3.2.3 Exercises on quadratic forms: conditions for semidefiniteness
- Determine whether each of the following quadratic forms in two variables is positive or negative definite or semidefinite, or indefinite.
- x^{2} + 2xy.
- −x^{2} + 4xy − 4y^{2}
- −x^{2} + 2xy − 3y^{2}.
- 4x^{2} + 8xy + 5y^{2}.
- −x^{2} + xy − 3y^{2}.
- x^{2} − 6xy + 9y^{2}.
- 4x^{2} − y^{2}.
- (1/2)x^{2} − xy + (1/4)y^{2}.
- 6xy − 9y^{2} − x^{2}.
- The matrix is
1 1 1 0 . - The matrix is
−1 2 2 −4 . - The matrix is
−1 1 1 −3 ; - The associated matrix is
4 4 4 5 . - The associated matrix is
−1 1/2 1/2 −3 . - The associated matrix is
1 −3 −3 9 . - The associated matrix is
4 0 0 −1 . - The associated matrix is
1/2 −1/2 −1/2 1/4 . - The associated matrix is
−1 3 3 −9 .
- Determine whether each of the following quadratic forms in three variables is positive or negative definite or semidefinite, or indefinite.
- −x^{2} − y^{2} − 2z^{2} + 2xy
- x^{2} − 2xy + xz + 2yz + 2z^{2} + 3zx
- −4x^{2} − y^{2} + 4xz − 2z^{2} + 2yz
- −x^{2} − y^{2} + 2xz + 4yz + 2z^{2}
- −x^{2} + 2xy − 2y^{2} + 2xz − 5z^{2} + 2yz
- y^{2} + xy + 2xz
- −3x^{2} + 2xy − y^{2} + 4yz − 8z^{2}
- 2x^{2} + 2xy + 2y^{2} + 4z^{2}
- The matrix is
−1 1 0 1 −1 0 0 0 −2 . - The matrix is
1 −1 2 −1 0 1 2 1 2 . - The matrix is
−4 0 2 0 −1 1 2 1 −2 . - The matrix is
−1 0 1 0 −1 2 1 2 2 . - The matrix is
−1 1 1 1 −2 1 1 1 −5 . - The matrix is
0 1/2 1 1/2 1 0 1 0 0 . - The matrix is
−3 1 0 1 −1 2 0 2 −8 , - The matrix is
2 1 0 1 2 0 0 0 4 .
- Consider the quadratic form 2x^{2} + 2xz + 2ayz + 2z^{2}, 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 . - Determine the values of a for which the quadratic form x^{2} + 2axy + 2xz + z^{2} is positive definite, negative definite, positive semidefinite, negative semidefinite, and indefinite.
The matrix is
1 a 1 a 0 0 1 0 1 . 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.
- Consider the matrix
a 1 b 1 −1 0 b 0 −2 . 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 + b^{2} < 0 (looking at determinant).
It is negative semidefinite if and only if a ≤ −1, −2a − b^{2} ≥ 0, and 2a + 2 + b^{2} ≤ 0. The second condition implies the first, so the matrix is negative semidefinite if and only if a ≤ −1 and 2a + 2 + b^{2} ≤ 0.
Otherwise the matrix is indefinite.
- Show that the matrix
1 0 1 1 0 1 0 0 1 0 1 0 1 0 0 1 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.)