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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.

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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


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).

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Yes, I do! Thanks for catching that. – Potato Jun 2 '12 at 1:37
Now, you get to choose where $y - \dim\operatorname{im}\alpha$ basis elements of $W$ go in the $z$-dimensional space $Y$. Think about entries of a matrix: how many do you get to pick? – Dylan Moreland Jun 2 '12 at 1:39
I really don't know how much linear algebra you've developed at this point, but I'm secretly identifying this subspace of yours with $\operatorname{Hom}(W/\operatorname{im}\alpha, Y)$, which makes the answer for the dimension very clear. – Dylan Moreland Jun 2 '12 at 1:44
Oh, so it's just dim(W/im a)*(dim Y)? – Potato Jun 2 '12 at 1:57
Yes, although showing that probably requires more argument. But we could prove it directly: if I extend the basis for $\operatorname{im}\alpha$ with new elements $\{c_j\}$ to get a basis for all of $W$, and take a basis $\{d_k\}$ for $Y$, what is a basis for this subspace of $\operatorname{Hom}(W, Y)$? – Dylan Moreland Jun 2 '12 at 4:16

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 $\dim \operatorname{Ker} \Xi=\dim \operatorname{Hom}(W,Y)-\dim \Xi(\operatorname{Hom}(W,Y))=(\dim W-\dim \alpha(V))~\dim Y$

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