TITLE: Eigenvalues of adjoint operator [General Case]
QUESTION [5 upvotes]: I am asked to prove that if $T: V \rightarrow V$ is a linear operator over a complex inner product space $(V,\langle,\rangle)$, then $\overline{\lambda}$ is an eigenvalue of $T^*$ where $\lambda$ is an eigenvalue of $T$.
The problem is easily reduced to the cases where either $V$ is a finite dimensional space or $T$ is a normal operator, but I am stuck with the general case.

REPLY [16 votes]: As nicely pointed out by uniquesolution, we must assume that $V$ is finite dimensional. However, there is no need to use determinants:
Suppose $T$ is a linear operator on a finite-dimensional complex vector space $V$ and $\lambda \in \mathbf{C}$. Note that $\lambda$ is not an eigenvalue of $T$ if and only if $T - \lambda I$ in invertible, which happens if and only if there exists an operator $S$ on $V$ such that
$$
S(T - \lambda I) = (T - \lambda I)S = I.
$$
Taking adjoints of all three sides above shows that the equations above are equivalent to
$$
(T^* - \bar{\lambda} I)S^* = S^*(T^* - \bar{\lambda} I) = I.
$$
Thus we see that $T - \lambda I$ is invertible if and only if $T^* - \bar{\lambda} I$ is invertible. In other words, $\lambda$ is an eigenvalue of $T$ if and only if $\bar{\lambda}$ is an eigenvalue of $T^*$.