Let $\dim(V) < \infty $ and suppose that $\tau \in \mathcal L (V)$ satisfies $\tau^2 = O $. Show that $2 rk(\tau ) \le dim (V) $

Question

Let $\dim(V) < \infty $ and suppose that $\tau \in \mathcal L (V) $ satisfies $\tau^2 = \mathcal O $ (Where $\mathcal O$ is the zero map):

show that $2 rk(\tau ) \le \dim (V) $.

I proved that but without using the condition "$\dim(V) < \infty $". I proved it as the follow:

By Rank Plus Nullity Theorem :

$$ \dim(V) = rk (\tau) + \operatorname{null}(\tau) $$ $ \operatorname{im}(\tau ) \subset \ker (\tau ) $. Indeed, if we take a $$w \in \operatorname{im}( \tau) \Rightarrow \exists v \in V : \tau (v)=w \Rightarrow 0 = \tau^2 (v) = \tau (w) \Rightarrow w \in \ker(\tau) $$

So $rk (\tau) \le \operatorname{null} (\tau) $ Then $$ \dim(V) \ge rk (\tau) + rk (\tau) =2 rk(\tau) $$

So I see like this is true for the vector spaces that have infinte dimension too. Is that true ?

This is the proof of The Rank Plus Nullity Theorem in the book which I study; " Advanced linear algebra by steven Roman":

The Rank Plus Nullity Theorem

Let $\tau\in\mathcal{L}(V,W)$. Since any subspace of$V$ has a complement, we can write $$V=\ker(\tau)\oplus\ker(\tau)^c$$

where ker$(\tau)^c$ is a complement of ker$(\tau)$ in $V.$ It follows that

$$\dim(V)=\dim(\ker(\tau))+\dim(\ker(\tau)^c)$$

Now, the restriction of $\tau$ to ker$(\tau)^c$

$$\tau^c{:}\ker(\tau)^c\to W$$

is injective, since

$$\ker(\tau^c)=\ker(\tau)\cap\ker(\tau)^c=\{0\}$$

Also, $\operatorname{im}(\tau^{c})\subseteq\operatorname{im}(\tau).$ For the reverse inclusion, if $\tau(v)\in\operatorname{im}(\tau)$ then since

$v=u+w$ for $u\in \ker(\tau)$ and $w\in \ker(\tau)^c$,we have

$$\tau(v)=\tau(u)+\tau(w)=\tau(w)=\tau^c(w)\in\mathrm{im}(\tau^c)$$

Thus $\operatorname{im}(\tau^c)=\operatorname{im}(\tau).$ It follows that

$$\ker(\tau)^c\approx\mathrm{im}(\tau)$$

From this, we deduce the following theorem.

$\textbf{Theorem 2. 8 Let }\tau \in \mathcal{L} ( V, W).$

1. Any complement of $\ker(\tau)$ is isomorphic to $\operatorname{im}(\tau )$

2. $( \textbf{The rank plus nullity theorem})$

$$\dim(\ker(\tau))+\dim(\operatorname{im}(\tau))=\dim(V)$$

or, in other notation,

$$\mathrm{rk}(\tau)+\mathrm{null}(\tau)=\dim(V)$$