Dimension of $\{\phi :\mathbb{R}^{11}\to\mathbb{R}^9\text{ linear}: U\subseteq\ker\phi\}$ Let $U\subseteq\mathbb{R}^{11}$ be of dimension $4$ and consider the vector space $$V=\{\phi :\mathbb{R}^{11}\to\mathbb{R}^9\text{ linear}: U\subseteq\ker\phi\}.$$
I am interested in the dimension of $V$. What is it?
If I choose a basis $(v_1,\ldots,v_4)$ of $U$ and extend it to a basis $(v_1,\ldots,v_{11})$ of $\mathbb{R}^{11}$, then a $\phi\in V$ is uniquely determined by the values on $(v_5,\ldots,v_{11})$. Does this say something about the dimension?
 A: Yes; a linear map in $\mathcal L\bigl(\mathbf R^{11},\mathbf R^9\bigr)$ is determined by the images of the vectors of a basis of $\mathbf R^{11}$, i.e. by the column vectors of a $9\times 11$-matrix relative to this basis. The condition of vanishing  on the first  $4$ vectors of the basis means it depends on the $9\times(11-4)=63$ other coefficients. 
This the dimension of this space of linear maps is $63$.
Another approach:
Linear maps vanishing on $U$ are in bijection with $\mathcal L\bigl(\mathbf R^{11}\mkern-4mu/U,\mathbf R^9\bigr)$. As $R^{11}\mkern-4mu/U\simeq R^7$, we obtain an isomorphism:
$$ \mathcal L\bigl(\mathbf R^{11}\mkern-4mu/U,\mathbf R^9\bigr)\simeq \mathcal L\bigl(\mathbf R^{7},\mathbf R^9\bigr)\simeq \mathbf R^{7\times9}. $$
A: It says everything about the dimension! In total, among those vectors you can freely choose (i.e. the different $\phi(v_i)\in \Bbb R^9$ for $ 5\leq i \leq 11$, and assuming some basis on $\Bbb R^9$ is already picked out for us), how many entries can you freely choose? That's the dimension.
A: You're on exactly the right track.
Select a basis $(w_1,\dots,w_9)$ of $\Bbb R^9$.  Define $\phi_{i,j} \in V$ by
$$
\phi_{ij}(v_k) = \begin{cases}
w_j & k = i \\
0 & k \neq i
\end{cases}
$$
Verify that $\{\phi_{i,j}: 5 \leq i \leq 11, 1 \leq j \leq 9\}$ forms a basis of $V$.
A: You can think about your problem as finding yet another kernel of a linear map. Concretely, your set up is the following. Let $V$ and $W$ be finite dimensional, and let $\hom(V,W)$ be the space of linear maps $V\to W$. Fix a subspace $U$ of $V$ and let $i:U\to V$ be the inclusion. There is a linear map
$$i^* : \hom(V,W) \to\hom(U,W)$$
the sends a linear map $f$ to the restriction $fi$ of $f$ to $U$. You want to find the kernel of $i^*$. I claim that $i^*$ is onto: indeed, given any linear map $g : U\to W$ there is at least way one of extending a basis of $U$ to one of $V$, and then defining $g$ arbitrarily on this extension to coincide with $g$ on $U$. Thus the dimension of the kernel of $i^*$, by rank-nullity, is equal to the difference
$$\dim \hom(V,W)-\dim\hom(U,W) = (\dim(V)-\dim(U))\dim(W)$$
which in your case is $(11-4)9=7\cdot 9 = 63$. 

The point here is that there is an exact sequence
$$0\to U\to V\to V/U\to 0$$
that yields an exact sequence
$$0\to \hom(V/U,W)\to \hom(V,W)\to \hom(U,W)\to 0$$
that identifies $\hom(V/U,W)$ as the kernel of $i^*$. This means that 
$$\dim\ker(i^*) = \dim \hom(V/U,W) = \dim(V/U)\dim(W)$$
as computed above. 
