Prove that if $ T^2 = 0 $, then $ \operatorname{Im} T \subseteq \operatorname{Ker}T $

Question

The exercise is as follows:

Let $ V $ be a linear space of dimension $ n $ and let $ T : V \rightarrow V $ be a linear transformation such that $ T^2 = 0 $. Prove that $\operatorname{Im}T \subseteq \ker T $ and $\dim\ker T \ge \frac{n}{2} $. Is my proof valid?

Let $ u \in\operatorname{Im}T $. Therefore, there exists $ v \in V $ such that $ T(v) = u $. $ T^2 = 0 \Rightarrow T^2(v) = 0 \Rightarrow T(T(v)) = 0 \Rightarrow T(u) = 0 \Rightarrow u \in \ker T$. And therefore, we get $\operatorname{Im}T \subseteq \ker T $.

Suppose $\dim\ker T < \frac{n}{2} $. $\operatorname{Im}T \subseteq \ker T $ and therefore $\dim\operatorname{Im}T \le\dim\ker T < \frac{n}{2} \Rightarrow\dim\operatorname{Im}T + \dim\ker T \le n $. But $\dim\operatorname{Im}T + \dim\ker T = n $ - contradiction.

Answer 1

From the first relation (in second part) $\dim\mathrm{Im}(T)+\dim\ker{T}<n$ not $\le$ and that's contradiction. The other steps are valid.