110.201 Quiz 3 Solutions: Friday
March 28, 2005
Problem 1
1. Yes, the map is linear. It is not an isomorphism; its kernel contains,
for example,
f
(
x
) =
x
, which means the map is not onetoone
2. The map is linear; this much is obvious. It is also clearly injective, and
since an injective linear map
T
:
U
→
V
between linear spaces of the
same finite dimension is an isomorphism, the map is an isomorphism.
3. The map is not linear.
T
(
kA
) =
k
2
T
(
A
) for all real
k
.
Problem 2
1. There are several ways to solve this problem.
One is the following:
Let
B
2
=
{
x
4
,
2
x
3

1
,
1

x
2
,
3
x

1
,
2
x
} ≡ {
f
1
, f
2
, f
3
, f
4
, f
5
}
. Observe
that
1
=
(3
/
2)
f
5

f
4
(1)
x
=
f
5
/
2
(2)
x
2
=
(3
/
2)
f
5

f
4

f
3
(3)
x
3
=
f
2
/
2
(4)
x
4
=
f
1
(5)
Therefore span
{
f
1
, f
2
, . . . , f
5
}
= span
{
1
, x, x
2
, x
3
, x
4
}
=
P
4
. It follows
immediately that the
f
i
are a basis of
P
4
, since there are only five of
them and dim
P
4
= 5. The changeofbasis matrix from
B
1
to
B
2
can
be found immediately by writing the elements of
B
2
in terms of the
standard basis:
1
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f
1
f
2
f
3
f
4
f
5
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 Spring '05
 CONSANI
 Linear Algebra, Algebra, Vector Space, basis, F3

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