The following lemma will be needed in the proof of Theorem 3.1, at a point where we will apply Lemma 2.3.
Proof: Let , be vectors such that if and only if , i=1, , n, and if and only if , j=1, , m. Then a vector is in if and only if there are constants so that . For this we need
so that d is in if and only if the vector
is in the span of the vectors
This characterization of the image of C under the projection and the fact that inner product is continuous is enough to yield the result.
We now develop a criterion slightly different from that of the previous section. We suppose that consists of k negative eigenvalues (counting multiplicity), that ker(A) is finite dimensional (this will be implied by the requirement that is trivial), and for some . Suppose that is spanned by . We may assume that , , are orthogonal to ker(A) by projecting onto . Define E to be
Proof: Again we need to show that A is strongly positive on . By orthogonalizing the matrix E as in Lemma 2.2, we may assume that if and that . Take an . Then x=y+z, where and . One easily sees that .
By arguing as in Theorem 2.1, one can show that there are constants , , so that , where v is orthogonal to , , . In addition, v is orthogonal to ker(A), since y and the 's are. Substituting,
Since , we have , so that holds for all j. Therefore
with the last inequality following from the fact that v is orthogonal to all non-positive eigenvectors of A. We must now relate the length of v to that of x. As before, we use Lemma 2.3. Since is trivial, there is a so that , where is independent of x.
For a similar inequality relating and , we must first show that the orthogonal projection of onto has only the trivial intersection with the span of . Suppose that some is also in the orthogonal projection of onto . There is a vector so that . Then for each j from 1 to k,
so that is the zero vector. To apply Lemma 2.3 we must also show that the image of under orthogonal projection onto is closed. This follows from Lemma 3.1. Therefore, there is some (not depending on x or y) so that . We now obtain that
for all , so that A is strongly positive on . As above, this is sufficient for the result to follow.
Acknowledgments. I would like to thank John Maddocks for suggesting that I extend the results of . I would also like to thank Thomas Schlumprecht for a useful conversation relating to Lemma 3.1.