Let $U,V$ be finite-dimensional vector spaces over $F$, and let $X \leq U$.
Let $B:U\times V \to F$ be a bilinear map. (Here, "bilinear map" does not imply that the map is symmetric or non-degenerate.)
Write $X^\perp=\{v\in V : B(X,v)=0\}$.
Show that $\dim X + \dim X^\perp \geq \dim V$.
I can show that equality holds if $B$ is non-degenerate. [Proof sketch: take a basis $\{u_i\}_{i=1}^k$ for $X$. Extend to a basis $\{u_i\}_{i=1}^n$ for $U$. Flip up to the dual basis $\{\varepsilon_i\}_{i=1}^n$ for $U^*$. Map over to the basis $\{v_i\}_{i=1}^n$ for $V$ (which, by non-degeneracy, is isomorphic to $U^*$ by the map $v \mapsto (u \mapsto B(u,v))$). Show that $\{v_i\}_{i=k+1}^n$ is a basis for $X^\perp$.]
I don't see how to proceed to show the (non-strict) inequality when the non-degeneracy condition is dropped. Perhaps the above argument can be modified, but I'm struggling to fill in the details when we don't necessarily have the isomorphism $V\cong U^*$ (which I suppose might be precisely why the inequality arises -- the map $v \mapsto (u \mapsto B(u,v))$ could have non-trivial kernel).
Hints appreciated :) Thanks.