Example 10-27. Find the eigenvalues and eigenvectors associated with the
matrix
A=
−1 4 6 − 6
1 4 0 2
−4 0 7 − 8
− 1 −2 0 0
Solution: The characteristic equation of Acan be calculated by evaluating the de-
terminant
C(λ) = |A−λI |=
∣∣
∣∣
∣∣
∣
− 1 −λ 4 6 − 6
1 4 −λ 0 2
− 4 0 7 −λ − 8
− 1 − 2 0 −λ
∣∣
∣∣
∣∣
∣
= (λ−1)(λ−2)(λ−3)(λ−4) = 0.
The eigenvalues are λ= 1, 2 , 3 , 4 and each eigenvector associated with Amust be a
nonzero solution vector which satisfies the equation
AX =λX
or
− 1 −λ 4 6 − 6
1 4 −λ 0 2
− 4 0 7 −λ − 8
− 1 − 2 0 −λ
x 1
x 2
x 3
x 4
=
0
0
0
0
.
Substitute successively the values λ= 1 , 2 , 3 , 4 into this equation and each time
solve for X to obtain the eigenvectors
X 1 =
1
− 1
2
1
, X
2 =
− 2
0
0
1
, X
3 =
− 1
− 1
1
1
, X
4 =
− 2
− 1
0
1
.
The modal matrix Q, is the matrix having these eigenvectors for its column vectors.
The modal matrix Q can be used to produce a diagonal matrix containing the
eigenvalues of Asuch that
Q−^1 AQ =D=diag (1 , 2 , 3 ,4).
The proof of this statement is left as an exercise. It can also be verified that
det(A) = 24 and (rank A) = 4.
As a final note, it should be pointed out that when one or more of the eigenvalues
of a matrix Aare repeated roots, then a set of nlinearly independent eigenvectors
