$\newcommand{\lbrac}[1]{\left( #1 \right)}$ Let $V$ be a complex inner product space, and let $T:V\to V$ be a linear operator over $V$ and $T^*$ its adjoint. Suppose $\lambda$ is an eigenvalue of $T$. Prove that $\overline{\lambda}$ is an eigenvalue of $T^*$.
Let $\mathcal{B}$ be an orthonormal basis for $V$, and let $A$ be such that $A = [T]_{\mathcal{B}}^{\mathcal{B}}$. Since $\lambda$ is an eigenvalue of $T$, we know that $\det\lbrac{A-\lambda I} = 0$. For the same basis $\mathcal{B}$, the matrix representation of $T^*$ relative to $\mathcal{B}$ is $\overline{A^t}$. We get that $$ \det\lbrac{\overline{A^t} - \overline{\lambda} I} = \det\lbrac{\overline{A^t - \lambda I}} = \det\lbrac{\overline{\lbrac{A - \lambda I}^t}} $$ We also notice that $$ \left[\overline{A}^t\right]_{i,j} = \left[\overline{A}\right]_{j,i} = \overline{\left[A\right]}_{j,i} = \overline{\left[A^t\right]}_{i,j} $$ and therefore $$ \det\lbrac{\overline{\lbrac{A - \lambda I}^t}} = \det\lbrac{\overline{\lbrac{A - \lambda I}}^t} = \det{\lbrac{\overline{A-\lambda I}}} = \overline{\det{\lbrac{A-\lambda I}}} $$ Since $\lambda$ is an eigenvalue of $A$, the latter is zero, and therefore, $\overline{\lambda}$ is a root of the characteristic polynomial of $\overline{A^t}$, and therefore, $\overline{\lambda}$ is an eigenvalue of $T^*$.
Is this proof OK? Plus, is there a way to prove this using the fact that for a given subspace $W$, $W$ is $T$-invariant iff $W^{\perp}$ is $T^*$-invariant?