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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.

Help please?

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 $X$?2011-12-07
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    @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.2011-12-07
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    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.2011-12-07
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    @MarcvanLeeuwen: Yes, your interpretation of the question is exactly right.2011-12-07

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