= 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 $\operatorname{im}\alpha$. If $v$ belongs to both spaces, write $v=\alpha w$; then $\alpha v=\alpha^2w=\alpha w=v$, whereas membership in the <kernel> gives $\alpha v=0$. Thus the intersection is zero and
$$
\boxed{U=\ker\alpha\oplus\operatorname{im}\alpha}.
$$
In a <basis> assembled from <bases> of these two spaces, the <idempotent linear map> has <matrix> $\operatorname{diag}(0_{d-r},I_r)$, where $d=\dim U$ and $r=\operatorname{rank}\alpha$. Therefore
$$
\boxed{\chi_\alpha(t)=t^{d-r}(t-1)^r}.
$$
This <basis> consists entirely of <eigenvectors>, so \b[$\alpha$ is diagonalizable over the reals], including the extreme cases $r=0$ and $r=d$.
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