Dimensionality of the set of linear maps between inner product spaces

Question

Suppose we have two inner product spaces $V$ and $W$ with dimensions $n$ and $m$, respectively. For some set $\{ \mathbf{v}_i\}_{i=1}^k \in V$ and some nonzero $\mathbf{w} \in W$, where $1 \leq k \leq n$, we let the subspace $U \subset L(V,W)$ be the set of linear maps $T$ between these inner products spaces such that $\langle T \mathbf{v}_i, \mathbf{w} \rangle = 0$.

Now, if that set $\{ \mathbf{v}_i\}_{i=1}^k$ happens to be linearly independent, then what can we say about the dimension of $U$? Is there a range of values that $U$ can take, or is there a unique value? Moreover, I wonder what does it mean, intuitively, for $U$ to have a dimension?

I'm not quite sure where to begin with any of these questions, so any prods in the right direction or heuristic explanations would be helpful.

Answer 1

Firstly, on your question about dimension of $U$: since $L(V,W)$ is a vector space, we can do usual vector space things (finding a basis, finding its dimension, etc.). This of course goes for its subspaces, including $U$.

Now, note that $\langle Tv_i, w \rangle =0$ is equivalent to $T(\text{Span}(\{v_i\}))\subseteq \{w\}^{\perp}$. We are searching for the subspace $U\subset L(V,W)$ of linear maps that send the subspace $\text{Span}(\{v_i\})\subset V$, having dimension $k$, to the subspace $\{w\}^{\perp}\subset W$, having dimension $m-1$. Note that such a map $T$ need not have rank equal to $m-1$, only $<m$; in fact, $T=0$ works fine.

The final steps are not hard, but I'll hide them in case you'd like to do them yourself.

The problem now is just combinatorial. Let $V_1 =\text{Span}(\{v_i\})\subset V$ and $V_2 = V_1^{\perp}$. There are $k(m-1)$ dimensions of linear maps sending $V_1$ into $\{w\}^{\perp}$ (since they have dimension $k$ and $m-1$, respectively), and $(n-k)m$ dimensions of linear maps sending $V_2$ into $W$ (for the same reason), so there are $k(m-1)+(n-k)m$ dimensions of maps from $V$ to $W$ doing both. That is, $\text{dim}\ U=k(m-1)+(n-k)m=mn-k$.

An afterthought (kept in a spoiler):

An intuitive way to view this might be that we removed $k$ dimensions of $L(V,W)$ by restricting the image of $V_1$ (having dimension $k$) by one dimension. This generalises easily: restricting the image any $k$-dim subspace of $V$ to some $l$-dim subspace of $W$ (and thus excluding $(m-l)$ dimensions) reduces the space of valid linear maps by $k(m-l)$ dimensions. This property "stacks": applying it to disjoint subspaces of $V$ applies the equivalent dimensionality reductions independently, and they sum. The proof of this is not hard and may be a fun exercise, depending on your idea of fun.

Answer 2

Hint1. If $B=(v_1,\dots,v_n)$ is a basis of $V$, prove that the map $$U\to(w^\perp)^k\times W^{n-k},\;T\mapsto(T(v_1),\dots,T(v_n))$$ is an isomorphism.

Hint2. Justify that $(w^\perp)^k\times W^{n-k}$ is isomorphic to $(\Bbb R^{m-1})^k\times(\Bbb R^m)^{n-k}=\Bbb R^d$, where $d=k(m-1)+(n-k)m=mn-k$.

Edit. Alternatively and more directly, if moreover $C=(w_1,\dots,w_m)$ is a basis of $W$ such that $w_2,\dots,w_m\perp w$, check that the canonical isomorphism $$L(V,W)\xrightarrow{\sim} M_{m,n}(\Bbb R),\;T\mapsto[T]^B_C$$ maps $U$ onto the subspace of matrices whose $k$ first entries on the first line are $0$ (the dimension of this subspace is obviously $mn-k$).