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
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