From Friedberg's Linear Algebra
Let $T$ be a linear operator on a finite-dimensional inner product space $V$. If $T$ has an eigenvector, then so doess $T$*.
Proof. Suppose that $v$ is an eigenvector of $T$ with corresponding eigenvalue $\lambda$. Then for any $x \in V$,
$0 = \langle 0,x \rangle = \langle(T – \lambda\ I)(v),x \rangle = \langle v,(T-\lambda\ I)^*(x)\rangle = \langle v, (T^*-\bar{\lambda} I)(x)\rangle$,
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 correspoinding eigenvalue $\bar{\lambda}.$
I'm unable to see why $T^* -\bar{\lambda}I$ is not one-to-one, and why $T^* -\bar{\lambda}I$ has a nonzero null space.
Best Answer
Not onto: Assume it's onto, then exists $x'\in V$ s.t. $0=\langle v,(T^*-\bar\lambda I)(x')\rangle=\langle v,v\rangle,$ so $v=0,$ which contradicts $v\not=0.$
Not one-one: Since $T^*:V\to V, I_V:V\to V,$ clearly $(T^*-\bar\lambda I):V\to V,$ so $$\dim(V)=Nullity(T^*-\bar\lambda I)+Rank(T^*-\bar\lambda I),$$ but $Rank(T^*-\bar\lambda I)\lt\dim(V)$ since we just observed, so $Nullity(T^*-\bar\lambda I)\gt0$.