Eigenvalues of adjoint operator [General Case] 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.
 A: 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 $T^*$.
A: The result is true if $V$ is finite-dimensional, because in that case, $\lambda\in\mathbb{C}$ is an eigenvalue of $T$ if and only if $\det(T-\lambda I)=0$, the determinant of the adjoint is the complex conjugate determinant:
$$\det(T-\lambda I)=\overline{\det(T^*-\overline{\lambda}I)}$$
so $\det(T^*-\overline{\lambda}I)=0$, whence $\overline{\lambda}$ is an eigenvalue of $T^*$.
If $V$ is infinite dimensional, the result no longer holds. Take for example $V=l_2$, an infinite dimensional Hilbert space over the complex numbers, and let
$$T(a_1,a_2,a_3,\dots )=(a_2,a_3,\dots)$$
$T$ is the left shift, and every $\lambda$ such that $|\lambda| <1$ is
an eigenvalue of $T$ (just pick any $a_1\neq 0$ and take $v=(a_1,\lambda a_1,\lambda^2 a_1,\dots)$). The adjoint of the left shift is the right shift
$$T^*(a_1,a_2,a_3,\dots)=(0,a_1,a_2,\dots)$$
which has no eigenvalues at all.
