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 $x \in V,$
\begin{align}
0&= \langle 0, x \rangle\\
&= \langle (T-\lambda I)(v),x \rangle\\
&= \langle v, (T-\lambda I)^{*}(x) \rangle\\
&= \langle v, (T^{*}-\bar{\lambda} I)(x) \rangle,
\end{align}
and hence $v$ is orthogonal to the range of $T^{*}-\bar{\lambda}I$. So $T^{*}-\bar{\lambda}I$ is not onto and hence is not one-to-one. Thus $T^{*}-\bar{\lambda}I$ has a nonzero null space, and any nonzero vector in this null space is an eigenvector of $T^{*}$ with corresponding eigenvalue $\bar{\lambda}$.
I don't follow the proof very well, especially on the onto and nullspace part. Can someone explain?
Best Answer
The computations shows that there are a nonzero vector $v \in V$ which orthogonal to every vector in the image of $U := T^*-\overline{\lambda}I$, right?
If $U$ was onto, then the image of $U$ is the whole $V$, but the unique vector which is orthogonal to every vector in $V$ is the zero vector, and $v$ is nonzero, you see? Hence, $U$ cannot be onto.
Now, since $U$ is not onto, it follows that $U$ is not one-to-one (in finite dimensions, they are equivalent for linear operators) and hence there are a nonzero vector $w \in V$ such that $U(w) = 0$ (if not, then the only vector such that $U(x)=0$ is $x=0$, and this implies that $U$ is one-to-one).
This $w$ is an eigenvector of $T^*$ with eigenvalue $\overline{\lambda}$, done.