$\textbf{Exercise 10.}$ Suppose $V$ and $W$ are finite-dimensional and $U$ is a subspace of $V$. Let $\begin{equation*} \mathcal{S} = \{T \in \mathcal{L}(V, W) : U \subseteq \operatorname{null} T\}. \end{equation*}$ $\begin{align*} \text{(a) } & \text{Show that } \mathcal{S} \text{ is a subspace of } \mathcal{L}(V, W). \\ \text{(b) } & \text{Find a formula for } \dim \mathcal{S} \text{ in terms of } \dim V, \dim W, \text{ and } \dim U. \end{align*}$ $\textbf{Solution 10.}$ Define $\Phi: \mathcal{L}(V, W) \to \mathcal{L}(U, W)$ by $\Phi(T) = T|_U$. (a) Because $\Phi$ is a linear map and $\operatorname{null} \Phi = \mathcal{S}$, we can conclude that $\mathcal{S}$ is a subspace of $\mathcal{L}(V, W)$. (b) Exercise 13 in Section 3A implies that $\operatorname{range} \Phi = \mathcal{L}(U, W)$. Thus the fundamental theorem of linear maps shows that $\begin{align*} (\dim V)(\dim W) &= \dim(\mathcal{L}(V, W)) \\ &= \dim \operatorname{null} \Phi + \dim \operatorname{range} \Phi \\ &= \dim \mathcal{S} + \dim \mathcal{L}(U, W) \\ &= \dim \mathcal{S} + (\dim U)(\dim W). \end{align*}$ Thus $\begin{equation*} \dim \mathcal{S} = (\dim V - \dim U)(\dim W). \end{equation*}$ --- ![[sol-5.pdf#page=9]]