= Solution
Use the convention of part (a): the <inner product> is linear in its first argument. If $Tv=\lambda v$ with $v\ne0$, the <adjoint operator> identity and $T^*=-T$ give
$$
\lambda\langle v,v\rangle=\langle Tv,v\rangle
=\langle v,T^*v\rangle=-\langle v,Tv\rangle
=-\overline\lambda\langle v,v\rangle.
$$
<Positive definiteness> makes $\langle v,v\rangle>0$, so
$$
\boxed{\lambda=-\overline\lambda,\qquad\operatorname{Re}\lambda=0.}
$$
Thus every <eigenvalue> of a <skew-adjoint operator> is purely imaginary, including zero.
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