Linear functional and transposition. This is a question from Hoffman.
Let $V$ be a finite-dimensional 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 $T^tf=cf$.
I know that essentially I need to prove that there exist an $f$ such that $fT(\beta)=cf(\beta)$ for all $\beta$ in $V$. 
What puzzles me is that there doesn't seem to be anymore information on the linear transformation $T$ other than the fact that there is a non-zero vector $\alpha$ in $V$ such that $T\alpha=c\alpha$; we have no idea of what we will get when we let $T$ operate on vectors other than $\alpha$ in $V$. 
It therefore seems that there is a certain $f$ that can ignore the effect of $T$ somehow,but I can't find that $f$.

Let $V$ be a space with dimension $n$, and let $U=T-cI$. so we have$$null(U)=n-Rank(U).$$and also for $U^t$ we have $$null(U^t)=n-Rank(U^t).$$because$$null(U^t)=null(U)$$ and also $U\alpha=T\alpha-c\alpha=0$,we have $null(U)\neq 0$. So that $null(U^t)\neq 0$ and there exists at least one nonzero linear functional $f$ such that $U^t f=0$ and also for all $\beta\in V$ we have \begin{align}U^t f(\beta)&=f(U(\beta))=f(T-cI)(\beta)\\
&=f(T\beta-c\beta)=f(T\beta)-cf(\beta)\\
&=T^tf(\beta)-cf(\beta)=0.
\end{align}
