Prove if $0$ is the only eigenvalue of $T$ then $T$ is nilpotent I have seen a different answer to this so I would like to know if my approach is correct or not. If not why?
Suppose $V$ is a complex vector space, $T\in\mathcal{L}(V)$
Prove if $0$ is the only eigenvalue of $T$ then $T$ is nilpotent
Attempt: Suppose the only eigenvalue of $T$ is $0$.
Then $V$ has a basis $v_1,...,v_m$ of generalized eigenvectors in the generalized eigenspace for $0$.
Thus $$T^{\text{dim}V}(v_i)=0 \text{ for each } i=1,...,m$$
Since $v_1,...,v_m$ is a basis of $v$, $$T^{\text{dim}V}(v)=0$$
for each $v \in V$. So $T$ is nilpotent.
 A: 
Lemma 1 All eigenvalues = $0$ implies strict upper-triangularity

By induction. If $V = \mathbb{C}$ then $v \neq 0 \implies T(v)= 0 \implies T = 0 $. Assume it true for $1,...,n-1$ then $T$ has all eigenvalues equal to $0$ implies there exists a non-zero $v$ such that $T(v)= 0$. Complete $v$ up to a basis $v,w_1,...,w_{n-1}$ and consider the matrix of $T$ in this new basis. We see that the first column is zero and therefore $T= \begin{pmatrix}0 &T_{1,2}  & \dots & T_{1,n} \\ 0&T_{2,2}  & \dots & T_{2,n}  \\ \vdots & \vdots & \ddots & \vdots 
 \\ 0 &T_{n,2} & \dots &  T_{n,n}\end{pmatrix}$ in this new basis. Let $T'= \begin{pmatrix} T_{2,2}  & \dots & T_{2,n}  \\  \vdots & \ddots & \vdots 
 \\ T_{n,2} & \dots &  T_{n,n}\end{pmatrix}$, it is easy to see that $T$ has eigenvector $w = 0v+\alpha_1w_1+...+\alpha_{n-1}w_{n-1}$ if and only $T'$ has eigenvector $w$ in the subspace of dimension $n-1$ generated by the $w_i$ (for some eigenvalue $\lambda$). Therefore $T'$ has all eigenvalues $0$ and by induction $T'$ is strict upper-triangular in some basis. Perform a coordinate change to that basis.
The proof is completed by proving the lemma

Lemma 2 Strict upper-triangularity implies nilpotent 

By induction. Base case $n=1$ then $T= \begin{pmatrix}0\end{pmatrix}$ is nilpotent with exponent 1. Assume it true for $1,...,n-1$. Let $T= \begin{pmatrix}0 &T_{1,2}  & \dots & T_{1,n} \\ 0& 0  & \dots & T_{2,n}  \\ \vdots & \vdots & \ddots & \vdots 
 \\ 0 & 0  & \dots &  0 \end{pmatrix}$. It is easy to see that this is equivalent to the condition
\begin{equation} j\leq i \implies T_{i,j} =0  \end{equation}
and therefore  therefore that \begin{equation} j  \leq i + 1 \implies (\forall k) \  T_{i,k}T_{k,j} =0 \end{equation} by taking cases:


*

*$k\leq i \implies T_{i,k}T_{k,j} = 0 T_{k,j}= 0$

*$k>i  \implies k\geq i+1 \geq j  \implies  T_{i,k}T_{k,j} =T_{i,k} 0 = 0$
and therefore \begin{equation}j  \leq i + 1 \implies   T^2_{i,j} = \sum_kT_{i,k}T_{k,j} = 0 .\end{equation} Therefore we have that $T^2= \begin{pmatrix}0_{1 \times n-1} &T'  
 \\ 0 & 0_{n-1 \times 1}   \end{pmatrix}$ where $T'$ is strict upper-triangular. Let $k$ be the witness to the strict upper-triangularity, i.e. $(T')^{k}=0,$ then it is easy to see that $T^{2k} = 0$. QED
