= Solution
First suppose $E=E^*$. For $u\in U=\operatorname{im}E$, choose $v$ with $Ev=u$. For $w\in W=\ker E$,
$$
\langle u,w\rangle=\langle Ev,w\rangle=\langle v,Ew\rangle=0.
$$
Thus $U\subseteq W^\perp$. The <direct sum of vector spaces> gives $\dim U=\dim V-\dim W$, while the <orthogonal complement> has $\dim W^\perp=\dim V-\dim W$. Equal finite dimensions imply $U=W^\perp$.
Conversely, suppose $U=W^\perp$. Decompose arbitrary vectors as $v=u+w$ and $v'=u'+w'$, with $u,u'\in U$ and $w,w'\in W$. Since $E$ is the indicated <projection>, $Ev=u$ and $Ev'=u'$. Orthogonality makes the cross terms vanish, so
$$
\langle Ev,v'\rangle=\langle u,u'\rangle=\langle v,Ev'\rangle.
$$
Therefore $E$ satisfies the defining identity of its own <adjoint operator>. By uniqueness from part (a), $E=E^*$. Hence
$$
\boxed{E=E^*\quad\Longleftrightarrow\quad U=W^\perp.}
$$
This is the characterization of an <orthogonal projection> among projections along a complementary subspace.
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