[proofplan] We decompose each vector $v \in V$ as $(v - \alpha(v)) + \alpha(v)$ and show the first summand lies in $\ker(\alpha)$ and the second in $\mathrm{im}(\alpha)$. Then we show the intersection is trivial using $\alpha^2 = \alpha$. Finally, we verify that $\alpha$ acts as the identity on $\mathrm{im}(\alpha)$, confirming it is the projection. [/proofplan] [step:Show $V = \ker(\alpha) + \mathrm{im}(\alpha)$ by decomposing each vector] For any $v \in V$, write $v = (v - \alpha(v)) + \alpha(v)$. The second summand $\alpha(v) \in \mathrm{im}(\alpha)$ by definition. For the first summand, compute: \begin{align*} \alpha(v - \alpha(v)) = \alpha(v) - \alpha^2(v) = \alpha(v) - \alpha(v) = \mathbf{0}, \end{align*} using $\alpha^2 = \alpha$. So $v - \alpha(v) \in \ker(\alpha)$. [/step] [step:Show $\ker(\alpha) \cap \mathrm{im}(\alpha) = \{\mathbf{0}\}$ using idempotency] Suppose $v \in \ker(\alpha) \cap \mathrm{im}(\alpha)$. Then $v = \alpha(u)$ for some $u \in V$, and $\alpha(v) = \mathbf{0}$. Combining: \begin{align*} \mathbf{0} = \alpha(v) = \alpha(\alpha(u)) = \alpha^2(u) = \alpha(u) = v. \end{align*} So $v = \mathbf{0}$, and the intersection is trivial. Therefore $V = \ker(\alpha) \oplus \mathrm{im}(\alpha)$. [/step] [step:Verify $\alpha$ is the projection onto $\mathrm{im}(\alpha)$ along $\ker(\alpha)$] For any $w \in \mathrm{im}(\alpha)$, write $w = \alpha(u)$ for some $u$. Then $\alpha(w) = \alpha^2(u) = \alpha(u) = w$. So $\alpha|_{\mathrm{im}(\alpha)} = \mathrm{id}_{\mathrm{im}(\alpha)}$. For any $v \in \ker(\alpha)$, $\alpha(v) = \mathbf{0}$. So $\alpha$ acts as the identity on $\mathrm{im}(\alpha)$ and as zero on $\ker(\alpha)$: it is the projection onto $\mathrm{im}(\alpha)$ along $\ker(\alpha)$. [/step]