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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.

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up vote 1 down vote accepted

The condition $B(X,v) = 0$ is equivalent to $B(x_i,v)=0$ for all $i$, where the $x_i$ form a basis for $X$. Each condition $B(x_i, v) = 0$ reduces the dimension by at most one, so $\dim X^{\perp} \ge \dim V - \dim X$.

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Excellent, thank you very much. – Iain Jan 28 '12 at 16:59

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