## Idempotent linear transformations are precisely subspace projections

Let $V$ be an $n$-dimensional vector space over a field $F$. Let $\varphi : V \rightarrow V$ be a linear transformation, and suppose $\varphi^2 = \varphi$.

1. Prove that $\mathsf{im}\ \varphi \cap \mathsf{ker}\ \varphi = 0$.
2. Prove that $V = \mathsf{im}\ \varphi \oplus \mathsf{ker}\ \varphi$.
3. Prove that there is a basis $B$ of $V$ such that $M^B_B(\varphi)$ is diagonal, with only 1 and 0 as diagonal entries.

1. Suppose $x \in \mathsf{im}\ \varphi \cap \mathsf{ker}\ \varphi$; say $x = \varphi(y)$. Now $0 = \varphi(x) = \varphi^2(y)$ $= \varphi(y) = x$.
2. Let $v \in V$. Now $\varphi(v - \varphi(v)) = \varphi(v) = \varphi^2(v)$ $= \varphi(v) - \varphi(v)$ $= 0$, so that $v - \varphi(v) \in \mathsf{ker}\ \varphi$. That is, $v = \varphi(v) + (v - \varphi(v)) \in \mathsf{im}\ \varphi + \mathsf{ker}\ \varphi$. Using part (1), we have $V = \mathsf{im}\ \varphi \oplus \mathsf{ker}\ \varphi$.
3. Choose a basis $E_I$ for $\mathsf{im}\ \varphi$ and a basis $E_K$ for $\mathsf{ker}\ \varphi$. Using this previous exercise, $B = E_I \cup E_K$ is an ordered basis for $V$, and moreover $M^B_B(\varphi) = \left[ \begin{array}{c|c} M^{E_I}_{E_I}(\varphi) & 0 \\ \hline 0 & M^{E_K}_{E_K}(\varphi) \end{array} \right]$. (Where we restrict $\varphi$ as necessary). Clearly this matrix is diagonal, and moreover, the upper left block is $I$ while the lower right block is 0.