We continue to examine some more aspects of Hermitian matrices and some general properties of all matrices. We will stick in this exposition to simple 2 x 2 matrices, but understand everything treated can be generalized to larger array square matrices. Consider first the case of a positive matrix.
Here, a Hermitian matrix is positive if all eigenvalues > (=) to 0. Let's consider this example for which we are to check whether it's positive: M =
(1 -φ .........i)
(-i.........1 - φ)
Write out:
M = (1 - φ)^2 - (-i)(i) = (1 - φ) (1 - φ) - (1) = φ^2 - 2φ + 1 - 1 = 0
Or:
φ^2 - 2φ = 0 so: φ( φ - 2) = 0
whence: φ1 = 0 and φ2 = 2
Thus, the condition is met that the eigenvalues (φ1, φ2) are equal to or greater than 0 so the matrix is positive.
We now consider any general symmetric matrix, i.e. such that A = t^A in the context of when such a matrix is positive definite. The conditions for this are:
i) All the eigenvalues are positive (e.g. φi > 0)
ii) All the determinants are positive (e.g. ac - bd > 0)
iii)The pivots ([ac - bd]/ a) > 0
iv) t^x A x > 0
Example: State whether matrix W =
(2.....6)
(6......19)
is positive definite.
Solution: We check each of the conditions (i - iv) to see if they are met.
First, the eigenvalues must be > 0. We check the calculations - see, e.g.
http://brane-space.blogspot.com/2011/12/revisiting-linear-algebra.html
to see : φ1 = 0.096 and φ2 = 20.904
so both meet the condition.
Check that the determinant D > 0:
D = (2) (19) - (6)(6) = 38 - 36 = 2
so the condition is met.
Check that the pivot P > 0.
P = ([ac - bd]/ a) = 2/ 2 = 1
so, the condition is met.
Check to see if: t^x A x > 0
We let: x =
[x1]
[x2]
so that: t^x = [x1 ...x2]
Then we obtain the operation: t^x W x =
[x1 ...x2] [2x1 + 6x2]
...............[6x1 + 19x2]
Yielding the quadratic form: 2x1^2 + 12x1 x2 + 19 x2^2
where: a = 2, 2b = 12 so b = 12/2 = 6 and lastly c = 19
Hence, it meets the condition since a, b, c are all positive.
Then the matrix is positive definite.
Problems:
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 )
2)
(3.......1)
(0.......3)
3)
(2.....6)
(6.....18)
4)
(2i.....1)
(2.......i)
5)
(-½ i.....i)
(2i....... i)
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