Hoffman's book question: the transpose of a linear transformation Please, I need a hint to solve this question.
Let $V$ be the vector space over the field $F$ and let $T$ be a linear operator on $V$.Let $c$ be a scalar and suppose there is a non-zero vector $\alpha$ in $V$ such that $T \alpha = c \alpha $. Prove that there is a non-zero linear functional $f$ on $V$ such that $T^tf=cf$.
 A: Hint: What is the relation between the eigenvalues of $T: V \rightarrow V$ and the eigenvalues of $T^*:V^*\rightarrow V^*$?
(Edit: more details so that this is slightly more educational.) 
Eigenvalues have a lot to do with this problem. The statement that there exists $c$ and nonzero $v$ such that $Tv = cv$ is the same as saying that $c$ is an eigenvalue. Similarly, the problem is essentially asking you to show that $c$ is also an eigenvalue of $T^*$. In finite dimensions, recall that $\det A$ = $\det A^*$. Also it may be checked that
$$
(T-xI)^* = T^* - \overline{x}I
$$
where on the RHS, $I$ denotes the identity on $V^*$. This tells us that if $c$ is a solution to $$\det T-xI = 0,$$ then $\overline{c}$ is a solution to $$\det T^*-xI = 0.$$
So we get that $\overline{c}$ is an eigenvalue of $T^*$. In the real case we are finished. In the complex case (Hoffman/Kunze always considers subfields of the complex numbers IIRC), we first find a nonzero linear functional $f_0$ such that $T^* f_0 = \overline{c} f_0$. Then, taking the complex conjugate of $f_0$ yields an eigenvector for $c$.
