19.2 The eigenvalue problem 535
The left side of this equation is called the characteristic (or secular)determinantof
the matrix A. It is a polynomial of degree nin λ, and its roots are the values of λfor
which (19.9) is true; that is, the values of λfor which the matrix equation (19.6) has
nonzero solutions. Equation (19.9) is called the characteristic equationof the matrix A.
The values of λfor which the matrix eigenvalue equation(19.6) has nonzero
solutions are called the eigenvaluesor characteristicvalues of the matrix A(they
are also called latent roots or proper values). A matrix of order nhas neigenvalues,
λ
1, λ
2, =, λ
n, and these form the eigenvalue spectrumof A.
EXAMPLE 19.2Solve the characteristic equation of the matrix
The characteristic equation of Ais
This equation was solved in Example 17.8:
D 1 = 1 −(λ 1 + 1 1)(λ 1 − 1 1)(λ 1 − 1 2) 1 = 10 when λ 1 = 1 −1,+1, and+2.
Corresponding to each eigenvalue,λ 1 = 1 λ
k, there is a solutionx 1 = 1 x
kof the eigenvalue
equation such that
Ax
k1 = 1 λ
kx
kk 1 = 1 1, 2, 3, =,n (19.11)
These solutions are called the eigenvectorsor characteristic vectorsof A(they
are also called latent vectors, proper vectors, or poles). The vectors are obtained by
solving the system of homogeneous equations (19.8) for each value of λin turn (as in
Example 17.9). The problem of finding the eigenvalues and eigenvectors is called the
matrix eigenvalue problem(or algebraic eigenvalue problem).
EXAMPLE 19.3Find the eigenvectors of the matrix Ain Example 19.2.
The secular equations are (see also Example 17.8)
(1) (− 21 − 1 λ)x 1 + y 1 + z = 10
(2) − 11 x +(4 1 − 1 λ)y+ 5 z = 10
(3) −x 1 + y 1 + 1 (−λ)z 1 = 10
det(AI−==)
−−
−−
−−
λ =
λ
λ
λ
D
211
11 4 5
11
0
A=
−
−
−
211
11 4 5
110