= Solution
Induction from $M^2=M$ gives $M^j=M$ for every integer $j\geq1$. Separate the zeroth term of the <matrix exponential> series:
$$
\begin{aligned}
e^{\alpha M}
&=I+\sum_{j=1}^\infty\frac{\alpha^jM^j}{j!}
=I+M\sum_{j=1}^\infty\frac{\alpha^j}{j!}\\
&=I+(e^\alpha-1)M.
\end{aligned}
$$
Therefore the <exponential of an idempotent linear map> is
$$
\boxed{e^{\alpha M}=I+(e^\alpha-1)M=(I-M)+e^\alpha M.}
$$
The last form shows its action directly: it multiplies the <image of a linear map> of $M$ by $e^\alpha$ and fixes its <kernel of a linear map>. No Hermiticity assumption on the <idempotent matrix> is needed.
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