Suppose that $∀a \in K, w \in W$, $⟨a,w⟩ = 0$. Prove there is $U ⊂ K$ with dim at least $\dim K - \dim V + \dim W$ with $⟨u,v⟩ = 0$ $∀v\in V, u \in U$

Question

Suppose $K$ and $V$ are finite dimensional vector spaces and let $\left< \cdot , \cdot \right>$ be a bilinear form on $K \times V$. If there exists a subspace $W \subset V$ such that for all $a \in K$ and $w \in W$, $\left< a, w \right> = 0$, Show that there exists a subspace $U \subset K$ with dimension at least $\dim K - (\dim V - \dim W)$ such that $\langle u, v \rangle = 0$ whenever $u \in U$ and $v \in V$.

The reason was explained as the fact that "row rank of $\left< \cdot , \cdot \right>$ is equal to its column rank." However, I am failing to see how matrices come into play for this question. Any help would be greatly appreciated.

Answer 1

Matrices are useless here. Let $U$ be the kernel of the linear map $$f:K\to(V/W)^*,$$ $$a\mapsto[v+W\mapsto\langle a,v\rangle].$$ By the rank-nullity theorem, $$\dim U=\dim K-\dim {\rm im}f.$$ But $$\dim{\rm im}f\le\dim((V/W)^*)=\dim(V/W)=\dim V-\dim W.$$ The conclusion follows.

Answer 2

Let $G$ be the entire vector space. So both $V$ and $K$ are subspace of $G$.

Decompose $G$ as $G=K\oplus K^\perp$, and decompose $V$ as $V=W_1\oplus W_1^\perp$, where $W_1$ is the largest subspace of $V$ such that $W_1\perp K$. So we have $W_1\subseteq K^\perp, W_1^\perp\subseteq K$. Next, we decompose $K$ as:

$$K=U\oplus W_1^\perp$$

such that $U\perp W_1^\perp$. By this construction, we have $U\subseteq K\perp W_1$, therefore, $U\perp V$.

$$\dim U=\dim K-\dim W_1^\perp=\dim K-(\dim V-\dim W_1)$$

Since $W_1$ is the largest subspace of $V$ such that $W_1\perp K$, we have $\dim W_1 \ge \dim W$

$$\dim U=\dim K-(\dim V-\dim W_1)\ge \dim K-(\dim V-\dim W)$$