## Monday, December 5, 2011

### Solutions to Linear Algebra Problems

I now give the solutions to the linear algebra problems:

Find the characteristic polynomials and the eigenvalues for each of the following matrices:

1) A =
(1.. ..i)
(-i.......1)

Solution:

We have P_A(t) =

(t -1.........i)
(-i.........t-1)

So: P_A(t) = (t - 1)^2 - (-i)(i) = t^2 -2t +1 -1 = 0

Then: P_A(t0 = t^2 - 2t = t(t - 2)

The eigenvalues E1,2 are:

E1 = 0, E2 = 2

2) A =
(1.. …. .2)
(2.......-2)

P_A(t) =

(t - 1......2)
2........t +2)

P_A(t) = (t - 1)((t + 2) - 4

P_A(t) = t^2 + t - 2 -4 = t^2 + t - 6

But: t^2 + t - 6 = (t + 3) (t - 2)

So: E1 = -3, and E2 = 2

3) A =
(3.. ……2)
(-2...... 3)

Then:

P_A(t) =

(t - 3........2)
(-2......t - 3)

P_A(t)= (t - 3)(t - 3) - (2)(-2)

P_A(t) = t^2 - 6t + 13

We use the quadratic formula, where a = 1, b = -6, c = 13:

E1,2 = {- b +/- [b^2 – 4 ac]^ ½} / 2a

E1,2 = {6 +/- [(-6)^2 – 4 (13)]^ ½} / 2(1)

E1,2 = 6 +/- [-16]^ ½} / 2(1)

E1,2 = {6 +/- (4i)} / 2

E1 = (6 + 4i)/2 = 3 + 2i

E2 = (6 - 4i)/ 2 = 3 - 2i