$H$ be a complex Hilbert space , $T: H \to H$ be a compact operator , if $c$ is a eigen-value of $T$ , is $\bar c$ an eigenvalue of $T^*$?

Let $H$ be a complex Hilbert space , $T: H \to H$ be a compact operator , if $c$ is an eigen-value of $T$ , then is $\bar c$ an eigenvalue of $T^*$ ?

Not necessarily. Consider the "damped" left shift operator $L'$ on $l^2(\mathbb{N})$, which we can define as $$L' = \sum_{n=1}^{\infty} \lambda_n\langle e_{n+1}, \cdot\rangle e_n$$ where $\lambda_n > 0$ decreases to $0$ (we define $e_i = \{\delta_{ni}\}_{n=1}^{\infty}$ to be the standard orthonormal basis for $l^2(\mathbb{N})$). As opposed to the typical left shift operator, $L'$ is compact. Any sequence $x = \{x_n\}_{n=1}^{\infty}\in l^2(\mathbb{N})$ such that $x_1\neq 0$ and $x_n = 0$ for $n\geq 2$ is an eigenvector of $L'$ with eigenvalue $0$. Now, we will show that $0$ is not an eigenvalue of $(L')^*$. We note that $$(L')^* = \sum_{n=1}^{\infty} \lambda_n\langle e_n, \cdot\rangle e_{n+1}$$ We can see that for $(L')^*x = 0$ (where $x = \{x_n\}_{n=1}^{\infty}$), we must have $x_n = 0$ for all $n\in \mathbb{N}$. Therefore, $0$ is not an eigenvalue of $(L')^*$.
• Don't we need $\sum |\lambda_n|$ to be convergent for $L$ to be compact ? Anyway thanks for the answer ; do you know a counterexample with non-zero eigenvalue ? – user228169 May 1 '17 at 8:29
• I don't think that's required. I'm pretty sure that the partial sums are norm-convergent to $L'$, which implies that $L'$ is the norm limit of a sequence of finite-rank operators (which are consequently compact) and is therefore compact. Check Theorem 12 here. – Michael Lee May 1 '17 at 8:33
• I think that it might be true for $c\neq 0$. If you use the singular value decomposition again with one of your terms as $c\langle e_n, \cdot\rangle e_n$ for unit eigenvector $e_n$, then the SVD of the adjoint will clearly have a term $\overline{c}\langle e_n, \cdot\rangle e_n$. – Michael Lee May 1 '17 at 8:59