Problem: Let $V$ be a vector space over $\mathbb{R} $ (any field can be taken) and $T:V\to V$ be linear. Let $T\circ T = T$. Show that \[ V = \ker(T) \oplus \operatorname{im} (T). \]
Solution: Let $v\in V$. Note that \begin{align*} T\circ T = T & \implies T(T(v)) = T(v) \\ & \implies T(T(v)) - T(v) = 0 \\ & \implies T \left( T(v)-v \right) = 0. \end{align*} Thus, $T(v)-v\in \ker(T)$. So, we have \[ v = (T(v) - v) + T(v) \implies V = \ker(T) + \operatorname{im}(T). \]
To show it is a direct sum, we need to show that $\ker(T)\cap \mathop{\mathrm{im}}(T) = \{0\}$. Let $ w \in \ker(T)\cap \mathop{\mathrm{im}}(T)$. Then, \[ w \in \ker (T) \implies t(W) = 0. \] Since, $w\in \mathop{\mathrm{im}}(T) $ so there exists $v\in V$ such that $T(v) = w$. Now note that \[ T(w) = 0 \implies T(T(v)) = 0 \implies T(v) = 0 \implies w =0. \] Hence, we showed that \[ \textcolor{blue}{\boxed{V = \ker(t) \oplus \mathop{\mathrm{im}}(T) }} \]