# Eigenvalues of compact operators and his adjoint.

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 $\lambda = 0$, $v = e_1$.

-
That $\nu$ is a eigenfunction of $T^*$ fails even for finite dimensional vector spaces. –  Fabian Feb 8 '12 at 18:50
you're right the second point was silly asking. –  Giorgio Feb 8 '12 at 19:11
The result on eigenvalues is essentially the Fredholm alternative (assuming $\lambda$ is real), because $T^*$ is also compact. (Of course it doesn't hold in general when $T$ is noncompact.) –  Jonas Meyer Feb 19 '12 at 3:47

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 $\begin{bmatrix}0\\1\end{bmatrix}$).
I want to mention what wasn't stated explicitly: The fact that $T^*-\lambda I$ not being invertible and $\lambda\neq 0$ implies that $\lambda$ is an eigenvalue for $T^*$ is where compactness of $T^*$ is used. This is sometimes called the Fredholm alternative. –  Jonas Meyer Feb 19 '12 at 3:48