A question about Linear Transformations from Golan's Linear Algebra book I have found the following problem from Golan's Linear Algebra.. book.

Let $\alpha,\beta:V \to W$ be two linear transformations between two vector spaces $V$ and $W$ defined over the same field $F$. If for each $v\in V$ there exists a scalar $c_v\in F$ (depending on $v$) such that $\alpha(v)=c_v \beta(v)$, then prove that there exists a scalar $c\in F$ such that $\alpha=c \beta$.

I first tried to define the map $f:V \to F$ by $f(v)=c_v$ and I guessed that this map will be linear, but I found that the map fails to be linear. My next attempt is to show that this map $f$ is constant but I am unable to justify this. I am confused how to approach this problem. A slight hint is required.
 A: This problem can be broken into two steps.
The first step is to define a linear operator $\phi$ on the subspace $\def\Im{\operatorname{Im}}W'=\Im(\beta)\subseteq W$. which can be done as follows. Clearly the requirement $\alpha(v)=c_v \beta(v)$ for all $v$ implies $\ker\beta\subseteq\ker\alpha$, so $\alpha$ factors through a unique map $\overline\alpha:V/\ker\beta\to W$. Also $\Im(\alpha)\subseteq\Im(\beta)=W'$, so one can define $\phi$ to be the composite map $W'\to V/\ker\beta\to W'\subseteq W$ where the first is the (inverse of the) isomorphism given by the (first?) isomorphism theorem, and the second map is given by$~\overline\alpha$.
Now the hypothesis gives that all nonzero vectors of $W'$ are eigenvectors of$~\phi$ (to be precise $\beta(v)$ is eigenvector for eigenvalue$~c_v$), and the second step is to show that this implies that $\phi$ is a multiple of the identity (that is, all the $c_v$ are equal). This step is dealt with here. One easy way to see it is that using the general fact that a sum of distinct eigenspaces is always direct; if there were at least two (nonzero) eigenspaces then their sum would contain non-eigenvectors. So there is at most one eigenspace, and all of $W'$ must be that eigenspace.
A: Here's a way to handle the problem: 
First, consider the case that $\beta = 0$.  Then, $\alpha = 0$ must hold trivially.  So, we assume that this is not the case.
Then, there exists a basis $\{v_j\}_{j \in J}$ of the kernel of $\beta$.  Extend this to a basis $\{v_j\}_{j \in I}$ of $V$, so that $J \subsetneq I$.  We note that $\{\beta(v_j)\}_{j \in I \setminus J}$ forms a basis of the image of $\beta$.
Define $c_j = c_{v_j}$.  By the above argument, $\alpha$ and $\beta$ are both $0$ on $\ker(\beta)$.  So, fix a $k \in I \setminus J$.  We note that for $j \in I \setminus J$,
$$
\alpha(v_j + v_k) = c_j \beta v_j + c_k \beta v_k = c_j(\beta (v_j + v_k)) + (c_k - c_j) \beta v_k
$$
However, by our assumption, $\alpha(v_j + v_k)$ is a multiple of $\beta(v_j + v_k) = \beta v_j + \beta v_k$.  This necessarily implies that $(c_k - c_j) = 0$.
Moreover, if $j \in J$, we have
$$
\alpha(v_j) = c_j \beta v_j = c_j 0 = 0 = c_k 0 = c_k \beta v_j
$$
Thus, for any $j \in I$, $\alpha(v_j) = c_k \beta(v_j)$.  Defining $c = c_k$, we note that for all $j \in I$, $\alpha(v_j) = c \beta(v_j)$.
Thus, $(\alpha - c\beta)v_j = 0$ for any $j \in I$.  Since a linear transformation is determined by how it acts on a basis, we conclude that $\alpha - c \beta = 0$.  That is, $\alpha = c\beta$, as desired.
