$T \tau = \lambda \cdot \tau$$ ( meaning that $$\tau \in V_{\lambda}$$ ).
$$( T + N – \lambda \cdot I_d )^r = ( T – \lambda \cdot I_d + N )^r$$.
Since $$N \circ T = T \circ N$$ Then, $$( T – \lambda \cdot I_d + N )^r \tau =( \sum_{k=0}^r { r \choose k } \cdot( T – \lambda \cdot I_d )^k \cdot N^{r-k} ) \tau = { r \choose 1 } ( T – \lambda \cdot I_d )N^{r-1} \tau + … + { r \choose r }( T – \lambda \cdot I_d )^r \tau = 0$$
[ Missing arguments ]

I don't know how to continue and use the fact that $$T$$ is diagonalizable. I used the fact that $$N$$ is nilpotent in the binomial expansion where $${ r \choose 0 }N^r \tau = 0$$ since $$N^r =0$$ but I'm not sure if that's all the usage of nilpotency I'll need since I can't continue. Can you please help?
( Thanks in advance! )

Since $$T$$ is diagonalizable, $$V$$ decomposes to a direct sum of eigenspaces with respect to $$T$$. You can further restrict $$N$$ to each eigenspace. $$N|_{V_{\lambda}}$$ are nilpotent, so you can find a basis of chains, with respect to $$N$$, for each eigenvalue $$\lambda$$. Note that each basis consists of eigenvectors with respect to $$T$$. Moreover, since $$V$$ is a direct sum of the eigenspaces, you can join the chain bases to get a basis for $$V$$.
Now for arbitrary eigenvectors $$v_{\mu}\in V_{\mu}$$ for each $$\mu\neq\lambda$$, look at the sum: $$\left(T+N-\lambda I\right)^{r}\sum_{\mu\neq\lambda}v_{\mu}=\sum_{\mu\neq\lambda}\left(T+N-\lambda I\right)^{r}v_{\mu}$$ Decompose every $$v_{\mu}$$ into a linear combination of the $$\mu$$-chain basis, and plug it into your binomial expression. Explain yourself why the total sum is non-zero (Here the fact that we picked chain bases comes into play).
Note that for $$v_{\lambda}\in V_{\lambda}$$ we get $$\left(T+N-\lambda I\right)^{r}v_{\lambda}=0$$
To finish, take an arbitrary $$v\in V$$ such that $$\left(T+N-\lambda I\right)^{r}v=0$$, decompose $$v$$ into a sum of eigenvectors and use what we showed to prove $$v$$ lies entirely in $$V_{\lambda}$$.