If $\lambda$ is an eigenvalue of a compact operator $T: \mathcal{H} \to \mathcal{H}$, then $\bar{\lambda}$ is an eigenvalue of $T^{\ast}$

Question

I tried to manipulate the kernel of $T - \lambda$ and tried to manipulate inner products of things. But still don't know how to use ``compactness" of $T$ to prove this fact.

Can someone give me a hint?

Answer 1

Remark that if $\lambda$ is a non zero element of the spectrum of $T$ it is equivalent to $\lambda$ is an eigenvalue, and it is equivalent to $T-\lambda$ is not invertible which is equivalent to $(T-\lambda)^*=T^*-\bar\lambda$ is not invertible, we deduce that $\bar\lambda$ is an element of the spectrum of $T^*$ thus $\bar\lambda$ is an eigenvalue of $T^*$ since $T^*$ is also compact.

https://en.wikipedia.org/wiki/Spectral_theory_of_compact_operators