Linear functionals and dual spaces

Question

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 that there is a non-zero vector $\alpha$ such that $T\alpha=c\alpha$. Prove that there is a non zero linear functional $f$ on $V$ such that $T^{t}f=cf$.

My attempt:

Let $T\alpha_i=c\alpha_i$ for $1\le i\le n$. Let $\{\alpha_1,\alpha_2,\dots,\alpha_n\}$ be a basis of $V$ and $\{f_1,f_2,\dots,f_n\}$ be the corresponding dual basis. We choose $f_1$ as the required linear functional. Then $T^tf(u)=f(Tu)$ with $u=c_1\alpha_1+c_2\alpha_2+\dots+c_n\alpha_n$, so $$Tu=c_1T(\alpha_1)+c_2T(\alpha _2)+\dots+c_nT(\alpha_n)$$ $$f(Tu)=f(c_1T(\alpha_1)+c_2T(\alpha _2)+\dots+c_nT(\alpha_n))$$ $$=cc_1f(\alpha_1)=cf(u)$$

But this holds true only if $T$ maps the other basis elements $\alpha_2,\alpha _3,\dots,\alpha_n$ to vectors spanned by these basis elements excluding $\alpha_1$. Can anyone clarify?

Answer 1

Thanks to Daniel Fischer for pointing out that my first answer was incorrect.

OK so I have now a correct solution. Replace $T$ by $T-cI$ (just for simplicity and clarity) thus I assume that $c=0$. Now we want a functional $f$ so that $$T^tf=0$$ that means that $$T^tf(\beta)=f(T(\beta))=0$$ for all $\beta\in V$.

So we need $$\text{image} T\subseteq \ker f$$

Since $T(\alpha)=0$ we see that

$$\dim \text{image} T \leq \dim V-1$$ and therefore it is always possible to chose such an $f$.