SOLUTION TO ASSIGNMENT 3  MATH 251, WINTER 2007
1. Let
T
:
V
→
V
be a nilpotent linear operator. Prove that if
n
= dim(
V
) then
T
n
≡
0. Show that for
every
n
≥
2 there exists a vector space
V
of dimension
n
and a nilpotent linear operator
T
:
V
→
V
such
that
T
n

1
6≡
0.
Proof.
We prove the result by induction on
n
. Suppose that
n
= 1, we then have a linear map
T
:
F
→
F
.
T
cannot be surjective (else it is an isomorphism), hence its image must be the zero subspace, which shows
that
T
=
T
1
= 0. Now for the induction step.
Again,
T
:
V
→
V
cannot be surjective, because it would then be an isomorphism and in particular
T
n
is an isomorphism for every
n
. Thus,
W
= Im(
T
) is of dimension at most
n

1. Note that
T
(
W
)
⊂
T
(
V
) =
W
so we may view
T
as a map from
W
to itself.
Now
T
n

1
(
W
) =
{
0
}
by induction and so
T
n
(
V
) =
T
n

1
(
T
(
V
)) =
T
n

1
(
W
) =
{
0
}
.
To show this result is optimal consider the linear transformation on
F
n
given by
T
(
α
1
, . . . , α
n
) = (0
, α
1
, . . . , α
n

1
)
.
One finds that
T
n

1
(
α
1
, . . . , α
n
) = (0
, . . . ,
0
, α
1
) and so
T
n

1
is not the zero transformation, though
T
is
nilpotent. (Clearly
T
n
= 0.)
/
2. (a) Find a linear map
T
:
R
3
→
R
3
whose image is generated by (1
,
2
,
3) and (3
,
2
,
1). Here ‘find’ means
represent by a matrix with respect to the standard basis.
(b) Find a linear map
T
:
R
4
→
R
3
whose kernel is generated by (1
,
2
,
3
,
4)
,
(0
,
1
,
0
,
1).
Proof.
(a) Let us define
T
by
T
(1
,
0
,
0) = (1
,
2
,
3)
, T
(0
,
1
,
0) = (3
,
2
,
1)
, T
(0
,
0
,
1) = (0
,
0
,
0), or in terms of
a matrix
T
(
x
1
, x
2
, x
3
) =
1
3
0
2
2
0
3
1
0
.
Clearly the image is spanned by (1
,
2
,
3)
,
(3
,
2
,
1).
(b) We claim that
B
=
{
(1
,
2
,
3
,
4)
,
(0
,
1
,
0
,
1)
,
(0
,
1
,
0
,
0)
,
(0
,
0
,
1
,
0)
}
is a basis for
R
4
. It is enough to
show that
S
spans
R
4
(then it must be a minimal spanning set).
For that, it is enough to show that
e
1
, . . . , e
4
are in Span(
B
). For
e
2
, e
3
we have it by construction. We also
e
4
as (1
,
2
,
3
,
4)

4(0
,
1
,
0
,
1) +
2(0
,
1
,
0
,
0)

3(0
,
0
,
1
,
0). Once we know
e
2
, e
3
, e
4
are in the span it also follows that (1
,
2
,
3
,
4)

2(0
,
1
,
0
,
0)

3(0
,
0
,
1
,
0)

4(0
,
0
,
0
,
1) is in the span.
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 Spring '08
 Goren
 Linear Algebra, Vector Space, linear transformation, Standard basis, nilpotent linear operator

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