Let $V$ be a finite-dimensional vector space over the field $\mathbb{F}$ 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 $T^{t}f=cf.$
Here $T^{t}$ is the transpose of $T$.
This question has been asked before but it seems like I have a false proof for this problem.
Suppose for all $f\in V^{*}, T^tf\ne cf$ .i.e, $T^tf(\beta)\ne cf(\beta)$ for all $\beta \in V.$
In particular $T^tf(\alpha)\ne cf(\alpha)\implies fT(\alpha)\ne cf(\alpha)\implies f(c\alpha)\ne cf(\alpha),$ a contradiction!
This proof looks legit to me but I haven't used any tools mentioned in the section and looking at other proofs I hardly doubt if mine indeed is a proof. What is the flaw here?
What you write after "i.e." is false. In particular, $T^{t}f(0) = cf(0)$ no matter what you do. Instead, $T^{t}f \neq cf$ if and only if there exists some $\beta \in V$ such that $T^{t}f(\beta) \neq cf(\beta)$.