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Let $T$ be a linear operator on a finite dimensional inner product space $V$. If $T$ has an eigenvector, then so does $T^*$.
Proof. Suppose that $v$ is an eigenvector of $T$ with corresponding eigenvalue $\lambda$. Then for any $x \in V$,
$$
0 = \langle0,x\rangle = \langle(T-\lambda I)v,x\rangle
= \langle v, (T-\lambda I)^*x\rangle = \langle v,(T^*-\bar{\lambda} I)x\rangle
$$
This means that $(T^*-\bar{\lambda} I)$ is not onto. WHY?
(Of course the proof is not completed in here)
Assume $T^*-\overline{\lambda}I$ is onto and take $v$ such that $(T-\lambda I)v=0$.
By surjectivity, there exists $x\in V$ such that $(T^*-\overline{\lambda}I)x=v$. For this specific $x$, we get $$ 0=(v,(T^*-\overline{\lambda}I)x)=(v,v)=\|v\|^2\quad \Rightarrow \quad v=0. $$ Therefore $T^*-\overline{\lambda}I$ surjective implies $T-\lambda I$ injective.
By rank-nullity and symmetry, it follows that $T-\lambda I$ is not injective if and only if $T^*-\overline{\lambda}I$ is not injective. That is $\lambda $ is an eigenvalue for $T$ if and only if $\overline{\lambda}$ is an eigenvalue for $T^*$.
If you want to know why $T^{*}$ is not onto if $T$ is not onto, just observe that the matrix of $T^{*}$ is just the conjugate transpose of the matrix of $T$ and we know that a matrix and its conjugate transpose have the same rank (which is equal to the dimension of the range space).
