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Let $\phi$ be a bilinear form on the vector space that has basis $\{x_i\}\cup\{y_j\}$.

If we have $x\in \operatorname{Span}\{x_i\}$ and $y\in\operatorname{Span}\{y_i\}$ are such that $\phi\left(x\,\,\,,\,\,\,\operatorname{Span}\{x_i\}\right)+\phi\left(y\,\,\,,\,\,\,\operatorname{Span}\{y_i\}\right)=0$,

[Edit: As suggested by @awllower, I should indeed explain my notation: By taking $\operatorname{Span}\{v_i\}$, $v=x, y$ as a variable of the bilinear form, I mean that for all $v\in \operatorname{Span}\{v_i\}$ the relation is true.]

Given that $\phi$ is non-degenerate on $\operatorname{Span}\{x_i\}$

Can I conclude that $x=0=y$?

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Per chance the notation ought to be improved, as the bilinear form takes not as its variable the span of some vectors... – awllower Dec 6 '11 at 14:20
And is the basis dis-joint? – awllower Dec 6 '11 at 14:21
@awllower: Thanks, yes, the basis is disjoint. And by the span I mean for all vectors $v\in \operatorname{Span}\{v_i\}$ where $v=x\,\,\,\text{or}\,\,\,y$. – Ralph Dec 6 '11 at 14:40

No, you can't. You could have $\phi(y,z)=0$ for all $z$ despite $y\ne0$, and then your condition would be fulfilled with $x=0$ and $y\ne0$.

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Maybe the use of the matrix multiplication can elaborate further, as there exists $A$ such that $A*y_i$ are all 0 without $A$=0. – awllower Dec 8 '11 at 9:31

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