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 [6]. I would also like to thank Thomas Schlumprecht for a useful conversation relating to Lemma 3.1.
