Restriction of linear operator and nonsingularity I want to prove that $\mathcal{V} = \tau(\mathcal{V}) \bigoplus NS(\tau)$, if the restriction of $\tau$ to $\tau(\mathcal{V})$ is nonsingular. 
I have difficulty in comprehending the meaning of this statement. Since for the linear operator $\tau \in \mathcal{L}(\mathcal{V}, \mathcal{V})$, we have the following theorem: dim($\tau(\mathcal{V})$) + dim($NS(\tau)$) = dim($\mathcal{V}$).
Shouldn't this mean that dim($\tau(\mathcal{V}) \cap NS(\tau)$) = ${0}$, and therefore it proves the given statement?

Notice that the usual notations are:
$$\tau(\mathcal V)=\operatorname {Im}(\tau)\quad\text{and}\quad NS(\mathcal V)=\ker(\tau)$$
and as you said by the rank-nullity theorem we have
$$\dim(\tau(\mathcal{V})) + \dim(NS(\tau)) = \dim(\mathcal{V})\tag2$$
and by the hypothesis since the restriction of $\tau$ to $\tau(\mathcal V)$ is nonsingular then for every $x\in\tau(\mathcal V)\setminus\{0\},\;\; \tau(x)\ne0$ hence
$$\tau(\mathcal V)\cap NS(\tau)=\{0\}\tag 2$$
hence by $(1)$ and $(2)$ we have the desired result.
