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.

Tue Aug 20 10:23:22 CDT 1996