which implies x 1 =x 2. This gives the eigenvector x=col(x 1 , x 2 ) = col(x 2 , x 2 )where the
component x 2 must be some nonzero constant. Selecting the value x 2 = 1 gives the
eigenvector x=col(1 ,1). This shows that corresponding to the eigenvalue λ= 5 there
is the eigenvector x=col(1 ,1). Note also that any nonzero constant times col(1 ,1) is
also and eigenvector.
Example 10-25.
Find the eigenvalues and eigenvectors associated with the matrix
A=
2 2 2
0 0 4
0 −3 7
Solution The eigenvalues and eigenvectors of the matrix Aare determined by solving
the matrix equation
(A−λI )x=
2 −λ 2 2
0 −λ 4
0 −3 7 −λ
x 1
x 2
x 3
=
0
0
0
(10 .32)In order for this system to have a nonzero solution for the column vector x, Cramer’s
rule requires that
det(A−λI ) = 0
or ∣∣
∣∣
∣∣2 −λ 2 2
0 −λ 4
0 −3 7 −λ∣∣
∣∣
∣∣=λ(^3) + 9λ (^2) − 26 λ+ 24 = 0
Solving this equation for λone finds the factored form
(λ−2)(λ−3)(λ−4) = 0 with roots λ= 2, λ = 3, and λ= 4
which are called the eigenvalues of the matrix A. The eigenvector corresponding to
the eigenvalue λ= 2 is found be substituting the value λ= 2 into the equation (10.32)
to obtain
2 −2 2 2
0 − 2 4
0 −3 7 − 2
x 1
x 2
x 3
=
0 2 2
0 −2 4
0 −3 5
x 1
x 2
x 3
=
0
0
0