Solving this equation for the values of λgives the eigenvalues (λ 1 , λ 2 ,... , λ n)associated
with the matrix A. Substituting an eigenvalue λinto the equation (10.29) enables
one to solve for the corresponding eigenvector.
Example 10-24.
Find the eigenvalues and eigenvectors associated with the matrix A=
(
1 4
2 3
)
Solution The eigenvalues and eigenvectors of the matrix Aare determined by solving
the matrix equation Ax =λx or
(A−λI )x=
(
1 −λ 4
2 3 −λ
)(
x 1
x 2
)
=
(
0
0
)
(10 .31)
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 ∣
∣∣
∣
1 −λ 4
2 3 −λ
∣∣
∣∣= (1 −λ)(3 −λ)−8 = λ^2 − 4 λ−5 = 0
Solving this equation for λgives
(λ+ 1)(λ−5) = 0 with roots λ=− 1 and λ= 5
which are called the eigenvalues of the matrix A. The eigenvector corresponding to
the eigenvalue λ=− 1 is found be substituting λ=− 1 into the equation (10.31) to
obtain (
1 −(−1) 4
2 3 −(−1)
)(
x 1
x 2
)
=
(
2 4
2 4
)(
x 1
x 2
)
=
(
0
0
)
which gives the equation 2 x 1 +4x 2 = 0 or x 1 =− 2 x 2. This result specifies how the first
component of the eigenvector is related to the second component of the eigenvector
and gives x=col(x 1 , x 2 ) = col(− 2 x 2 , x 2 )for the eigenvector. Note that x 2 must be some
nonzero constant in order that the eigenvector be nonzero. For convenience select the
value x 2 = 1 to obtain the eigenvector x=col(− 2 ,1). Note that any nonzero constant
times an eigenvector is also an eigenvector. To summarize what has just been done,
one can say the solution of the matrix equation (10.31), using the value λ=− 1 ,
tells us that col(− 2 ,1) is an eigenvector of the matrix Aand any constant times the
eigenvector is also an eigenvector. In a similar fashion, substitute the value λ= 5
into the equation (10.31) to obtain
(
1 − 5 4
2 3 − 5
)(
x 1
x 2
)
=
(
−4 4
2 − 2
)(
x 1
x 2
)
=
(
0
0
)
