# Problem proving: $V = \ker T \oplus \operatorname{im}T$

Let $V$ be a finite linear subspace and $T$ be a linear transformation defined like this: $T:V \to V$ such that $\ker T^2 \subseteq \ker T$

Prove that: $V = \ker T \oplus \operatorname{im}T$

What I did is:

It's known that: $V = \operatorname{im}(T) + \ker(T)$ so all I need to prove is that:

$$\operatorname{im}(T) \cap \ker(T) = \{0\}$$

So I said that because $V$ is finite:

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

According to the dimensions theorem:

\begin{align*} \dim(\operatorname{im}(T) + \ker(T)) &= \dim(\operatorname{im}(T)) + \dim(\ker(T)) - \dim(\operatorname{im}(T) \cap \ker(T))\\\\ \dim(\operatorname{im}(T) \cap \ker(T)) &= \dim(\operatorname{im}(T)) + \dim(\ker(T)) - \dim(\operatorname{im}(T) + \ker(T)) \\\\ \dim(\operatorname{im}(T) \cap \ker(T)) &= 0 \end{align*}

But for some reason I didn't use the fact that $\ker T^2 \subseteq \ker T$, so I must have been wrong here.

• Why do you think you know that $V=\ker(T)+\mathrm{im}(T)$? – Zev Chonoles Jul 17 '15 at 16:25
• @Omnomnomnom: I'm pretty sure it's your inclusion that always holds: $T(x) = 0 \implies T(T(x)) = 0$ – john Jul 17 '15 at 16:30