Eigenvalue of an operator implies eigenvalue of the dual?

Question

I helped some students today with linear algebra, which I took last year. They asked me a question from their homework to which I couldn't find an answer:

Let $V$ be a finitely generated vector space over $\mathbb F$. We are given an operator $T\in End(V)$, such that it has some nontrivial eigenvector $v$ with some eigenvalue $c\in\mathbb F$. Show that exists a functional $l\in V^*$ such that $T^*(l)=cl$ (when $T^*:V^* \rightarrow V^*$ is defined by $T^* (f)=f\circ T$).

I tried first to complete $\{v\}$ to a basis of $V$, induction, and some more possible paths, but none of them seem to work, because even in the two-dimensional scenario, I still fail to see the property which $f$ must fulfill. What am I missing?

Answer 1

Fix a basis $B$ of $V$. Asserting that $c$ is an eigenvalue of $T$ is equivalent to the assertion that $c$ is a root of the characteristic polynomial of the matrix $A=[T]_B^B$. Now, let $B^*$ be the dual basis of $B$. Then $[T^*]_{B^*}^{B^*}=A^T$ and $A$ and $A^T$ have the same characteristic polynomials.