May 11, 2007
Here is a brief overview of the main ideas of the statement and proof of
the theorem on rational normal form. Let
V
be a ﬁnite dimensional vector
space and let
T
be an operator on
V
.
A
T
invariant subspace
W
of
V
is said to be
cyclic
if there exists a
vector
w
∈
W
such that
W
is spanned by (
w, Tw, T
2
w,
···
). If this is the
case, and if
m
is the smallest integer such that
T
m
w
belongs to the span of
(
w, Tw, T
2
w,
···
, T
m

1
w
), then in fact
β
w
:= (
w, Tw, T
2
w,
···
, T
m

1
w
) is a
basis for
W
. The matrix for
T
W
with respect to this basis is the companion
matrix of the characteristic polynomial
f
T
W
of
T
W
; furthermore in this case
f
T
W
is also the minimal polynomial of
T
W
.
Theorem:
If
T
is an operator on a ﬁnite dimensional vector space
V
,
then
V
can be written as a direct sum of cyclic subspaces
W
i
. Moreover,
these subspaces can be chosen so that the characteristic polynomial of each
T
W
i
is a power of some irreducible factor of the minimal polynomial
p
T
of
T
.
Here is one way to attack the proof.
Let
p
T
:=
p
e
1
1
···
p
e
r
r
be the factorization of
p
T
into irreducible factors.
Lemma:
Let
V
i
:= Ker
p
e
i
i
(
T
). Then
V
=
V
1
⊕
V
2
··· ⊕
V
r
, and each
V
i
is nonzero.
Proof: Let
f
=
p
e
1
1
and let
g
=
Q
i>
1
p
e
i
i
. Since
f
and
g
have no common
factors, there are polynomials
p
and
q
such that
pf
+
qg
= 1. Now if
v
∈
V
, let
v
g
:=
p
(
T
)
f
(
T
)(
v
) and let
v
f
:=
q
(
T
)
g
(
T
)(
v
). Then
v
=
v
f
+
v
g
.
Moreover,
f
(
T
)(
v
f
) =
f
(
T
)
g
(
T
)
q
(
T
)(
v
) = 0, so
v
f
∈
Ker
(
f
(
T
)). Similarly
v
g
∈
Ker
(
g
(
T
)). Thus
V
= Ker(
f
(
T
)) + Ker(
g
(
T
)). Moreover, if
v
∈
Ker(
f
(
T
))
∩
Ker(
g
(
T
)), then
v
=
p
(
T
)
f
(
T
)(
v
) +
q
(
T
)
g
(
T
)(
v
) = 0. The
same equation shows that
q
(
T
) is inverse to
g
(
T
) on Ker(
f
(
T
)). Note that if