Next: Negative eigenvalues Up: Sufficient Conditions for Multiply Previous: Introduction.

# Non-positive eigenvalues

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.

Next: Negative eigenvalues Up: Sufficient Conditions for Multiply Previous: Introduction.

Tom Vogel
Tue Aug 20 10:23:22 CDT 1996