# Nonsingular bilinear map

This is related to a previous question I asked. But I realize that my logic there is total bonkers. And it would be great if someone could help me out a bit.

$B:V\times V\to F$ is a bilinear form where $V$ is a finite-dimensional vector space.

$X\leq V$ is a subspace which is also the annihilator of another subspace $Y\leq V$ wrt $B$.

I am given that $B|_X$ is nonsingular.

I wish to show that it follows that $B$ itself is nondegenerate.

Thanks.

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The term you should use is nondegenerate, as far as I know; nonsingular applies to linear maps (and certainly one doesn't want to say $B$ vanishes only on $(\vec0,\vec0)$). Being degenerate means some nonzero vector annihilates every other vector. Now how would such a vector in $V$, if it existed, be related to $X$? –  Marc van Leeuwen Dec 7 '11 at 10:33
@MarcvanLeeuwen: Thanks, I have edited the word. :) If there is a vector $x\in X$ that kills off all the others wrt $B$ then we have $B(x,X)=0$, right? But I still don't quite see how to prove the statement. –  Zachmann Dec 7 '11 at 10:37
What you are trying to prove is the nonexistence of a vector **in the whole space $V$** that kills off everyone, knowing the nonexistence of a vector $x\in X$ like the one you describe. Given the negative character of this statement, you seem to begin "at the wrong end" of the argument. –  Marc van Leeuwen Dec 7 '11 at 11:13
@MarcvanLeeuwen: Yes, your interpretation of the question is exactly right. –  Zachmann Dec 7 '11 at 11:16

If $B$ is not symmetric, then in general a subspace has both a "left annihilator" and a "right annihilator". You didn't specify, so I'll just pick one of the two:

For $W \subset V$, $W^{\perp} = \{v \in V \ | \ B(v,W) = 0 \}$.

Now let $0 \neq v \in V$. We want to find $v' \in V$ with $B(v,v') \neq 0$.

Suppose first that $v \in X$. Then by nonsingularity of $X$, there exists $v' \in X$ such that $B(v,v') \neq 0$.

Now suppose $v \in V \setminus X$. Since $X = Y^{\perp}$, $v \notin Y^{\perp}$, i.e., there exists $v' \in Y$ with $B(v,v') \neq 0$.

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Thank you, Pete. –  Zachmann Dec 7 '11 at 12:07

I suppose you've figured it out by now but just in case...

Assume $X$ is given to be the left annihilator of $Y$: $$X=Y^{\perp,\mathrm{left}} = \{v \in V \ | \ B(v,Y) = 0 \}.$$ Now if there existed a nonzero $v\in V$ with $\forall w\in V: B(v,w)=0$, then certainly $v\in Y^{\perp,\mathrm{left}}=X$, and certainly $B(v,x)=0$ for all $x\in X$, contradicting the nondegeneracy of $B|_X$.

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Thank you, Marc. –  Zachmann Dec 7 '11 at 12:07