Dimension of $\{\phi :\mathbb{R}^{11}\to\mathbb{R}^9\text{ linear}: U\subseteq\ker\phi\}$

Question

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?

Answer 1

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 $V$.

Answer 2

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.

Answer 3

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 $(11-4)9=7\cdot 9 = 63$.

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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.

Answer 4

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: $$ \mathcal L\bigl(\mathbf R^{11}\mkern-4mu/U,\mathbf R^9\bigr)\simeq \mathcal L\bigl(\mathbf R^{7},\mathbf R^9\bigr)\simeq \mathbf R^{7\times9}. $$