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

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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
up vote 3 down vote accepted

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 $\begin{bmatrix}0\\1\end{bmatrix}$).

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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

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