As in [6], we define 
to be the vector such
that 
.
Proof: Certainly 
, where 
and 
. Take 
so that
. Defining 
to be the orthogonal
projection of onto the range of A, one finds that
, concluding the proof.
Now define the matrix B by
Proof: This is essentially Lemma 1 of [3], obtained
by writing out the matrix multiplication. Note that 
is
orthogonal to ker(A).
Proof: If the lemma is false, then there exist sequences 
and 
so that 
and 
.
By taking a subsequence, we may assume that 
converges (strongly) to
. Therefore 
converges to z as well. Since Y is closed,
, so converges to the zero vector. This contradicts
the assumption that .
Proof: Let 
be the span of the vectors 
, and 
its orthogonal complement.
is said to be strongly positive on the
space if for some 
there holds 
for all 
. This condition is
sufficient to imply that y is a strong local constrained minimum
(see [1, chapter 3,]), so this is what we need to prove.
Let x be an arbitrary element of . I will first show that
there exist unique scalars 
, 
, 
so that
is perpendicular to 
, ,
. We seek , , to solve
so that it certainly suffices to show that the matrix
is invertible. If not, then some non-trivial linear combination of the
columns is the zero vector, implying that there is a non-trivial linear
combination 
of the vectors 
, , which is
perpendicular to , , . This leads to a
contradiction as follows. Since is a non-trivial vector which is
orthogonal to all of the non-positive eigenvectors of A,
. On the other hand, a non-trivial linear combination
satisfies
since 
is the 
eigenvalue of
B.
Since x is orthogonal to all of the 
's,
for i=1, , n. (Of course if
, 
for all i.) It follows that
for all i. Let 
. Then
holds for all i.
Now, v is certainly perpendicular to ker(A), since v is
perpendicular to all eigenspaces corresponding to non-positive
eigenvalues. Using this fact and the A-orthogonality of the 
's,
equation 2.2 becomes
Substituting this into
we obtain
with the last inequality due to the fact that the first k eigenvalues of
B are negative. Since v is orthogonal to , ,
, it follows that
and we therefore must bound 
from below in terms of 
. We will do this by applying Lemma 2.3, where Y will
be and Z is the span of 
. To
apply this lemma, it is necessary to show that the intersection of
and the span of 
is trivial.
Suppose that 
, i.e., that
for j=1, , n. By
equation 2.1, 
for all i.
Since the 's are chosen to be orthogonal to ker(A), we have
holding for all i. By the A-orthogonality of
the 's and the assumption that the first k eigenvalues of B
are negative, it follows that 
. Thus the intersection of
and the span of is the zero
vector.
Since 
, Lemma 2.3 yields a
positive 
so that 
if 
. By
normalizing we obtain 
for general
, where does not depend x. Substituting this into
equation 2.3 we obtain that
for . This strong positivity on implies that y
is a strong constrained local minimum, as mentioned above.
