APPM 3310: Matrix Methods — Exam #2 Solutions — April 2, 2010
Problem 1. (25 points) Suppose we have A=
−1 0 2
1 1 −1
0 1 1
.
(a) (9 pts) Find an orthogonal basis for range(A).
(b) (8 pts) What is the dimension of the kernel of A?
(c) (8 pts) Find a basis for coker(A).
Solution:
(a) A=
−1 0 2
1 1 −1
0 1 1
→
−102
011
011
→
−102
011
000
.So range(A) =
span
−1
1
0
,
0
1
1
. Now using Gram Schmidt, starting with ~v1=
−1
1
0
, we have
~v2=~w2−h~w2, ~v1i
k~v1k2~v1=
0
1
1
−1
2
−1
1
0
=
1/2
1/2
1
so an orthogonal basis of range(A) is
−1
1
0
,
1/2
1/2
1
.
(b) By the Fundamental Theorem of Linear Algebra dim(ker A) = m−rank(A)=3−2 = 1
(c) Solve ATy= 0,
−1 1 0 0
0 1 1 0
2−1 1 0
→
−1 1 0 0
0 1 1 0
0 1 1 0
→
−1 1 0 0
0 1 1 0
0 0 0 0
→
−x+y= 0
y+z= 0
z=z
→
x=−z
y=−z
z=z
thus coker(A) = span
−1
−1
1
.
Problem 2. (25 points)
(a) (15 pts) Find the Gram matrix generated by the vectors p1(x) = xand p2(x) = x2with
respect to the L2inner product on [0,1]. Show all work.
(b) (10 pts) Is the Gram matrix described in part (a) positive definite? Why or why not?
Justify your answer.
Solution:
(a) Note hx, xi=R1
0x2dx =x3
3
1
0= 1/3 and hx, x2i=R1
0x3dx =x4
4
1
0= 1/4 and hx2, x2i=
R1
0x4dx =x5
5
1
0= 1/5 so
K=hx, xi hx, x2i
hx2, xi hx2, x2i=1/3 1/4
1/4 1/5
