Exercise on a subspace of $\operatorname{Hom}(W,Y)$ The following problem is exercise 170 in Golan's linear algebra book. I have posted a solution attempt in the answers.
Problem: Let $V$, $W$, and $Y$ be vector spaces finitely-generated over a field $F$ and let $\alpha:V\rightarrow W$ be a linear transformation. Show that the set of all linear transformations $\beta: W\rightarrow Y$ satisfying the condition that $\beta\circ \alpha$ is the 0-transformation is a subspace of $\operatorname{Hom}(W,Y)$, and compute its dimension.
 A: Let $V$, $W$, and $Y$ have dimensions $x$, $y$, and $z$. Suppose $\beta_1$ and $\beta_2$ are two linear transformations satisfying the given condition. Then 
$\alpha\circ(\beta_1(w)+\beta_2(w))=\alpha\circ\beta_1(w)+\alpha\circ\beta_2(w)=0$
and $\alpha\circ(c\beta_1)=c(\alpha\circ\beta)=0$, so the specified set of $\beta$ transformations is a linear subspace. 
We know that $\dim\ker\alpha+\dim \text{im } \alpha= x$. Consider some basis $\{b_i\}$ for $\text{im } \alpha$. We can extending this to a basis for $W$. All of the basis vectors for $\text{im } \alpha$ need to be mapped to 0, but we are free to choose what the remaining vectors get mapped too. That means we need the dimension of $Hom(W/im \ a, Y)$ which is just dim(W/im a)*(dim Y).
A: It can be defined an homomorfism $\Xi:\operatorname{Hom}(W,Y)\rightarrow
\operatorname{Hom}(V,Y)$ via $\beta\to \beta\circ \alpha$. $\operatorname{Ker} \Xi=\{\beta/ \beta\circ\alpha=0\}$ is a su subspace of $\operatorname{Hom}(W,Y)$.
Moreover $\dim \operatorname{Hom}(W,Y)=\dim \operatorname{Ker} \Xi +\dim \Xi(\operatorname{Hom}(W,Y))$. There exist a base $\{v_1,\ldots,v_n\}$ of $V$ with  $\alpha(V)=\langle \alpha(v_1),\ldots,\alpha(v_{n-r})\rangle $ and $\langle v_{n-r+1},\ldots,v_n \rangle=\operatorname{Ker}~\alpha$. Let $\{u_1,\ldots,u_s\}$ base of $Y$. Let define $\phi_{i~j}$ by  $ \phi_{i~j}(v_k)=\delta_{i~k}u_j$ for $i=1,\ldots,n-r~~$ and $j=1,\ldots,s$, and $\delta_{i~k}$ the  formalism of Krocnecker. Then $\{\phi_{i~j}\}$ is a base of $\Xi(\operatorname{Hom}(W,Y))$. Thus $\dim \operatorname{Ker} \Xi=\dim \operatorname{Hom}(W,Y)-\dim \Xi(\operatorname{Hom}(W,Y))=(\dim W-\dim \alpha(V))~\dim Y$
