# Mathematical methods for economic theory

Martin J. Osborne

## 3.2.3 Exercises on quadratic forms: conditions for semidefiniteness

1. Determine whether each of the following quadratic forms in two variables is positive or negative definite or semidefinite, or indefinite.
1. x2 + 2xy.
2. x2 + 4xy − 4y2
3. x2 + 2xy − 3y2.
4. 4x2 + 8xy + 5y2.
5. x2 + xy − 3y2.
6. x2 − 6xy + 9y2.
7. 4x2 − y2.
8. (1/2)x2 − xy + (1/4)y2.
9. 6xy − 9y2 − x2.

1. The matrix is
 1 1 1 0
.
The determinant is −1 < 0, so the quadratic form is indefinite.
2. The matrix is
 −1 2 2 −4
.
The first-order principal minors are −1 and −4; the determinant is 0. Thus the quadratic form is negative semidefinite (but not negative definite, because of the zero determinant).
3. The matrix is
 −1 1 1 −3
;
the leading principal minors are −1 and 2, so the quadratic form is negative definite.
4. The associated matrix is
 4 4 4 5
.
The leading principal minors are 4 > 0 and (4)(5) − (4)(4) = 4 > 0. Thus the matrix is positive definite.
5. The associated matrix is
 −1 1/2 1/2 −3
.
The leading principal minors are −1 < 0 and (−1)(−3) − (1/2)(1/2) > 0. Thus the matrix is negative definite.
6. The associated matrix is
 1 −3 −3 9
.
The principal minors are 1 > 0, 9 > 0, and (1)(9) − (−3)(−3) = 0. Thus the matrix is positive semidefinite.
7. The associated matrix is
 4 0 0 −1
.
The determinant is −4 < 0. Thus the matrix is indefinite.
8. The associated matrix is
 1/2 −1/2 −1/2 1/4
.
The determinant is (1/2)(1/4) − (−1/2)(−1/2) < 0. Thus the matrix is indefinite.
9. The associated matrix is
 −1 3 3 −9
.
The principal minors are −1 < 0, −9 < 0, and (−1)(−9) − (3)(3) = 0. Thus the matrix is negative semidefinite.

2. Determine whether each of the following quadratic forms in three variables is positive or negative definite or semidefinite, or indefinite.
1. x2 − y2 − 2z2 + 2xy
2. x2 − 2xy + xz + 2yz + 2z2 + 3zx
3. −4x2 − y2 + 4xz − 2z2 + 2yz
4. x2 − y2 + 2xz + 4yz + 2z2
5. x2 + 2xy − 2y2 + 2xz − 5z2 + 2yz
6. y2 + xy + 2xz
7. −3x2 + 2xy − y2 + 4yz − 8z2
8. 2x2 + 2xy + 2y2 + 4z2
1. The matrix is
 −1 1 0 1 −1 0 0 0 −2
.
The first-order minors are −1, −1 and −2, the second-order minors are 0, 2, and 2, and the determinant is 0. Thus the matrix is negative semidefinite.

2. The matrix is
 1 −1 2 −1 0 1 2 1 2
.
The first-order principal minors are 1, 0, and 2, so the only possibility is that the quadratic form is positive semidefinite. However, the first second-order principal minor is −1. So the matrix is indefinite.
3. The matrix is
 −4 0 2 0 −1 1 2 1 −2
.
The first-order principal minors are −4, −1, and −2; the second-order principal minors are 4, 4, and 1, and the third-order principal minor is 0. Thus the matrix is negative semidefinite.
4. The matrix is
 −1 0 1 0 −1 2 1 2 2
.
The leading principal minors are −1, 1, and 7, so the quadratic form is indefinite.
5. The matrix is
 −1 1 1 1 −2 1 1 1 −5
.
The leading principal minors are −1, 1, and 0, so the matrix is not positive or negative definite, but may be negative semidefinite. The first order principal minors are −1, −2, and −5; the second-order principal minors are 1, 4, and 9; the third-order principal minor is 0. Thus the matrix is negative semidefinite.
6. The matrix is
 0 1/2 1 1/2 1 0 1 0 0
.
Thus the form is indefinite: one of the first-order principal minors is positive, but the second-order one that is obtained by deleting the third row and column of the matrix is negative.
7. The matrix is
 −3 1 0 1 −1 2 0 2 −8
,
with leading principal minors −3, 2, and −4. So the form is negative definite.
8. The matrix is
 2 1 0 1 2 0 0 0 4
.
The leading principal minors are 2, 3, and (2)(8) − (1)(4) = 12 > 0. Thus the matrix is positive definite.

3. Consider the quadratic form 2x2 + 2xz + 2ayz + 2z2, where a is a constant. Determine the definiteness of this quadratic form for each possible value of a.
The matrix is
 2 0 1 0 0 a 1 a 2
.
The first-order minors are 2, 0, and 2, the second-order minors are 0, 3, and −a2, and determinant −2a2. Thus for a = 0 the matrix is positive semidefinite, and for other values of a the matrix is indefinite.

4. Determine the values of a for which the quadratic form x2 + 2axy + 2xz + z2 is positive definite, negative definite, positive semidefinite, negative semidefinite, and indefinite.
The matrix is
 1 a 1 a 0 0 1 0 1
.
The leading principal minors are 1, −a2, and −a2.

Thus if a ≠ 0 the matrix is indefinite.

If a = 0, we need to examine all the principal minors to determine whether the matrix is positive semidefinite. In this case, the first-order principal minors are 1, 0, and 1; the second-order principal minors are 0, 0, and 0; and the third-order principal minor is 0. Thus the quadratic form is positive semidefinite.

Conclusion: If a ≠ 0 the matrix is indefinite; if a = 0 it is positive semidefinite.

5. Consider the matrix
 a 1 b 1 −1 0 b 0 −2
.
Find conditions on a and b under which this matrix is negative definite, negative semidefinite, positive definite, positive semidefinite, and indefinite. (There may be no values of a and b for which the matrix satisfies some of these conditions.)
The matrix is not positive definite or positive semidefinite for any values of a and b, because two of the first-order principal minors are negative. Necessary and sufficient conditions for it to be negative definite are
• a < 0
• a − 1 > 0, or a < −1 (looking at first second-order principal minor)
• 2a + 2 + b2 < 0 (looking at determinant).
Thus it is negative definite if and only if a < −1 and 2a + 2 + b2 < 0.

It is negative semidefinite if and only if a ≤ −1, −2a − b2 ≥ 0, and 2a + 2 + b2 ≤ 0. The second condition implies the first, so the matrix is negative semidefinite if and only if a ≤ −1 and 2a + 2 + b2 ≤ 0.

Otherwise the matrix is indefinite.

6. Show that the matrix
 1 0 1 1 0 1 0 0 1 0 1 0 1 0 0 1
is not positive definite.
The second order principal minor obtained by deleting the second and fourth rows and columns is 0, so the matrix is not positive definite. (Alternatively, the third-order leading principal minor is 0, and the principal minor obtained by deleting the second and third rows and columns is 0.)