We left off with some further matrix problems which are now solved:
Determine which of the following matrices is positive definite:
1)
(1......2)
(2.....1)
Check the eigenvalues, viz.
(1-φ ......2)
(2.......1- φ) = (1-φ )^2 - 4 = 0
Or: φ ^2 - 2φ - 3 = 0
And: (φ - 3) (φ + 1) = 0 so φ1 = 3, φ = -1
Since: φ2 is less than zero the matrix is not positive definite
2)
(1....-1)
(-1.....2)
Again, we first find the eigenvalues using:
(1-φ ......-1)
(-1.......2- φ) = (1-φ ) (2- φ) - 1 = 0
Or: 1 - 3φ + φ^2 - 1 = 0
Leading to: φ^2 - 3φ = 0 or φ (φ - 3) = 0
and this yields: φ1 = 0, and φ2 = 3
Only φ2 > 0 so the matrix is not positive definite
3)
(3.....2)
(2.....1)
Solve for the eigenvalues:
(3-φ ......2)
(2......1- φ) = (3-φ ) (1- φ) -4 = 0
And:
1 - 4φ + φ^2 - 4 = 0 or φ^2 - 4φ - 1 = 0
Use the quadratic formula to solve, viz. φ1, φ2 = {-b +/- [b^2 - 4ac]^½} / 2a
where: b = -4, a = 1 and c = -1
then:
φ1, φ2 = {4 +/- [4^2 - 4(1)]^½} / 2 = {4 +/- [16+ 4]^½} / 2
φ1, φ2 = {4 +/- [20]^½} / 2 = 2 +/- 2 [5]^½
So: φ1 = 2 + 2 [5]^½ and φ2 = 2 - 2 [5]^½
Since: φ2 = 2 - 2 [5]^½ = 2 - 2(2.236) = 2 - 4.472 = -2.472
is less than 0, the matrix is not positive definite.
4)
(1..........2..........3)
(2.........0...........1)
(3 .........1...........1)
Set out the 3 x 3 matrix to obtain the eigenvalues so:
(1- φ..........2..........3)
(2.........- φ.............1)
(3 .........1...........1 - φ)
Proceed to find the eigenvalues using:
(1- φ ) (-2....... 1)
.............. (1 ......1 - φ) -
(2) (2....... 1)
..... (3 ....1 - φ) +
(3) (2....... - φ)
.... (3 .........1 )
Leading to the eigenvalue equation:
1 - φ( -φ + φ^2 -1 ) - 2(-2φ -1) + 3(2 + 3φ)= 0
Or:
φ^3 - 2φ^2 - 13φ - 7 = 0
Which is a cubic equation. One root, call it φ1 = -2.33, is less than zero, hence the matrix cannot be positive definite.
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