We now examine the solutions to the previous problems. Once more:
Check each of the following matrices to determine if positive or positive definite or positive semi-definite (the case where D = ac - bd = 0)
1)
(cos π/2.........-sin π/2)
(sin π/2.........cos π/2 )
Solution: We know:
cos π/2 = 0, - sin π/2 = -1, sin π/2 = 1 so the matrix can be written:
(0...-1)
(1....0)
And the eigenvalues φ are found from:
(-φ ......-1)
(1..........- φ) =
φ^2 + 1 = 0 or φ^2 = -1 whence: φ1 = i, φ2 = -i
So they are imaginary. The determinant D = 0 - (-1)(1) = 1
Check that the pivot P > 0 when P = ([ac - bd]/ a) = (0 - (-1))/0 = 1/0
But 1/0 = oo so this test is indeterminate. (Note also we cannot say the eigenvalues > 0)
2)
(3.......1)
(0.......3)
The Determinant is D = ac - bd = (3)(3) - 0 = 0
The eigenvalues must be > 0. We check with the calculations, using :
(3-φ ......1)
(0.........3- φ) =
(3- φ) ^2 = 0 or 9 - φ ^2 = 0
so: φ1 = 3, and φ2 = - 3
The matrix fails the tests for being positive, positive definite because one eigenvalue is negative.
3)
(2.....6)
(6.....18)
The determinant D = (2)(18) - (6)(6) = 36 - 36 = 0
Hence the matrix is positive semi-definite, and also since the eigenvalues:
φ1 = 0, φ2 = 20
4)
(2i.....1)
(2.......i)
The determinant D = (2i)i - (1)(2) = 2 (i)^2 - 2 = 2(-1) - 2 = -4
Hence, not positive or positive definite, or positive semi-definite
5)
(-½ i.....i)
(2i....... i)
The determinant D = (-½ i) i - i((2 i) = [-½ (-1)] - (2)(-1) = ½ + 2 = 5/2
Check that the pivot P > 0. P = ([ac - bd]/ a) = ( 2½)/ (-½ i) = -1.25i
SO the test fails for this matrix, or th result is indeterminate.
Additional Problems:
Determine which of the following matrices is positive definite:
1)
(1......2)
(2.....1)
2)
(1....-1)
(-1.....2)
3)
(3.....2)
(2.....1)
4)
(1..........2..........3)
(2.........0...........1)
(3 .........1...........1)
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