On proving $f\circ T=cf$ for a functional $f$ where $c$ is an eigevalue of $T$ Let $T: V\to V$ be a linear transformation on a finite dimensional vector space $V$ over a field $F$ and let $c\in F$ be such that  $T(v)=cv$ for some $0\ne v\in V$. 
Then how to show that there is a linear transformation (not identically zero) $f:V \to F$ such that $f\circ T=cf$ ? 
My try: Let $W$ be the eigenspace corresponding to eigenvalue $c$. Then one can extend a basis of $W$ to a basis of $V$ . I thought of defining $f$ to be $c$ at $v$ and zero at other basis vectors ... but unfortunately I that doesn't seem to work. 
Please help. 
 A: I assume that $f$ is a non-zero functional in this context.
Note that $\lambda$ is an eigenvalue of $T$ if and only if $(T - c I)$ fails to be invertible.  It suffices to prove that if $(T - \lambda I)$ fails to be invertible, then the adjoint map, i.e. the induced map $(T - c I)^*:V^* \to V^*$ defined by $f \mapsto f \circ (T - c I)$, also fails to be invertible. By contrapositive, this amounts to showing that if $(T - c I)^*$ is inverible, then $(T - c I)$ is invertible.
So, suppose that $c$ is not an eigenvalue of $T^*$, which is to say that there exists no non-zero $f \in V^*$ with $f \circ T = c f$, which is to say that $(T - c I)^*$ is invertible.  That is, there exists a map $\Gamma: V^* \to V^*$ such that $\Gamma \circ (T - c I)^* = I$, where $I$ denotes the identity map (in this case, the identity over $V^*$).  The identification of $V^{**}$ with $V$ allows us to state that there exists a unique map $G:V \to V$ such that $\Gamma = G^*$.  With that and using the fact that $(A \circ B)^* = B^* \circ A^*$, we have
$$
\Gamma \circ (T - c I)^* = I_{V^*} \implies\\
G^* \circ (T - cI)^* = (I_V)^* \implies\\
[(T - cI) \circ G]^* = (I_V)^* \implies\\
(T - cI) \circ G = I_V.
$$
Thus, $G$ is the inverse of $(T - cI)$, which means that $c$ is not an eigenvalue of $T$.
The conclusion follows.
