Chapter 7. The Eigenvalue Problem

7. The Eigenvalue Problem, December 17, 2009 9
use Mathematica to give me the inverse matrix (see the file WorkBook7 The
eigenvalue problem.nb). Mathematica tells me that
A −1 =
⎛
⎜
⎝
160.333 −146.667 −46.6667
−146.667 133.333 43.3333
−46.6667 43.3333 13.3333
Acting with A −1 on φ gives (see WorkBook7.nb)
End of Example
⎞
⎟
⎠ (47)
ψ = A −1 φ = {−773.3, 705.0, 228.0} (48)
Exercise 1 Prove that if a matrix H is Hermitian then so is its inverse.
Note a very important fact. If A −1 exists then Aψ = φ has a solution.
This also means that the solution of Aψ = 0 is ψ = 0 (because A −1 Aψ =
A −1 0 = 0 and A −1 Aψ = Iψ = ψ). The solution ψ = 0 is of no interest,
and we say that if A −1 exists then Aψ = 0 does not have a solution.
How does this relate to the eigenvalue problem The eigenvalue equation
writtenintheform
(A − λI)ψ = 0 (49)
is a homogenous system of equations. If the operator
( − λÎ)−1 (50)
exists then (A − λI)ψ = 0 does not have a solution.
Therefore (A − λI)ψ = 0 has a solution if and only if (A − λI) −1 does
not exist, which happens exactly when
det(A − λI) =0 (51)
Note that this condition does not involve the unknown vector ψ; it depends
only on λ (which we don’t know) and on the matrix elements A ij (which we
do know). This is an equation for λ. One can prove, based on the properties
of determinants, that the left-hand side of Eq. 51 is a polynomial of order
N in λ, which is called the characteristic polynomial of the matrix A. The
fundamental theorem of algebra tells us that Eq. 51 must therefore have