A Proof of the Sylvester Criterion for Quadratic Forms via Optimality Conditions for Quadratic Functions

We give a proof of the so-called Sylvester criterion for quadratic forms (for real symmetric matrices), based on elementary optimality properties of quadratic functions.


Introduction
Some time ago we have surveyed on this journal (Giorgi (2017)) several proofs of the so-called Sylvester criterion for quadratic forms. In the present paper we give a new proof of the said criterion, proof based on basic optimality properties of quadratic functions. A similar approach has been considered also by Hestenes (1966Hestenes ( , 1975, however we think that our treatment is simpler and more suitable for didactic purposes. In our proof we shall make no reference to other properties of quadratic forms or symmetric matrices. We consider, without loss of generality, a real symmetric matrix A of order n and x ∈ R n as a column vector. We recall that the expression Q(x) = x Ax = n i=1 n j=1 a i j x i x j is a quadratic form associated to the matrix A. We recall that Q(x) (or its associated symmetric matrix A) is The reason for choosing a symmetric matrix is that, if A is not symmetric, then 1 2 x (A + A )x = x Ax for any x ∈ R n and obviously 1 2 (A + A ) is symmetric. For simplicity we shall obtain the Sylvester criterion only for definite quadratic forms. For the semidefinite case the reader may consult Chabrillac and Crouzeix (1984), Debreu (1952), Gantmacher (1959) and Takayama (1985).
We recall that the k-th order leading principal minor (or the k-th order NW-principal minor or the k-th order successive principal minor) of the square matrix A, of order n, not necessarily symmetric, denoted by ∆ k , k = 1, ..., n, is the determinant of the square submatrix of A, of order k, consisting of the first k rows and the first k columns of A : a 11 a 12 a 13 a 21 a 22 a 23 a 31 a 32 a 33 The Sylvester criterion for establishing the sign of Q(x) (or of its associated symmetric matrix A) is the following one.
Theorem 1. Let be given the symmetric matrix A, of order n.
1. Q(x) (or A) is positive definite if and only if all its k-th order leading principal minors ∆ k , k = 1, ..., n, are positive.

Q(x) (or A)
is negative definite if and only if all its k-th order leading principal minors ∆ k , k = 1, ..., n, have the sign of (−1) k , i. e.

The Main Results
Let us consider the following quadratic function f : where x ∈ R n , A real and symmetric matrix of order n, with det(A) 0, b ∈ R n and c ∈ R. We have the following results, as a consequence of basic optimality properties of (1).
Theorem 2. The stationary point of (1) x 0 = −A −1 b has the following properties.
1. x 0 is the unique stationary point of the quadratic function (1).
On the grounds of this relation and by a well-known property of the determinants, we have a n1 · · · a nn n j=1 a n j x 0 x ∈ R n , by Taylor's expansion formula we obtain relation (2).
Theorem 3. Let be given the real symmetric matrix A, of order n and with det(A) 0. Then, the following conditions are equivalent.
i) A is positive definite.
ii) For every b ∈ R n and every c ∈ R, the point x 0 = −A −1 b is a strict minimizer of f, defined by relation (1).
iii) For every b ∈ R n and every c ∈ R, the point x 0 = −A −1 b is a minimizer of f, defined by relation (1).

iv) A is positive semidefinite.
Proof. The implications i) =⇒ ii) and iii) =⇒ iv) result from relation (2). The implication ii) =⇒ iii) is trivial. It remains to prove the implication iv) =⇒ i). This implication is well known in the theory of quadratic forms and says, in other words, that the class of semidefinite quadratic forms, with det(A) 0, which are not definite, is empty. Let A be positive semidefinite and suppose that there exists a vector y ∈ R n {0} such that y Ay = 0. The vector y is therefore a local minimum point of the function g : R n −→ R defined by g(x) = x Ax, x ∈ R n . We have therefore ∇g(y) = 0, i. e. 2 n j=1 a i j y j = 0, ∀i = 1, ..., n.
Being y 0, it results det(A) = 0, in contradiction with the assumptions. Therefore A is positive definite.
We are now ready to prove Theorem 1. We begin by proving the point 1.
b) Sufficiency. This part of the proof will be performed by induction. For n = 1 the assertion is satisfied. Let us suppose that the assertion is true for n = k; we prove that the same assertion is true for n = k + 1.
Let us consider the real symmetric matrix B = a i j , of order (k + 1), such that it holds

Let us put
By the induction assumption A is positive definite. By Theorem 3 the point x 0 = −A −1 b is a global minimum point of the function f : R k −→ R defined by (1). Taking Theorem 2, point 2, into account, we have Let be y = y 1 , ..., y k+1 ∈ R k+1 {0} and put z = y 1 , ..., y k . If y k+1 = 0, then we have z 0 and hence y By = z Az > 0. If y k+1 0, then the pointz = 1 y k+1 z satisfies, by relation (3), the following relation Therefore y By > 0, ∀y ∈ R k+1 {0} , i. e. B is positive definite.
The proof of Theorem 1, part 2, is very easy: the symmetric matrix A is negative definite if and only if −A is positive definite and the k-th leading principal minor of −A is (−1) k times the corresponding leading principal minor of A.
Remark 1. The necessity part of Theorem 1 can be proved in a more direct way if we make reference to (well known!) properties of quadratic forms. For example, if we recall that A (symmetric) is definite positive if and only if all its (real!) eigenvalues λ 1 , λ 2 , ..., λ n are positive and that det(A) = λ 1 λ 2 ...λ n > 0, we have at once that ∆ 1 > 0, ∆ 2 > 0, ..., ∆ n > 0.
Another classical result establishes that the symmetric matrix A is definite positive if and only if there exists a non-singular matrix P, such that A = P P.
Remark 2. Also the equivalence iv) ⇐⇒ i) of Theorem 3 is immediate if we make reference to the characterization of the sign of Q(x) by means of eigenvalues of A : if A is positive semidefinite, without being positive definite, one at least of its eigenvalues is zero, but then det(A) = 0, in contradiction with the assumptions of Theorem 3.
Remark 3. It is well known that the examination of the sign of the leading principal minors is not sufficient to check if a symmetric matrix is positive (negative) semidefinite. For example the matrix 0 0 0 −1 has nonnegative leading principal minors, but it is not positive semidefinite. Indeed, it is negative semidefinite. See, e. g., Chabrillac and Crouzeix (1984), Debreu (1952).
For other considerations on the relations between quadratic forms and quadratic functions, the reader may consult Han and Mangasarian (1984).