A special case of a nilpotent linear operator Let $V$ be a vector space of dimension $n$ over a field $F$ and let $T:V\rightarrow V$ be a non-zero linear operator s.t. $T\cdot T=0$. Assume $\dim \mathcal{R}(T)=r$ and $W$ is a subspace of $V$ s.t. $V=\mathcal{N}(T)\oplus W$. Here $\mathcal{R}(T)$ and $\mathcal{N}(T)$ denote the range (or image) and the nullspace (or kernel) of $T$, respectively.
I want to show the following 


*

*$2r\leq n$

*If $\{\mathbf{w}_1,\dots,\mathbf{w}_r\}$ is a basis of $W$ then
$\{T(\mathbf{w}_1),\dots,T(\mathbf{w}_r)\}$ is a linearly independent
subset of $\mathcal{N}(T)$


Any ideas?
 A: Consider a basis for $\mathcal{N}(T)$, $W$. Rank-nullity tells us that $\dim \mathcal{N}(T)=n-r$, $\dim W=r$, so we can say this is some $\{v_1,\ldots,v_{n-r},w_1,\ldots, w_r\}$. The images of the $\{w_i\}$, $\{Tw_i\}$, are a linearly independent subset of $V$, and they all are in the kernel, as $T^2=0$. So $\text{span}\{Tw_i\}$ is a vector subspace of $\mathcal{N}(T)$, so $r\le n-r$. So $2r\le n$.
The linear independence comes from the fact that the $\{w_i\}$ form a basis for the complement of the kernel. Assume that $\sum_{i=1}^r c_iTw_i=0$. Then $\sum_{i=1}^r T(c_iw_i)=0$, so $T(\sum_{i=1}^r c_iw_i)=0$. Then $\sum_{i=1}^r c_iw_i\in\mathcal{N}(T)$, but the direct sum decomposition says that $W\cap \mathcal{N}(T)=\{0\}$. So in fact $\sum c_iw_i=0$, and since the $w_i$ are a basis for $W$, each of the $c_i$ must be 0.
A: Since $T \ne 0$ and $T^2 = 0$ we can deduce that the minimal polynomial of $T$ is equal to $\mu_T(x) = x^2$. Therefore, $T$ has Jordan blocks of size $1$ and $2$, all pertaining to the only eigenvalue $0$.
On the other hand, $\dim \mathcal{N}(T) = n - \dim\mathcal{R}(T) = n-r$, which is the number of Jordan blocks.
Assume there are $k$ blocks of size $1$. Then there are $n-r-k$ blocks of size $2$. The sum of sizes of all blocks must be $n$ so
$$n = k + 2(n-r-k) = 2n-2r - k \implies k = n-  2r$$
Since trivially $k \ge 0$, we conclude $n \ge 2r$.

Assume that $\alpha_1 Tw_1 + \cdots + \alpha_rTw_r= 0$ for some scalars $\alpha_1, \ldots, \alpha_r$. We have:
$$0 = \sum_{i=1}^r \alpha_i Tw_i = T\left(\sum_{i=1}^r \alpha_i w_i\right) \implies \sum_{i=1}^r \alpha_i w_i \in W \cap \mathcal{N}(T) = \{0\}$$
So $\sum_{i=1}^r \alpha_i w_i = 0$. Now linear independence of $\{w_1, \ldots, w_r\}$ implies $\alpha_1 = \cdots = \alpha_r = 0$. 
