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Solutions to practice problems for Test 3
1. Label the following statements as
true
or
false
.
true
(a) Every vector space that is spanned by a ﬁnite set has a ﬁnite basis.
false
(b) A vector space cannot have more than one basis.
true
(c) If a vector space has a ﬁnite basis, then the number of vectors in
every basis is the same.
false
(d) The dimension of
P
n
is
n
.
true
(e) There exists an isomorphism from
P
3
to
R
4
.
true
(f) There exists an isomorphism from
R
n
to
R
m
if and only if
n
=
m
.
true
(g) If
F
is a linear transformation from
R
5
to
R
3
and the dimension
of the kernel of
F
is 2, then
F
is onto.
false
(h) Suppose that
V
is a ﬁnitedimensional vector space, that
S
1
is a
linearly independent subset of
V
, and that
S
2
is a subset of
V
that spans
V
. Then
S
1
can contain more vectors than
S
2
.
true
(i) If
S
is a basis for the vector space
V
, then every vector in
V
can
be written as a linear combination of vectors in
S
in only one way.
true
(j) If
V
is a vector space having dimension
n
, and if
S
is a subset of
V
with
n
vectors, then
S
is linearly independent if and only if
S
spans
V
.
true
(k) If
X
and
Y
are two bases for a vector space
V
, then the change
of basis matrix
Y
I
X
is invertible and (
Y
I
X
)
−
1
=
X
I
Y
.
true
(l) If
F
is an isomorphism from a vector space
V
to a vector space
W
, then
X
is a basis of
V
if and only if
F
(
X
) is a basis of
W
.
false
(m) For any scalar
k
and square matrix
A
, det(
kA
) =
k
det(
A
).
true
(n) For any
n
×
n
matrices,
A
and
B
, (det
A
)(det
B
) = det(
AB
).
true
(o) For any
n
×
n
matrices,
A
and
B
, det(
AB
) = det(
BA
).
2. Use a matrix representation to ﬁnd bases for the kernel and the image of the
linear transformation
T
:
P
2
→
P
3
deﬁned by
T
(
a
+
bx
+
cx
2
) = (
a
+2
b
−
2
c
)+(2
a
+2
b
)
x
+(
−
a
+
b
−
4
c
)
x
2
+(3
a
+2
b
+2
c
)
x
3
Solution.
Let
X
= (1
,x,x
2
) be the standard basis of
P
2
and let
Y
=
(1
,x,x
2
,x
3
) be the standard basis on
P
3
. Then
Y
T
X
=
[
K
Y
(
T
(1))
K
Y
(
T
(
x
))
K
Y
(
T
(
x
2
))
]
=
[
K
Y
(1 + 2
x
−
x
2
+ 3
x
3
)
K
Y
(2 + 2
x
+
x
2
+ 2
x
3
)
K
Y
(
−
2
−
4
x
2
+ 2
x
3
)
]
=
1
2
−
2
2
2
0
−
1
1
−
4
3
2
2
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View Full Document The reduced row echelon form of
Y
T
X
is
1
0
2
0
1
−
2
0
0
0
0
0
0
A basis for the column space of
Y
T
X
is
B
=
1
2
−
1
3
,
2
2
1
2
So a basis for the image of
T
is
K
−
1
Y
(
B
) =
K
−
1
Y
1
2
−
1
3
,K
−
1
Y
2
2
1
2
= (1 + 2
x
−
x
2
+ 3
x
3
,
2 + 2
x
+
x
2
+ 2
x
3
)
The general solution to the equation
Y
T
X
⃗x
=
⃗
0 is
x
1
=
−
2
t, x
2
= 2
t, x
3
=
t.
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This document was uploaded on 06/18/2011.
 Spring '09
 Vector Space

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