We are now ready to prove the sufficiency of Sylvester’s criterion, which will complete the work we started in my last post. Set . If , then there is nothing to prove, so suppose that . Denote the sum of the principal minors of size by . By our hypothesis that the principal minors are nonnegative, . Since there exists a nonsingular principal minor, we see that .

To complete the proof, we will show that has only nonnegative eigenvalues by examining the roots of the characteristic polynomial .

**Lemma 4. **The characteristic polynomial of a linear operator between finite-dimensional vector spaces can be expressed as

where , for , is the sum of all the principal minors.

*Proof. *Recall that a linear operator on a finite-dimensional vector space induces a linear transformation on the exterior algebra

Since the exterior algebra is a -dimensional vector space, the linear transformation is a scalar multiple of the identity. We define the determinant to be the scalar. Taking , for ,

Define a -dimensional vector space and an operator defined by

In other words, is the operator whose matrix with respect to the basis is the principal submatrix with index set . The determinant of is the scalar of the rank- operator

Equivalently, is the principal minor with index set . Substituting this result into our last expression for , we see that

,

which shows that

Note that is the sum of all the principal minors, so

Since the matrix has rank , , for . Hence, the characteristic polynomial simplifies to

Define a polynomial by . Since is Hermitian, all the eigenvalues of are real; equivalently, all the roots of the polynomials and are real. Since , the polynomial only has nonzero roots. By Descartes’ rule of signs, the number of negative roots of is equal to the number of sign changes of the nonzero coefficients of the polynomial

But has no sign changes, since all the are nonnegative. We conclude that all the roots of are positive, which implies that all roots of the characteristic polynomial are nonnegative. Equivalently, all the eigenvalues of are nonnegative, which implies that is positive-semidefinite.

### Like this:

Like Loading...

*Related*