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