$ N$ is nilpotent and $ T $ is diagonalizable transformation s.t. $ N \circ T = T \circ N $. Show $ \ker q_{\lambda} ( T + N ) = V_{\lambda} $ 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! )
 A: I will try to describe the essence of what I think.
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 $V_{\lambda}$.
