 Solution

For every $u\in U$, write $u=(u-\alpha u)+\alpha u$. The first term lies in $\ker \alpha$, since ${\alpha }^2=\alpha$, and the second lies in $\mathrm{im}\alpha$. If $v$ belongs to both spaces, write $v=\alpha w$; then $\alpha v={\alpha }^{2}w=\alpha w=v$, whereas membership in the kernel gives $\alpha v=0$. Thus the intersection is zero andIn a basis assembled from bases of these two spaces, the idempotent linear map has matrix $\mathrm{diag}(0_{d-r},I_r)$, where $d=\dim U$ and $r=\mathrm{rank}\alpha$. ThereforeThis basis consists entirely of eigenvectors, so $\alpha$ is diagonalizable over the reals, including the extreme cases $r=0$ and $r=d$.