$H$ 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 $T^*$?

Question

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 $T^*$ ?

Answer 1

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 $(L')^*$.