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

Question

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 $\dim (\textsf V)$.

Apply rank-nullity to the map $\text{ker},T\oplus\text{im},T\to V$. — Angina Seng, Jul 26 '19 at 15:26
@LordSharktheUnknown I was in a hurry and didn't have time to seek through other best possible duplicates. But here the correct one: https://math.stackexchange.com/questions/1861598/is-it-true-that-if-t-is-a-linear-operator-on-a-finite-dimensional-vector-space?rq=1 — Kumar, Jul 27 '19 at 13:47

score 1 · Accepted Answer · edited Jul 26 '19 at 23:59

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., $T^2 = T$).

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

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