Let $T: H \to H$ be a compact operator with $H$ a Hilbert space. Let then $\lambda \neq 0$ be an eigenvalue of $T$ with eigenfunction $v$.
- Is then $\lambda$ an eigenvalue for the adjoint $T^*$ either?
- Is then $v$ an eigenfunction for $T^*$?
I know the above statements fail for $\lambda = 0$ and the counterexample is given by $T: l^2 \to l^2$, $e_i \mapsto e_{i+1}/2^{i-1}$ which has no eigenvalues while its adjoint has the couple $\lambda = 0$, $v = e_1$.
I guess you are assuming that $\lambda$ is real (in general, the spectrum of $T^*$ consists exactly of the conjugates of the spectrum of $T$).
So, if $\lambda$ is real, then $\lambda$ is an eigenvalue of $T$ if and only if it is an eigenvalue of $T^*$ (because $T-\lambda I$ is invertible if and only if $(T-\lambda I)^*=T^*-\bar{\lambda}I)$ is invertible).
And, as Fabian, said, the eigenvectors of $T$ are usually not eigenvectors of $T^*$: let $T=\begin{bmatrix}1&1\\0&1\end{bmatrix}$; then $v=\begin{bmatrix}1\\0\end{bmatrix}$ is an eigenvector with eigenvalue 1, but $$ T^*v=\begin{bmatrix}1&0\\1&1\end{bmatrix}\begin{bmatrix}1\\0\end{bmatrix} =\begin{bmatrix}1\\1\end{bmatrix}, $$ so $v$ is not an eigenvector if $T^*$.
(note that $1$ is still an eigenvalue of $T^*$, with eigenvector $\begin{bmatrix}0\\1\end{bmatrix}$).