In mathematics, Sylvester’s criterion is a necessary and sufficient criterion to determine whether a Hermitian matrix is positive-definite. Sylvester's criterion states that a n × n Hermitian matrix M is positive-definite if and only if all the following matrices have a positive determinant: In other words, all of the leading principal minors must be positive. By using appropriate permutations of rows and columns of M, it can also be shown that the positivity of any nested sequence of n principal minors of M is equivalent to M being positive-definite. An analogous theorem holds for characterizing positive-semidefinite Hermitian matrices, except that it is no longer sufficient to consider only the leading principal minors as illustrated by the Hermitian matrix A Hermitian matrix M is positive-semidefinite if and only if all principal minors of M are nonnegative.

Proof for the case of positive definite matrices

Suppose M_n is n\times n Hermitian matrix. Let be the leading principal minor matrices, i.e. the k\times k upper left corner matrices. It will be shown that if M_n is positive definite, then the principal minors are positive; that is, \det M_k>0 for all k. M_k is positive definite. Indeed, choosing we can notice that Equivalently, the eigenvalues of M_k are positive, and this implies that \det M_k>0 since the determinant is the product of the eigenvalues. To prove the reverse implication, we use induction. The general form of an Hermitian matrix is where M_n is an n \times n Hermitian matrix, \vec{v} is a vector and d is a real constant. Suppose the criterion holds for M_n. Assuming that all the principal minors of M_{n+1} are positive implies that, \det M_n>0, and that M_n is positive definite by the inductive hypothesis. Denote then By completing the squares, this last expression is equal to where (note that M_n^{-1} exists because the eigenvalues of M_n are all positive.) The first term is positive by the inductive hypothesis. We now examine the sign of the second term. By using the block matrix determinant formula on (*) we obtain Consequently,

Proof for the case of positive semidefinite matrices

Let M_n be an n x n Hermitian matrix. Suppose M_n is semidefinite. Essentially the same proof as for the case that M_n is strictly positive definite shows that all principal minors (not necessarily the leading principal minors) are non-negative. For the reverse implication, it suffices to show that if M_n has all non-negative principal minors, then for all t>0, all leading principal minors of the Hermitian matrix M_n+tI_n are strictly positive, where I_n is the nxn identity matrix. Indeed, from the positive definite case, we would know that the matrices M_n+tI_n are strictly positive definite. Since the limit of positive definite matrices is always positive semidefinite, we can take t \to 0 to conclude. To show this, let M_k be the kth leading principal submatrix of M_n. We know that is a polynomial in t, related to the characteristic polynomial p_{M_k} via We use the identity in Characteristic polynomial to write where is the trace of the jth exterior power of M_k. From Minor (linear algebra), we know that the entries in the matrix expansion of (for j > 0) are just the minors of M_k. In particular, the diagonal entries are the principal minors of M_k, which of course are also principal minors of M_n, and are thus non-negative. Since the trace of a matrix is the sum of the diagonal entries, it follows that Thus the coefficient of t^{k-j} in q_k(t) is non-negative for all j > 0. For j = 0, it is clear that the coefficient is 1. In particular, q_k(t) > 0 for all t > 0, which is what was required to show.