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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