Math 136
Term Test 2 Solutions
1.
Short Answer Problems
[1] a) What is the deﬁnition of a basis
B
of a vector space
V
?
Solution: A basis is a linearly independent spanning set.
[1] b) What is the deﬁnition of the dimension of a vector space
V
?
Solution: The dimension of a vector space is the number of vectors in any basis for
V
.
[2] c) Find a basis for
R
3
that includes the vectors
1
2
3
,
3
2
1
. Justify.
Solution: Since
1
2
3
,
3
2
1
is clearly linearly independent and the dimension of
R
3
is 3,
we just need to ﬁnd one vector in
R
3
which does not lie in the span of the vectors. We
know that (1
,
2
,
3)
×
(3
,
2
,
1) = (

4
,

8
,

4) does not lie in the plane spanned by (1
,
2
,
3)
and (3
,
2
,
1), hence a basis for
R
3
that includes these vectors is
1
2
3
,
3
2
1
,

4

8

4
[2] d) Let
~a
= (1
,
2)
∈
R
2
, and
L
:
R
2
→
R
2
be deﬁned by
L
(
~x
) = 2
~x

(
~x
·
~a
)
~a.
Determine if
L
is linear.
Solution: Let
~x,~
y
∈
R
2
and
k
∈
R
, then
L
(
k~x
+
~
y
) = 2(
k~x
+
~
y
)

([
k~x
+
~
y
]
·
~a
)
~a
= 2
k~x
+ 2
~
y

(
k
[
~x
·
~a
] + [
~
y
·
~a
])
~a
=
k
[2
~x

(
~x
·
~a
)
~a
] + 2
~
y

(
~
y
·
~a
)
~a
=
kL
(
~x
) +
L
(
~
y
)
Hence
L
is linear.
[2] e) Determine the standard matrix of a linear mapping
L
:
R
2
→
R
3
whose range
is span
{
(1
,
2
,
1)
}
and whose nullspace is span
{
(

1
,
2)
}
.
Solution: Since the range is span
{
(1
,
2
,
1)
}
all columns must be multiples of (1
,
2
,
1). Since
the nullspace is span
{
(

1
,
2)
}
, all columns must be multiples of (2
,
1). Hence, the standard
matrix could be any scalar multiple of
2 1
4 2
2 1
.
1
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2.
Let
T
:
R
3
→
R
2
be deﬁned by
T
(
x
1
, x
2
, x
3
) = (
x
1
+
x
3
,
0).
[2] a) Prove that
T
is linear.
Solution: Let
~x
= (
x
1
, x
2
, x
3
),
~
y
= (
y
1
, y
2
, y
3
) and
k
∈
R
. Then
T
(
k~x
+
~
y
) =
T
(
k
(
x
1
, x
2
, x
3
) + (
y
1
, y
2
, y
3
))
=
T
(
kx
1
+
y
1
, kx
2
+
y
2
, kx
3
+
y
3
)
= (
kx
1
+
y
1
+
kx
3
+
y
3
,
0)
=
k
(
x
1
+
x
3
,
0) + (
y
1
+
y
3
,
0)
=
kT
(
~x
) +
T
(
~
y
)
[3] b) Find the standard matrix of
T
.
Solution: We have
T
(1
,
0
,
0) = (1
,
0)
T
(0
,
1
,
0) = (0
,
0)
T
(0
,
0
,
1) = (1
,
0)
Hence
[
T
] =
±
T
(1
,
0
,
0)
T
(0
,
1
,
0)
T
(0
,
0
,
1)
²
=
³
1 0 1
0 0 0
´
.
[2] c) Find a spanning set for the nullspace of
T
.
Solution: The nullspace of
T
is the set of all (
x
1
, x
2
, x
3
) such that
T
(
x
1
, x
2
, x
3
) = (0
,
0).
Hence, we must have
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 Spring '08
 All
 Linear Algebra, Algebra, Vector Space

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