Onto and one-to-one 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)

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).
