MTH 235 : Linear Algebra
Sample problems  Midterm 2
1. Prove that
T
is a linear transformation, and find bases for the kernel (null space) and
the range of
T
. Is
T
onetoone? Is
T
onto?
(a)
T
:
R
3
→
R
2
defined by
T
(
a
1
, a
2
, a
3
) = (
a
1

a
2
,
2
a
3
).
Solution.
Let
α
= (
a
1
, a
2
, a
3
) and
β
= (
b
1
, b
2
, b
3
) be two vectors in
R
3
and
c
∈
R
any scalar.
T
(
cα
+
β
) =
T
(
ca
1
+
b
1
, ca
2
+
b
2
, ca
3
+
b
3
)
= ((
ca
1
+
b
1
)

(
ca
2
+
b
2
)
,
2(
ca
3
+
b
3
))
= (
c
(
a
1

a
2
) + (
b
1

b
2
)
, c
(2
a
3
) + 2
b
3
)
=
c
(
a
1

a
2
,
2
a
3
) + (
b
1

b
2
,
2
b
3
)
=
cT
(
α
) +
T
(
β
)
If
{
α
1
, α
2
, α
3
}
is a basis for
R
3
, then the range is the span of
{
T
(
α
1
)
, T
(
α
2
)
, T
(
α
3
)
}
.
Take the standard ordered basis,
{
(1
,
0
,
0)
,
(0
,
1
,
0)
,
(0
,
0
,
1)
}
of
R
3
.
T
(1
,
0
,
0) = (1
,
0)
T
(0
,
1
,
0) = (

1
,
0)
T
(0
,
0
,
1) = (0
,
2)
Hence,
{
(1
,
0)
,
(0
,
2)
}
is a basis for
R
T
. In particular, dim
R
T
= 2 = dim
R
2
so
T
is onto.
By the ranknullity theorem, dim
N
T
= 3

2 = 1.
It suffices
to provide one nonzero vector in the kernel. We will use linearity of
T
. Note
T
(1
,
0
,
0) +
T
(0
,
1
,
0) = (1
,
0) + (

1
,
0) = (0
,
0). In particular,
{
(1
,
1
,
0)
}
is a
basis for
N
T
. Since there is a nontrivial kernel,
T
is not onetoone.
1
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(b)
T
:
R
2
→
R
3
defined by
T
(
a
1
, a
2
) = (
a
1
+
a
2
,
0
,
2
a
1

a
2
).
Solution.
Let
α
= (
a
1
, a
2
) and
β
= (
b
1
, b
2
) be two vectors in
R
2
and
c
∈
R
any
scalar.
T
(
cα
+
β
) =
T
(
ca
1
+
b
1
, ca
2
+
b
2
)
= ((
ca
1
+
b
1
) + (
ca
2
+
b
2
)
,
0
,
2(
ca
1
+
b
1
)

(
ca
2
+
b
2
))
= (
c
(
a
1
+
a
2
) + (
b
1
+
b
2
)
,
0
, c
(2
a
1

a
2
) + (2
b
1

b
2
))
=
c
(
a
1
+
a
2
,
0
,
2
a
1

a
2
) + (
b
1
+
b
2
,
0
,
2
b
1

b
2
)
=
cT
(
α
) +
T
(
β
)
If
{
α
1
, α
2
}
is a basis for
R
2
, then the range is the span of
{
T
(
α
1
)
, T
(
α
2
)
}
. Take
the standard ordered basis,
{
(1
,
0)
,
(0
,
1)
}
of
R
2
.
T
(1
,
0) = (1
,
0
,
2)
T
(0
,
1) = (1
,
0
,

1)
Hence,
{
(1
,
0
,
2)
,
(1
,
0
,

1)
}
is a basis for
R
T
.
In particular, dim
R
T
= 2
6
=
dim
R
3
so
T
is not onto. By the ranknullity theorem, dim
N
T
= 2

2 = 0 so
that
N
T
=
{
0
}
and
T
is onetoone.
2. Let
T
:
V
→
W
be a linear transformation. Prove that ker
T
⊂
V
is a subspace.
Proof.
Let
α, β
∈
ker
T
, i.e.,
T
(
α
) = 0 =
T
(
β
). For
c
∈
F
, we would like to show
that
cα
+
β
∈
ker
T
, i.e.,
T
(
cα
+
β
) = 0. Since
T
:
V
→
W
is linear:
T
(
cα
+
β
) =
cT
(
α
) +
T
(
β
) =
c
·
0 + 0 = 0
3. Prove that a linear transformation
T
:
V
→
W
is onetoone if and only if ker
T
=
{
0
}
.
Proof.
“
⇒
” Suppose that the linear transformation
T
:
V
→
W
is onetoone.
Suppose that
α
∈
ker
T
, then by definition,
T
(
α
) = 0. In particular,
T
(0) = 0 =
T
(
α
).
Since
T
assumed to be onetoone,
α
= 0.
“
⇐
” Suppose ker
T
=
{
0
}
. If
T
(
α
) =
T
(
β
), then
T
(
α
)

T
(
β
) = 0. Since
T
linear,
T
(
α

β
) = 0. Since the kernel is trivial,
α

β
= 0 so that
α
=
β
.
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 Spring '10
 aa
 Linear Algebra

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