# Is this argument valid? — bilinear maps

Does this argument make sense?

So I have a bilinear form $B:V\times V\to F$. (BTW, is bilinear form equivalent to a bilinear map?) where $V$ is a finite-dimensional vector space.

I have a subspace $X\leq V$ which happens to be the annihilator of a subspace $Y\leq V$ wrt $B$.

Now, $B|_X$ is nonsingular and I claim that that implies that $B$ itself is nonsingular. (Is nonsingular the same as non-degenerate?)

So here is my argument:

$(x_1\,\,...\,\, x_k)^T[B](x_1 ...x_k)=M$ where $M$ is non-singular and $\{x_i\}$ is a basis for $X$ $\implies [B]\(x_1 ...x_k)=[(x_1\,\,...\,\, x_k)^T]^{-1}M$, which is non-singular.

Now $(y_1\,\,...\,\, y_m)^T[B](x_1 ...x_k)=0$ where $\{y_i\}$ is a basis for $Y$ $\implies (y_1... y_m)^T [(x_1\,\,...\,\, x_k)^T]^{-1}M=0$ $\implies Y=\{0\}$

$\implies \forall Y\leq V$ s.t. $X$ is its annihilator wrt $B$, $Y=\{0\}$

$\implies B$ is non-singular.

Thanks.

Alternative more elegant proofs are welcomed!

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A bilinear form is a bilinear map whose codomain is the scalar field. – Henning Makholm Nov 30 '11 at 21:26
@HenningMakholm: Thanks. – Zachmann Dec 1 '11 at 0:10

Reading the proof given in question, I get lost at $[(x_1\dots x_k)^T]^{−1}$. Presumably, $x_1,\dots,x_k$ are column vectors of size $\dim V$, but then $(x_1\dots x_k)^T$ need not be a square matrix. Anyway, turning everything into a mess of matrices and bases isn't the way to go.
Although this is not stated in the question, I will assume $B$ is symmetric, because otherwise "the annihilator of a subspace" does not make sense: one has left and right annihilators instead.
Proof. Suppose $B$ is degenerate, i.e., there is a nonzero vector $v$ such that $\{v\}^\perp = V$. Since $Y^\perp= X$, it follows that $v\in X$. But then $B_{|X}$ is also degenerate. $\Box$