# Mathematical methods for economic theory

Martin J. Osborne

## 3.2.3 Quadratic forms: conditions for semidefiniteness

### Two variables

First consider the case of a two-variable quadratic form
Q(xy) = ax2 + 2bxy + cy2.
If a = 0 then Q(x, 1) = 2bx + c. This expression is nonnegative for all values of x if and only if b = 0 and c ≥ 0, in which case ac − b2 = 0.

Now assume a ≠ 0. As before, we have

 Q(x, y) = a[(x + (b/a)y)2 + (c/a − (b/a)2)y2].
Both squares are nonnegative, so if a > 0 and ac − b2 ≥ 0 then this expression is nonnegative for all (xy). If these two conditions are satisfied then c ≥ 0.

We conclude that if a ≥ 0, c ≥ 0, and ac − b2 ≥ 0, then the quadratic form is positive semidefinite.

Conversely, if the quadratic form is positive semidefinite then Q(1, 0) = a ≥ 0, Q(0, 1) = c ≥ 0, and Q(−ba) = a(ac − b2) ≥ 0. If a = 0 then by the previous argument we need b = 0 and c ≥ 0 in order for the quadratic form to be positive semidefinite, so that ac − b2 = 0; if a > 0 then we need ac − b2 ≥ 0 in order for a(ac − b2) ≥ 0.

We conclude that the quadratic form is positive semidefinite if and only if a ≥ 0, c ≥ 0, and ac − b2 ≥ 0.

A similar argument implies that the quadratic form is negative semidefinite if and only if a ≤ 0, c ≤ 0, and ac − b2 ≥ 0.

Note that in this case, unlike the case of positive and negative definiteness, we need to check all three conditions, not just two of them. If a ≥ 0 and ac − b2 ≥ 0, it is not necessarily the case that c ≥ 0 (try a = b = 0 and c < 0), so that the quadratic form is not necessarily positive semidefinite. (Similarly, the conditions a ≤ 0 and ac − b2 ≥ 0 are not sufficient for the quadratic form to be negative semidefinite: we need, in addition, c ≤ 0.)

Thus we can rewrite the results as follows: the two variable quadratic form Q(x, y) = ax2 + 2bxy + cy2 is

• positive semidefinite if and only if a ≥ 0, c ≥ 0, and |A| ≥ 0
• negative semidefinite if and only if a ≤ 0, c ≤ 0, and |A| ≥ 0
where
A =
 a b b c
.
It follows that the quadratic form is indefinite if and only if |A| < 0. (Note that if |A| ≥ 0 then ac ≥ 0, so we cannot have a < 0 and c > 0, or a > 0 and c < 0.)

### Many variables

As in the case of two variables, to determine whether a quadratic form is positive or negative semidefinite we need to check more conditions than we do in order to check whether it is positive or negative definite. In particular, it is not true that a quadratic form is positive or negative semidefinite if the inequalities in the conditions for positive or negative definiteness are satisfied weakly. In order to determine whether a quadratic form is positive or negative semidefinite we need to look at more than simply the leading principal minors. The matrices we need to examine are described in the following definition.

Definition
The kth order principal minors of an n × n symmetric matrix A are the determinants of the k × k matrices obtained by deleting n − k rows and the corresponding n − k columns of A (where k = 1, ..., n).

Note that the kth order leading principal minor of a matrix is one of its kth order principal minors.

Example
Let
A =
 a b b c
.
The first-order principal minors of A are a and c, and the second-order principal minor is the determinant of A, namely ac − b2.

Example
Let
A =
 3 1 2 1 −1 3 2 3 2
.
This matrix has 3 first-order principal minors, obtained by deleting
• the last two rows and last two columns
• the first and third rows and the first and third columns
• the first two rows and first two columns
which gives us simply the elements on the main diagonal of the matrix: 3, −1, and 2. The matrix also has 3 second-order principal minors, obtained by deleting
• the last row and last column
• the second row and second column
• the first row and first column
which gives us −4, 2, and −11. Finally, the matrix has one third-order principal minor, namely its determinant, −19.
The following result gives criteria for semidefiniteness.
Proposition
Let A be an n × n symmetric matrix. Then
• A is positive semidefinite if and only if all its principal minors are nonnegative.
• A is negative semidefinite if and only if its kth order principal minors are nonpositive for k odd and nonnegative for k even.
Source
For a proof, see Gantmacher (1959), Theorem 4 on p. 307.
Example
Let
A =
 0 0 0 −1
.
The two first-order principal minors and 0 and −1, and the second-order principal minor is 0. Thus the matrix is negative semidefinite. (It is not negative definite, because the first leading principal minor is zero.)

### Procedure for checking the definiteness of a matrix

Procedure for checking the definiteness of a matrix
• Find the leading principal minors and check if the conditions for positive or negative definiteness are satisfied. If they are, you are done. (If a matrix is positive definite, it is certainly positive semidefinite, and if it is negative definite, it is certainly negative semidefinite.)
• If the conditions are not satisfied, check if they are strictly violated. If they are, then the matrix is indefinite.
• If the conditions are not strictly violated, find all its principal minors and check if the conditions for positive or negative semidefiniteness are satisfied.
Example
Suppose that the leading principal minors of the 3 × 3 matrix A are D1 = 1, D2 = 0, and D3 = −1. Neither the conditions for A to be positive definite nor those for A to be negative definite are satisfied. In fact, both conditions are strictly violated (D1 is positive while D3 is negative), so the matrix is indefinite.
Example
Suppose that the leading principal minors of the 3 × 3 matrix A are D1 = 1, D2 = 0, and D3 = 0. Neither the conditions for A to be positive definite nor those for A to be negative definite are satisfied. But the condition for positive definiteness is not strictly violated. To check semidefiniteness, we need to examine all the principal minors.