Question

How to prove a is positive semidefinite if and only if determinant of every principal submatrix is greater than or equal to 0?

Answer 1

Step-by-step explanation:

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Sylvester's criterion says that an n×n Hermitian matrix A is positive definite if and only if all its leading principal minors of are positive. If one knows that fact that every Hermitian matrix has an orthogonal eigenbasis, one can prove Sylvester's criterion easily by mathematical induction.

The base case n=1 is trivial. The forward implication is also obvious. So, we only need to consider the backward implication.

Suppose all leading principal minors of A are positive. In particular, det(A)>0. It follows that if A is not positive definite, it must possess at least two negative eigenvalues. As A is Hermitian, there exist two orthogonal eigenvectors x and y corresponding to two of these negative eigenvalues. Let u=αx+βy≠0 be a linear combination of x and y such that the last entry of u is zero. Then u∗Au=|α|2x∗Ax+|β|2y∗Ay<0. Hence the leading (n−1)×(n−1) principal submatrix of A is not positive definitive. By induction assumption, this is impossible. Hence A must be positive definite.