# Dimensions of vector subspaces

Given a bilinear map $B:X\times Y\to F$ where $X,Y$ are vector spaces and given $S\leq X$, why is $\dim S+\dim \operatorname{ann}(S)=\dim Y$ where $\operatorname{ann}(S)$ is the annihilator of $S$ viz. $\operatorname{ann}(S)=\{y\in Y|B(S,y)=0\}$? N.B. $B$ is may or may not be degenerate.

Added in repsonse to Jan's comment: Perhaps a more accurate statement would be $\dim S+\dim \operatorname{ann}(S)\geq\dim Y$? Is this right? How might I prove this?

Thanks.

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Just checking. :-) Are you the same as this user, or are you two different? If the two accounts are the same, then we can flag the moderators to merge these two accounts. – Srivatsan Nov 29 '11 at 15:20
Have you tried B=0 the zero bilinear form? – Jan Nov 29 '11 at 15:22
@Srivatsan: Nope, I don't know him :-) – Fernandez Nov 29 '11 at 15:38

For B degenerate this is wrong, for B nondegenerate the bilinear form gives you an isomorphism $Y \cong X^*$. This means for each basis $(x_i)$ of $X$ you find a dual basis $(y_j)$ of $Y$ such that $B(x_i,y_j)=\delta_ij$. Now do this with a basis of $X$ which contains a basis of $S$.