Part (d):
For an upper triangular matrix, the diagonal entries are the eigenvalues.
Problem
(
§
7.2 # 14)
.
Prove the converse of Exercise 13(d): If
T
is a linear operator
on an
n
dimensional vector space
V
and
(

1)
n
t
n
is the characteristic polynomial of
T
,
then
T
is nilpotent.
8
HOMEWORK 13 SOLUTIONS
Problem
(
§
7.2 # 17)
.
Let
T
be a linear operator on a finitedimensional vector space
V
such that the characteristic polynomial splits, and let
λ
1
, λ
2
, . . . , λ
k
be the distinct
eigenvalues of
T
. Let
S
:
V
→
V
be the mapping defined by
S
(
x
) =
λ
1
v
1
+
λ
2
v
2
+
· · ·
+
λ
k
v
k
,
where, for each
i
,
v
i
is the unique vector in
K
λ
i
such that
x
=
v
1
+
v
2
+
· · ·
+
v
k
. (This
unique representation is guaranteed by Theorem 7.3 (p. 486) and Exercise 8 of Section
7.1.)
(a)
Prove that
S
is a diagonalizable linear operator on
V
.
(b)
Let
U
=
T

S
. Prove that
U
is nilpotent and commutes with
S
, that is
SU
=
US
.
Proof.
Part (a):
Let
β
i
be a basis for
K
λ
i
(where
K
λ
i
is in terms of
T
as above). Then
put
β
=
β
1
∪ · · · ∪
β
k
. This is a basis for
V
, by Theorem 7.4, since the characteristic
polynomial of
T
splits.
But clearly, every vector in
K
λ
i
is an eigenvector of
S
(with
eigenvalue
λ
i
) and hence [
S
]
β
is a diagonal matrix. In particular,
S
is a diagonalizable
linear operator.
Part (b):
Let
U
=
T

S
. Let
β
i
(for each
i
) be as in part (a), except also assume
that it is a Jordan canonical basis for
K
λ
i
, and let
β
=
β
1
∪ · · · ∪
β
k
.
Then [
T
]
β
is
the Jordan canonical form for
T
, while [
S
]
β
is a diagonal matrix. Further, the diagonal
entries on both of these matrices agree, since they are just the eigenvalues of
T
, with the
correct multiplicities. Thus [
U
]
β
= [
T

S
]
β
= [
T
]
β

[
S
]
β
is an uppertriangular matrix,
with zeros down the diagonal.
In particular, from our work above, we know that the
characteristic polynomial of
U
is (

1)
n
t
n
, and hence
U
is nilpotent by Exercise 14 of this
section.
To show commutativity we need a lemma:
Lemma 1.
Let
A
=
λI
n
be the
n
×
n
identity matrix. Let
B
be the
n
×
n
matrix with
ones down the superdiagonal and zeros elsewhere. (In other words, a Jordan block for the
eigenvalue 0.) Then
A
and
B
commute.
Proof.
One easily calculates that
AB
=
λIB
=
λB
and
BA
=
BλI
=
λB
. This gives us
the lemma.
To finish the problem, notice that [
U
]
β
is just the matrix formed from [
T
]
β
by removing
the diagonal, and that [
S
]
β
consists of [
T
]
β
with everything above the diagonal removed.
So, breaking [
U
]
β
into Jordan blocks, we see that the eigenvalues along the diagonal in
[
S
]
β
match these blocks. Hence if we apply the lemma to each of these blocks, we see
that [
U
]
β
and [
S
]
β
commute on each block, and hence commute. This implies
U
and
S
commute.