Math 2270, Fall 2015
Instructor: Thomas Goller
17 November 2015
Test #4 Solutions
1
Inventions (10 points)
(a) Invent a matrix A and a nonzero vector ~x that is not an eigenvector of A. (2 points)


1 2
2
Possible solution: A =
, ~x =
0 0
3
(b) Invent a 2 ⇥ 2 matrix A such that
Possible solution: A =


1
is a 3-eigenvector of A. (2 points)
2
3 0
0 3
(c) Invent a matrix A with all entries nonzero that has 0 as an eigenvalue. (2 points)
Possible solution: A =

1 1
1 1
(d) Invent a matrix A that has a two-dimensional eigenspace. (2 points)
Possible solution: A =

0 0
0 0
(e) Invent a 2 ⇥ 2 matrix A with real eigenvalues that is not diagonalizable. (2 points)
Possible solution: A =

2 1
0 2
Math 2270, Fall 2015
2
Instructor: Thomas Goller
17 November 2015
Definition of Eigenvectors (10 points)
(a) Write down the equation that defines what it means for ~x to be a -eigenvector of a
matrix A. (2 points)
Solution: A~x = ~x
(b) Show that
Solution:


1
0

7
1
is an eigenvector of
2
0
7
1

7
=
2

7
=
2

7
and compute its eigenvalue . (2 points)
1
7
, so
2
=
1.
(c) Suppose ~x is a 2-eigenvector of A and a 5-eigenvector of B. Show that ~x is an eigenvector
of AB and compute its eigenvalue . (2 points)
Solution: (AB)~x = A(B~x) = A(5~x) = 10~x, so
= 10.
2
3
3 0 2
(d) Compute a basis for the 6-eigenspace of the matrix A = 4 6 6 45. (4 points)
3 0 8
2
3
2
3
22 3
3 0 2
3 0 2
x
3 3
Solution: A 6I = 4 6 0 45 ! 4 0 0 05. The solutions are ~x = 4 x2 5 =
3 0 2
0 0 802 3 2 39
x3
2 3
223
0
2 =
< 0
3
x2 415 + x3 4 0 5, so a basis for Nul(A 6I) is 415 , 405 .
:
;
0
1
0
3
Math 2270, Fall 2015
3
Instructor: Thomas Goller
17 November 2015
Diagonalization (10 points)
Let A =

3 1
.
2 2
(a) Compute the eigenvalues
Solution: det(A
and 2 = 1.
1,
2
I) = (3
of A. (2 points)
)(2
)
2=
2
5 +4 = (
4)(
1), so
1
=4
(b) Compute independent eigenvectors ~v1 , ~v2 of A. (4 points)


1 1
1
Solution: A 4I =
, so ~v1 =
.
0 0
1



1
2 1
2 1
A I=
!
, so ~v2 =
.
2 1
0 0
2
1
2
1
!
2

1
(c) Write down the diagonalization A = P DP
and D is diagonal. (2 points)
Solution:


3 1
1
=
2 2
1
1
2

4 0
0 1
1
3

, where the columns of P are eigenvectors
2 1
1 1
(d) Use your answer in (c) to write down a nice formula for A2015 . (2 points)
Solution:

3 1
2 2
2015

1
=
1
1
2

42015 0
0
1
1
3

2 1
1 1
Math 2270, Fall 2015
4
Instructor: Thomas Goller
17 November 2015
Matrix for the Derivative (10 points)
T
/ P2 be the linear transformation that takes the derivative. For example, T (5 +
Let P3
t2 3t3 ) = 2t 9t2 . Let E = {1, t, t2 , t3 } and E 0 = {1, t, t2 } denote the standard bases for P3
and P2 .
(a) Write down the E-coordinates of 5 + t2
points)
Solution: [5 + t2
2
3
5
607
7
3t3 ]E = 6
4 1 5, [2t
3
3t3 and the E 0 -coordinates of 2t
9t2 . (2
2
3
0
9t2 ]E 0 = 4 2 5.
9
(b) Compute the E 0 -coordinates of T (1), T (t), T (t2 ), and T (t3 ). (4 points)
2 3
2 3
0
1
4
5
4
Solution: T (1) = 0, so [T (1)]E 0 = 0 . T (t) = 1, so [T (t)]E 0 = 05.
0
0
2 3
2 3
0
0
2
2
3
2
3
4
5
4
T (t ) = 2t, so [T (t )]E 0 = 2 . T (t ) = 3t , so [T (t )]E 0 = 05.
0
3
(c) Write down the 3 ⇥ 4 matrix for T relative to the bases E and E 0 . (2 points)
2
3
0 1 0 0
Solution: 40 0 2 05
0 0 0 3
(d) Check your matrix is correct by multiplying it by the E-coordinates of 5 + t2
found in (a). You should get the E 0 -coordinates of 2t 9t2 . (2 points)
3
2 3
5
0 1 0 0 6 7
0
07 4 5
6
4
5
Solution: 0 0 2 0 4 5 = 2
1
0 0 0 3
9
3
2
3
2
3t3 you
Math 2270, Fall 2015
5
Instructor: Thomas Goller
17 November 2015
Discrete Dynamical System (10 points)
Let A be a 2 ⇥2 matrix with eigenvalues 1 = 1, 2 = 13 and corresponding eigenvectors
2
1
~v1 =
, ~v2 =
. Let B = {~v1 , ~v2 } be the eigenvector basis for R2 .
1
3
(a) Compute the change of basis matrix P . (3 points)
B

2
Solution: P =
1
B E
1
3
1
=
1
7

E
3 1
1 2

30
(b) Let ~x0 =
. Compute c1 and c2 such that ~x0 = c1~v1 + c2~v2 . (Hint:
50
(2 points)
Solution:

c1
=
c2
1
7

3 1
1 2

30
=
50
1
7


c1
= P ~x0 .)
c2
B E

140
20
=
70
10
(c) Use your answer in (b) to write down a nice formula for Ak ~x0 . (2 points)

2
Solution: A ~x0 = 20
+ 10 31k
1
k

1
3
(d) What vector does Ak ~x0 converge to as k gets large? (1 point)
Solution:

40
20
(e) Invent a vector ~y0 such that Ak ~y0 converges to ~0 as k gets large. (1 point)
Possible solution:

1
3
(f) Invent a vector ~z0 such that Ak ~z0 converges to ~z0 as k gets large. (1 point)
Possible solution:

2
1