The notion of a positive definite or negative definite linear transformation is very important in many applications. In particular it is used in versions of the second derivative test for functions of many variables. Here the main interest is the case of a linear transformation which is an n × n matrix but the theorem is stated and proved using a more general notation because all these issues discussed here have interesting generalizations to functional analysis.
Definition 14.12.1 A self adjoint A ∈ℒ
The following lemma is of fundamental importance in determining which linear transformations are positive or negative definite.
Lemma 14.12.2 Let X be a finite dimensional inner product space. A self adjoint A ∈ℒ
Proof: Suppose first that A is positive definite and let λ be an eigenvalue. Then for x an eigenvector corresponding to λ, λ
Now suppose all the eigenvalues of A are positive. From Theorem 14.11.4 and Corollary 14.11.7, A = ∑ _{i=1}^{n}λ_{i}u_{i} ⊗ u_{i} where the λ_{i} are the positive eigenvalues and
To establish the claim about negative definite, it suffices to note that A is negative definite if and only if −A is positive definite and the eigenvalues of A are
The next theorem is about a way to recognize whether a self adjoint n × n complex matrix A is positive or negative definite without having to find the eigenvalues. In order to state this theorem, here is some notation.
Definition 14.12.3 Let A be an n×n matrix. Denote by A_{k} the k ×k matrix obtained by deleting the k + 1,
The following theorem is proved in [9]. For the sake of simplicity, we state this for real matrices since this is also where the main interest lies.
Theorem 14.12.4 Let A be a self adjoint n×n matrix. Then A is positive definite if and only if det
Proof: This theorem is proved by induction on n. It is clearly true if n = 1. Suppose then that it is true for n− 1 where n ≥ 2. Since det

Now letting x ∈ ℝ^{n−1}, x≠0, the induction hypothesis implies

The dimension of
To show the converse, note that, as above,

Note that for x ∈ ℝ^{k},

From Lemma 14.12.2, this implies that all the eigenvalues of A_{k} are positive. Hence from Lemma 14.12.2, it follows that det
Corollary 14.12.5 Let A be a self adjoint n × n matrix. Then A is negative definite if and only if det
Proof: This is immediate from the above theorem by noting that, as in the proof of Lemma 14.12.2, A is negative definite if and only if −A is positive definite. Therefore, det