Show that $Z=\{T\in\operatorname{Hom}(V, V)|T(\alpha)=0,\ \forall\alpha\in W\}$ is a subspace of $\operatorname{Hom}(V, V)$ and compute its dimension. $\newcommand{\Hom}{\text{Hom}}$
The following exercise is from Brown's book, A Second Course in Linear Algebra.

Let $W$ be a subspace of $V$(over a field $F$) with $m=\dim W\leq\dim V=n<\infty $. Let $Z=\{T\in\Hom(V, V)\ |\ T(\alpha)=0,\ \forall\alpha\in W\}$. Show that $Z$ is a subspace of $\Hom(V, V)$ and compute its dimension.

My Attempt:
First, $Z\ne\varnothing$ since $O_V\in Z$. Now, let $S,T\in Z$,$\ $ $\alpha,\beta\in V$, and $a,b,x,y\in F$. We have
\begin{align}
(xS+yT)(a\alpha+b\beta) &= (xS)(a\alpha+b\beta)+(yT)(a\alpha+b\beta)\\
&= (xS)(a\alpha)+(xS)(b\beta)+(xT)(a\alpha)+(xT)(b\beta)\\
&= a(xS)(\alpha)+b(xS)(\beta)+a(xT)(\alpha)+b(xT)(\beta)\\
&= a(xS+yT)(\alpha)+b(xS+yT)(\beta).
\end{align}
Thus, $(xS+yT)\in Z$ and so $Z$ is a vector subspace of $\Hom(V, V)$.
But for the next part of the problem, I'm really not sure about how to compute $\dim Z$.
For every $T\in Z$, $\dim(\operatorname{Im}T)=n-m$. This implies that $Z\cong\mathbb{M}_{(n-m)\times n}$ (i.e. $Z$ and $\mathbb{M}_{(n-m)\times n}$ are isomorphic). So, $\dim Z = n(n-m)$.
The book does not have any solutions, so I don't know if my calculations are correct. Any corrections/alternate solutions will be appreciated. TIA.
 A: First comment: to check linearity, you do not need two constants, one suffices. So, you can ignore $y$ and $b$ for instance (i.e set them equal to $1$). Next, you didn't even prove $xS+yT\in Z$, all you showed is that $xS+yT\in \text{Hom}(V,V)$. You need to show that for every $\alpha\in W$, we have $(xS+T)(\alpha)=0$. If you manage to do this, that is what tells you $xS+T\in Z$.
Next, your claim that $\dim(\text{Im}(T))=n-m$ is wrong. All you know is that $W\subseteq\ker T$ (that's the definition of $Z$), so
\begin{align}
\dim\text{Im}(T)=\dim V-\dim\ker T\leq \dim V-\dim W=n-m.
\end{align}
Anyway, even if $\dim T=n-m$, I don't see how you jumped to your next conclusion (yes, the dimension you gave is correct but I don't see how you arrived there).

This can be generalized slightly. Suppose we have finite-dimensional vector spaces $V_1,V_2$ and a subspace $W\subset V_1$. Then, $Z:=\{T\in\text{Hom}(V_1,V_2)\,:\, T|_W=0\}$ is a subspace of $\text{Hom}(V_1,V_2)$. Regarding dimensions, fix a complementary subspace $W'$ of $W$ in $V_1$, so $V_1=W\oplus W'$. Then, $Z\cong \text{Hom}(W',V_2)$ (why)? So, what's the dimension? (Note that this approach of choosing a complement is essentially what's in the comments).
The more abstract route is if you know about quotient spaces (though in that case, this problem might already be trivial... but I'll just mention this for completeness). The universal property of quotient spaces tells us that $Z\cong \text{Hom}(V_1/W, V_2)$, so the dimension can easily be computed from here.
