M115A-AssignmentSevenExtraSolutions - MATH 115A -...

MATH 115A - Assignment Seven Extra Solutions
Paul Skoufranis
February 26, 2012
For Question 3 of Section 5.1, for each matrix A ∈ M
n × n
( F ) , determine all eigenvalues of A and for each
eigenvalue λ of A , find the set of eigenvectors corresponding to λ . If possible, find a basis of F n consisting
of eigenvectors of A and determine an invertible matrix Q and a diagonal matrix D such that Q
- 1
AQ = D .
§ 5.1 Question 3a) A =
1 2
3 2
for F = R .

§ 5.1 Question 3b) A =


0 - 2 - 3
- 1 1 - 1
2 2 5


for F = R .
Solution : We desire to find the eigenvalues and eigenvector of A . First we compute the eigenvalues.
To find the eigenvalues, we compute the characteristic polynomial:
χ
A
( λ ) = det




λ 2 3
1 λ - 1 1
- 2 - 2 λ - 5




= λ ( λ - 1)( λ - 5) + (2)(1)( - 2) + (3)(1)( - 2) - (3)( λ - 1)( - 2) - (2)(1)( λ - 5) - λ (1)( - 2)
= ( λ
3
- 6 λ
2
+ 5 λ ) - 4 - 6 + (6 λ - 6) - (2 λ - 10) + 2 λ
= λ
3
- 6 λ
2
+ 11 λ - 6
Therefore, to find the eigenvalues of A , we need to factor χ
A
. One way is to notice that χ
A
(1) = 0 (we
would try 1 as 1 divides - 6 (see the Rational Roots Theorem)) and then use long division of polynomials.
Another way is to notice that
χ
A
( λ ) = ( λ
3
- 6 λ
2
+ 5 λ ) + (6 λ - 6)
= λ ( λ - 5)( λ - 1) + 6( λ - 1)
= ( λ - 1)( λ ( λ - 5) + 6)
= ( λ - 1)( λ
2
- 5 λ + 6) = ( λ - 1)( λ - 2)( λ - 3)
Thus the eigenvalues of A are 1, 2, and 3.
Now we shall compute the eigenspaces. We notice that
E
3
= ker




3 2 3
1 2 1
- 2 - 2 - 2




= ker




1 0 1
0 1 0
0 0 0




= span ( { (1 , 0 , - 1) } )
where the second equality comes from a simple row reduction. Therefore { (1 , 0 , - 1) } is a basis for E
3
.
Next we notice that
E
2
= ker




2 2 3
1 1 1
- 2 - 2 - 3




= ker




1 1 0
0 0 1
0 0 0




= span ( { (1 , - 1 , 0) } )
where the second equality comes from a simple row reduction. Therefore { (1 , - 1 , 0) } is a basis for E
2
.