Chapter 7. The Eigenvalue Problem

7. The Eigenvalue Problem, December 17, 2009 11
This is the characteristic polynomial of the matrix A. It is a third-order
polynomial because we used a three-dimensional basis set and therefore A is
a3× 3 matrix.
The roots of this equation are the eigenvalues of A. They are (see Section
2ofWorkBook7)
⎫
λ 1 = −1.00391 ⎪⎬
λ 2 =0.0032479
(54)
⎪⎭
λ 3 =9.20007
The eigenvalues are real numbers because A is Hermitian.
Exercise 2 Use Mathematica to check that det A = λ 1 λ 2 λ 3 .
How do we find the eigenvectors For the eigenvector ψ(1) corresponding
to the eigenvalue λ 1 ,insertλ 1 in the equation Aψ = λψ to obtain
Aψ(1) − λ 1 ψ(1) = 0, (55)
We have modified the notation so that ψ(1) = {ψ 1 (1), ψ 2 (1), ψ 3 (1)} is the
eigenvector corresponding to the eigenvalue λ 1 .IfIinsertλ 2 in Aψ = λψ, I
get a different equation
Aψ(2) − λ 2 ψ(2) = 0 (56)
for a different eigenvector, ψ(2), corresponding to the eigenvalue λ 2 . Similarly,
Aψ(3) − λ 3 ψ(3) = 0 (57)
gives the third eigenvector ψ(3). Because I have three values for λ, Ihave
three distinct equations for ψ, and their solutions are ψ(1), ψ(2), and ψ(3).
Each symbol ψ(1), ψ(2), ψ(3) represents a vector in R 3 .Thecomponents
of ψ(α), α =1, 2, 3, are denoted by
With this notation,
⎛
⎜
(A − λ 1 I)ψ(1) = ⎝
ψ(α) ={ψ 1 (α), ψ 2 (α), ψ 3 (α)} (58)
⎞
3 − λ 1 2 4
⎟
2 1.2 − λ 1 3.1 ⎠
4 3.1 4− λ 1
⎛
⎜
⎝
ψ 1 (1)
ψ 2 (1)
ψ 3 (1)
⎞
⎟
⎠ = 0