Math 222 - Selected Homework Solutions from
Chapter 6.1
Instructor - Al Boggess
Fall 1998
Section 6.1
4. If A is a nonsingular matrix with eigenvalue , then Av = v for some
nonzero vector v. Applying A,1 to both sides of this equation, we
obtain
v = A,1Av = A,1v
Rearranging this equation, gives
A,1 v = ,1v
which implies that ,1 is an eigenvalue for A,1 with eigenvector v.
6. Suppose A2 = A and Av = v. Applying A to both sides of this last
equation we obtain
A2 v = Av = 2v so A2v = 2v
Using A2 v = Av, this equation becomes
Av = 2v
On the other hand Av = v. Therefore
v = 2 v
or ( , 2 )v = 0
Since v 6= 0, we conclude that , 2 = 0, which implies that = 0 or
1.
1
9. If is an eigenvalue for A, then
det(A , I ) = 0
Now the determinant of a matrix equals the determinant of its transpose. So
det(A , I ) = 0
as well. The transpose of I is itself, so this equation becomes
t
det(A , I ) = 0
t
Therefore is an eigenvalue for A .
A and A may not have the same eigenvectors. For example, let
!
1
2
A = ,1 ,2
t
t
A has eigenvalue 0 with eigenvector (,2; 1) , but A has eigenvalue 0
with eigenvector (1; 1).
t
2
t