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 $\dim (\textsf V)$.
Best Answer
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., $T^2 = T$).