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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