Does negative semidefinite intersection matrix of bunch of curves implies zero intersection with other curves? If we have the result as the title, then I can solve my real question. The original question was stated as follows.
In a paper I found the following lemma：
Let $S$ be a nonsingular projective surface, $R\in PicS$ a divisor with $R^2>0$ . Let 
$(E_{i})$ be the family of distinct curves such that $R\cdot E_{i}=0$.
Then the $E_{i}$ are numerically independent.
The proof just says that the result follows from Hodge Index Theorem. But I cant see how. HIT just assert that the intersection matrix $(E_i.E_j)$ is negative semidefinite. Why $E_i$ can't be numerically dependent?
Any hint is welcome. Thanks a lot.
 A: The answer already given here seems to assume the E_i are pairwise disjoint, which I didn't take to be the case from your question.
To get the result in general, I think you need to use the additional fact that an effective divisor can never be numerically trivial (since it will have positive intersection with an ample divisor.)  Then take a hypothetical numerical relation among the E_i.  By the above, it must have both positive and negative coefficients.  Group them to get a relation like
$\sum a_i E_i = \sum b_j E_j $ (modulo numerical equivalence)
with the a's and b's positive and all of the E_i and E_j distinct from each other.  If you call the left and right hand sides of this equation A and B, then you have that $A^2<0$ since it's a nonzero element and the intersection pairing is negative definite, but $A \cdot B \geq 0$ since all the curves intersect properly.  This contradicts $A\equiv B$. 
A: Hodge index tehorem for surfaces asserts that in the intersection quadratic form on $\mathrm{Num}(X)$, ther eis exactly one positive square. Hence, the restriction of this form to the orthogonal complement of $R$ must be negative definite. The classes of $E_i$ are pairwise orthogonal, and the quadratic form on the space defined by them in negative definite. Hence, these clacces are linearly independent.
