If $T\alpha=c\alpha$, then there is a non-zero linear functional $f$ on $V$ such that $T^{t}f=cf$ I am self-studying Hoffman and Kunze's book Linear Algebra. This is exercise $4$ from page $115$. It is in the section of The transpose of a Linear Transformation.

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 $T^{t}f=cf.$ ($T^{t}$ is the transpose of $T$.)

I tried to solve this question by induction on $\dim V$. I was able to show the base case, that is, when $\dim V=1$, but I got stuck in the  inductive step. If $\alpha $ is a non-zero vector, then we can find a base $\mathcal{B}=\{\alpha,\alpha_{1},\ldots\alpha_{m}\}$ of $V$. We can write $V=W_{1}\oplus W_{2}$, where $W_{1}=\langle \alpha \rangle$ and $W_{2}=\langle \alpha_{1},\ldots,\alpha_{m}\rangle.$ I can not show that $T(W_{2})\subset W_{2}.$ Anyway, $\alpha \notin W_{2}.$
EDIT: If $T$ is a linear transformation from $V$ to $W$, then $T^{t}$ is the linear transformation from $W^{\star}$ into $V^{\star}$ such that $$(T^{t}f)(\alpha)=g(T\alpha)$$ for every $f\in W^{\star}$ and $\alpha \in V$.
 A: Here's a basis-free construction. 
If $T \alpha = c \alpha$, for every $f \in V^*$ we have $(T^t f - c f)(\alpha) = f(T \alpha - c \alpha) = 0$.  Since $\alpha \ne 0$, there is some $g \in V^*$ with $g(\alpha) \ne 0$.  Thus the range of $T^t - c I$ is not all of $V^*$, which, since $V^*$ is a finite-dimensional space, means the kernel of $T^t - c I$ is not $\{0\}$, i.e. there is $f \in V^*$ with
$T^t f - c f = 0$.
Another way of putting it: any linear operator between finite-dimensional vector spaces has the same rank as its transpose (e.g. because $\text{Ker}(T) = \left(\text{Ran} T^t\right)^\perp$).  So $\text{Ker}(T - c I) \ne \{0\}$ iff $\text{rank}(T-cI) &lt \dim V$
iff $\text{rank}(T^t-cI) &lt \dim V^*$ iff $\text{Ker}(T^t - cI) \ne \{0\}$. 
A: Consider the orthogonal decomposition $V=\langle\alpha\rangle \oplus U\oplus\ker T\oplus W$, with $W=\big(T^{-1}\langle\alpha\rangle\big)^\perp$. Find a basis $\mathcal{B}_U=\{u_1,\cdots,u_k\}$ for $U$ such that $T(u_i)=c\alpha$ for each $i=1,2,\cdots,k$ (take an arbitrary basis and normalize). A desired linear map $f$ sends $\alpha\mapsto 1$, each $u_i\mapsto 1$ and all of $\ker T$ and $W$ to $0$.
