# Prove that if $\textsf V$ is finite-dimensional, then $\textsf V = \operatorname{im}(\textsf T) \oplus \ker(\textsf T)$

Given a linear transformation $$\textsf T : \textsf V \to \textsf V$$, suppose that $$\textsf V$$ is spanned by $$\operatorname{im} (\textsf T)$$ and $$\ker (\textsf T)$$. Prove that if $$\textsf V$$ is finite-dimensional, then $$\textsf V = \operatorname{im}(\textsf T) \oplus \ker(\textsf T)$$

I know that I need to show that $$\operatorname{im}(\textsf T) \cap \ker (\textsf T) = \{ 0 \}$$. Further, if $$\textsf T$$ is surjective the result follows directly from the rank-nullity theorem. However, I am stuck on the case where $$\textsf T$$ is not surjective.

This is not a duplicate of this question because here we are not raising $$\textsf T$$ to $$\dim (\textsf V)$$.

Define $$S: \text{Ker}(T) \times \text{Im}(T) \rightarrow V$$ as. $$S(v,w) = v + w$$. $$S$$ is well defined and, by hypothesis, sujective. By the rank-nullity theorem, $$\text{dim}[\text{Ker}(T)\times \text{Im}(T)] = \text{dim}\text{Ker}(S) + \text{dim}\text{Im}(S),$$ and thus $$S$$ is bijective.
Now, since every $$u \in V$$ is uniquely written as $$u=v+w$$, with $$v \in \text{Ker}(T)$$ and $$w \in \text{Im}(T)$$, we must have $$V = \text{Ker}(T) \oplus \text{Im}(T)$$.
Note: With this information, one can show that $$T$$ must be a projection (i.e., $$T^2 = T$$).