Solutions to Section 4.3 Homework Problems Problems 1-15 (odd ...

Solutions to Section 4.3 Homework Problems
Problems 1-15 (odd), 19, 21, 22, 23, 27, 33, 34, 37, and 38.
S. F. Ellermeyer
1. For the three vectors listed, the second one is not a multiple of the first, and
the third one cannot possibly be a linear combination of the first two because
there would be no way to create the 1 in the third entry using a linear
combination of the first two. Thus, according to Theorem 4 on page 247 of
the textbook, this set of vectors is linearly independent. Any linearly
independent set of three vectors in 3 is a basis for 3 . Thus, this set of
vectors is a basis for 3 .
3. Observe that
1 3 3
0 2 5
2 4 1
~
1 0 9
2
0 1 5
2
0 0 0
which shows that the given set of vectors is not linearly independent and
does not span 3 .
5. This is a set of four vectors in 3 , so it must be linearly dependent.
(Furthermore, it contains the zero vector.) Note that
1 2 0 0
3 9 0 3
0 0 0 5
~
1 0 0 0
0 1 0 0
0 0 0 1
so this set of vectors does span 3 (because every row in the above matrix
is a pivot row).
7. This set of vectors does not span 3 because it contains only two vectors. It
is a linearly independent set, because neither vector is a scalar multiple of
the other.
9.
A
1 0 3 2
0 1 5 4
3 2 1 2
~
,
1 0 3 2
0 1 5 4
0 0 0 0
shows that the equation Ax 03 has general solution
x
x1
x2
x3
x4
Thus, a basis for nulA is
3t 2s
5t 4s
t
s
t
3
5
1
0
s
,
2
4
0
1
.
1

3
5
1
0
,
2
4
0
1
11. The equation x 2y z 0 can be written as
The matrix
1 2 1
x
y
z
A 1 2 1
.
0.
has reduced row echelon form, and we see that the general solution of the
above equation is
x
x
y
z
Thus, a basis for nulA is
13. Since
a basis for nulA is
A~
2t s
t
s
2
1
0
6
5
2
1
0
,
t
1
0
1
1 0 6 5
0 1 5
2
3
2
0 0 0 0
,
5
3
2
0
1
The pivot columns of A form a basis for colA, so a basis for colA is
2
1
0
,
.
.
s
1
0
1
.
2

2
2
3
15. The space spanned by the given set of five vectors is the column space of
the matrix
Since
A
A~
,
4
6
8
1 0 3 1 2
0 1 4 3 1
3 2 1 8 6
2 3 6 7 9
1 0 3 0 4
0 1 4 0 5
0 0 0 1 2
0 0 0 0 0
and the pivot columns of A form a basis for colA, we see that a basis for
colA is
1
0
3
2
,
19. Since v1 3v2 5v3, we see that Spanv1,v2,v3 Spanv2,v3. Also, the
set v2,v3 is linearly independent because neither of these vectors is a
scalar multiple of the other. Therefore, the set v2,v3 is a basis for
H Spanv1,v2,v3.
21.
a. False. A set consisting of a single vector is in fact linearly
independent, unless that vector is the zero vector, in which case the
set is linearly dependent.
b. False. It might be the case that the set b1,b2,,bp is linearly
dependent (as in problem 19 above).
c. True.
d. False. A basis is a spanning set that is as small as possible.
e. False. Elementary row operations do not change linear
dependence relations that exist among the columns of a matrix.
22.
a. False. In order to be a basis for H, a set must both be linearly
0
1
2
3
,
1
3
8
7
.
.
.
3

independent and span H.
b. True.
c. True.
d. False.
e. False. The pivot columns of A form a basis for colA.
23. Any set of four vectors in 4 that spans 4 must also be linearly
independent.
27. A basis for the vector space described here is f,g where f is the function
ft sint and gt cost.
33. This set is linearly independent because neither function is a scalar multiple
of the other.
34. By inspection, we see that p1 p2 p3 z so the set p1,p2,p3 is linearly
dependent. We can throw out p3 and still have a set that spans
Spanp1,p2,p3. In other words, Spanp1,p2 Spanp1,p2,p3. Furthermore,
the set p1,p2 is linearly independent. Therefore, the set p1,p2 is a basis
for Spanp1,p2,p3.
37. Suppose that
c1t c2 sint c3 cos2t c4 sintcost 0 for all t .
Since the above equation must hold for t 0, then
c1 0 c2 sin0 c3 cos0 c4 sin0cos0 0,
which means that
c3 0.
Since the above equation must hold for t , then
c1 c2 sin c3 cos2 c4 sincos 0,
which means that
c1 c3 0,
which, since we have already decided that c3 0, means that
c1 0.
Now, our equation has become
c2 sint c4 sintcost 0 for all t .
Since the above equation must hold for t /6 and t /3, then
and
which means that
c2 sin 6
c2 sin 3
c4 sin 6
c4 sin 3
cos 6
cos 3
0
0
4

1
2 c2
3
2 c2
3
4 c4 0
3
4 c4 0.
The only solution of this system of equations is c2 c4 0.
We have now shown that if
c1t c2 sint c3 cos2t c4 sintcost 0 for all t ,
then it must be the case that c1 c2 c3 c4 0. This shows that the set of
functions f1,f2,f3,f4 defined by
f1t t
f2t sint
f3t cos2t
f4t sintcost
is a linearly independent set of functions.
5