Math 307: Problems for section 2.1
October 22, 2015
1.
Are the vectors
,

2
1
1
,

2
0
,

2
0
1
,

9
7
3
LAB/Octave to perform calculations, but explain your answer.
1
2
1
2
1
1
0
1

1
3
0
0
0
4
linearly independent? You may use MAT
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?
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
.
These are both of degree
n
, but their sum
p
1
(
x
) +
p
2
(
x
) = 1 is not a polynomial of degree
n
. So
the set is not closed under addition.