Math 307: Problems for section 2.1
October 4, 2009
1.
Are the vectors
1
2
1
2
1
,
1
0
−
2
1
1
,
1
−
1
3
−
2
0
,
0
0
−
2
0
1
,
0
4
−
9
7
3
linearly independent? You may use MAT-
LAB/Octave to perform calculations, but explain your answer.
Put the vectors in the columns of matrix and perform row reduction
V1 = [1 2 1 2 1]’;
V2 = [1 0 -2 1 1]’;
V3 = [1 -1 3 -2 0]’;
V4 = [0 0 -2 0 1]’;
V5 = [0 4 -9 7 3]’;
A=[V1 V2 V3 V4 V5];
rref(A)
ans =
1.00000
0.00000
0.00000
0.00000
1.00000
0.00000
1.00000
0.00000
0.00000
1.00000
0.00000
0.00000
1.00000
0.00000 -2.00000
0.00000
0.00000
0.00000
1.00000
1.00000
0.00000
0.00000
0.00000
0.00000
0.00000
Since the matrix has rank 4 there are only 4 independent vectors in the list, and the vectors are linearly
dependent. In fact a vector in the nullspace of
A
is
−
1
−
1
2
−
1
1
so
−
V
1
−
V
2 + 2
V
3
−
V
4 +
V
5 = 0 which
we can check:
-V1 - V2 + 2*V3 - V4 + V5
ans =
0
0
0
0
0
1

2.
Which of the following sets are subspaces of the vector space
V
? Why, or why not?
(a) The set
S
=
{
(
b
1
, b
2
, b
3
) :
b
1
= 0;
b
2
, b
3
∈
R
}
. (
V
=
R
3
)
(b) The set
S
=
{
(
b
1
, b
2
, b
3
) :
b
1
b
2
= 0
, b
3
∈
R
}
.
This is union of the plane
b
1
= 0
and the
plane
b
2
= 0
. (
V
=
R
3
)
(c) All infinite sequences
(
x
1
, x
2
, . . .
)
, with
x
i
∈
R
and
x
j
= 0
from some fixed point
onwards. (
V
=
R
∞
)
(d) All non-increasing sequences
(
x
1
, x
2
, . . .
)
, with
x
i
∈
R
and
x
j
+1
≤
x
j
for each
j
.
(
V
=
R
∞
)
(e) The set of all polynomial functions,
p
(
x
)
, where
p
(
x
) = 0
or
p
(
x
)
has degree
n
for some
fixed
n
≥
1
. (
V
is the vector space of all polynomials.)
(f) The set of odd continuous functions on the interval
[
−
1
,
1]
, i.e.,
f
∈
C
[
−
1
,
1]
such that
f
(
−
x
) =
−
f
(
x
)
. (
V
=
C
[
−
1
,
1]
)
(a)
S
is a subspace.
The zero vector, (0
,
0
,
0), is in
S
.
S
is also closed under addition and scalar
multiplication.
Let
x
1
= (0
, y
1
, z
1
) and
x
2
= (0
, y
2
, z
2
) be two arbitrary vectors in
S
and let
c
∈
R
. Then
x
1
+
x
2
= (0
, y
1
+
y
2
, z
1
+
z
2
) is also in
S
and
c
x
1
= (0
, cy
1
, cz
1
) is also in
S
.
(b)
S
is not a subspace. It is not closed under addition. Take for example the vectors (0
,
1
,
1) and
(1
,
0
,
1). They are both in
S
but their sum, (1
,
1
,
2), is not in
S
.
(c) This is a subspace. The zero vector clearly has
x
j
= 0 from some point onwards. If we add two
vectors which both have
x
j
= 0 from some point onwards then the sum will also have
x
j
= 0 from
the same point onwards. Also, if we multiply a vector with
x
j
= 0 from some point onwards by a
scalar, the result will still have
x
j
= 0 from the same point on. So the set is closed under addition
and scalar multiplication.
(d) This is not a subspace. It is not closed under scalar multiplication. Take for example the vector
(2
,
1
,
0
,
0
,
0
, . . .
). If we multiply this by (
−
1) we get (
−
2
,
−
1
,
0
,
0
,
0
, . . .
), which is no longer non-
increasing.
(e) This is not a subspace. Take for example the two polynomials
p
1
(
x
) =
x
n
+ 1 and
p
2
(
x
) =
−
x
n
.