# Quiz #7 Solutions

```Math 2270, Fall 2015
Instructor: Thomas Goller
10 November 2015
Quiz #7 Solutions
(1) Let A =

1
4
2
.
6
(a) Compute the eigenvalues
1,
2
of A. (2 points)
Solution:
1
det(A
I) =
so
7 and
1
=
2
4
2
6
= (1
)( 6
) 8=
2
+5
14 = ( +7)(
2),
= 2.
(b) Compute an eigenvector ~v1 that spans the 1 -eigenspace of A and an eigenvector ~v2
that spans the 2 -eigenspace of A. (2 points)



8 2
4 1
1
Solution: A + 7I =
!
, so ~v1 =
.
4 1
0 0
4



1 2
1 2
2
A 2I =
!
, so ~v2 =
.
4
8
0 0
1
(c) Use parts (a) and (b) to diagonalize A. (Hint: Write A = P DP
invertible and D is diagonal.) (1 point)
A=

1 2
4 1

7 0
0 2
1
9

1
4
2
5
47
2
The eigenvalue is 10.
2
8
7
, where P is
2
1
2 3
2
1
5
4
5
4
(2) Show that the vector 2 is an eigenvector of the matrix 7
3
2
eigenvalue? (2 points)
Solution:
1
2
8
7
3
3
15. What is its
6
32 3 2
3 2 3
2 3
3
1
5 4+9
10
1
5
4
5
4
5
4
5
4
1
2 =
7 + 16 3
= 20 = 10 25
6
3
2 + 14 + 18
30
3
Math 2270, Fall 2015
Instructor: Thomas Goller
d
dt
(3) Bonus problem: Consider the derivative P3
(1 bonus point)
(a) Use the standard basis for P3 to view
compute a matrix A for S.
Solution: Since dtd (1) = 0,
coordinates gives us
d
(t)
dt
d
dt
= 1,
2
0
60
A=6
40
0
/
10 November 2015
P3 , which is a linear transformation.
as a linear transformation R4
d 2
(t )
dt
= 2t, and
1
0
0
0
3
0
07
7.
35
0
0
2
0
0
d 3
(t )
dt
S
/ R4
, and
= 3t2 , taking E-
(b) Compute the eigenvalues and eigenvectors of A. Is A diagonalizable?
Solution:
1
det(A
so
= 0 and n1 = 4.
I) =
0
0
0
0
0
0
2
0
0
3
=
4
,
0
3
1 0 0
0 2 07
7
A
0 0 35
0 0 0
2 3
1
607
7
yields only one independent eigenvector ~v1 = 6
405, so A is not diagonalizable.
0
1
2
0
60
0I = 6
40
0