Math 308 - Differential Equations Fall 2003
Homework Assignment 3 Solutions (Problems 8, 9, and 10)
8. We have A=·a0
0a¸, so A−rI =·a−r0
0a−r¸and the characteristic polynomial is
det(A−rI ) = (a−r)2. Thus the only eigenvalue is r=a. Then
A−rI =·0 0
0 0¸
and therefore (A−rI )~v =~
0 for any vector ~v. So any nonzero vector ~v is an eigenvector.
We could also observe that for any vector ~v =·v1
v2¸,
A~v =·a0
0a¸·v1
v2¸=·av1
av2¸=a·v1
v2¸=a~v.
Thus, if ~v 6=~
0, ~v is an eigenvector associated with the eigenvalue a.
9. We have A=·a b
0d¸. We’ll have to assume that a6=d, otherwise we have a repeated
eigenvalue, which we haven’t covered yet. Then A−rI =·a−r b
0d−r¸and the character-
istic polynomial is det(A−rI ) = (a−r)(d−r). Thus the eigenvalues are
r1=aand r2=d.
Now find the eigenvectors.
For r1=a, we have A−r1I=·0b
0d−a¸, so we may take ~v1=·1
0¸.
For r2=d, we have A−r2I=·a−d b
0 0¸, so we may take ~v2=·b
d−a¸.
10. We have A=·a b
c0¸, so A−rI =·a−r b
c−r¸and the characteristic polynomial is
det(A−rI ) = (a−r)(−r)−bc =r2−ar −bc. Thus the eigenvalues are
r=a±√a2+ 4bc
2.
1
