Advances in Linear Algebra & Matrix Theory, 2012, 2, 48-50
doi:10.4236/alamt.2012.24008 Published Online December 2012 (http://www.SciRP.org/journal/alamt)
On the Minimal Polynomial of a Vector
Dabin Zheng, Heguo Liu
Faculty of Mathematics and Computer Science, Hubei University, Wuhan, China Email: dzheng@hubu.edu.cn, ghliu@hubu.edu.cn
Received October 26, 2012; revised November 30, 2012; accepted December 9, 2012
ABSTRACT
It is well known that the Cayley-Hamilton theorem is an interesting and important theorem in linear algebras, which was first explicitly stated by A. Cayley and W. R. Hamilton about in 1858, but the first general proof was published in 1878 by G. Frobenius, and numerous others have appeared since then, for example see [1,2]. From the structure theo- rem for finitely generated modules over a principal ideal domain it straightforwardly follows the Cayley-Hamilton theorem and the proposition that there exists a vector v in a finite dimensional linear space V such that v and a linear transformation of V have the same minimal polynomial. In this note, we provide alternative proofs of these results by only utilizing the knowledge of linear algebras.
Keywords: Finite Dimensional Linear Space; Linear Transformation; Minimal Polynomial
1. Introduction
Let
Fbe a field,
Vbe a vector space over
Fwith dimension n , and be a linear transformation of . It is known that becomes a
V V F x
-module according to the following definition:
F x V V
f x , v
f v .
For a fixed linear transformation and a vector , the annihilator of with respective to
vV
v is de-
fined to be
0ann v p x F x p
v .
Similarly, the annihilator of
Vwith respective to is defined to be
0 ,ann V p x F x p
v v V .
Since
F x is a principal ideal domain the ideals and can be generated by the unique monic polynomials, denote them by and
, respectively. Which are called the order ideals of and
Vin abstract algebras, respectively. They are also called the minimal polynomials of and
Vwith respective to
ann v
m x
v
ann V
mv x
in linear algebras, respectively. It is v clear that the minimal polynomial of zero vector (or zero transformation) is 1. By the structure theorem for finitely generated modules over a principal ideal domain [3,4], the module can be decomposed into a direct sum of finite cyclic submodules:
V
1
2
sV F x
F x
F x , (1) and
1, 2,,
sare vectors in
Vsuch that
i i ,
i
i1
ann d x d x d
x , s
(2) where i 1, 2,
, 1 . Let
x be the characteristic polynomial of . By (1) and (2) one has
m
x ann
s d
s x ;
1 1
s s
i i
i i
x ann d x
.
Furthermore, these results straightforwardly imply the following theorem:
Theorem 1 [3,4] With the notations as above, we have 1) [Cayley-Hamilton Theorem]
m
x
x , and so
0. 2) There exists a vector v V such that
mv x m x
.
2. Proofs Based on Linear Algebras
In this section we give an alternative proof of Theorem 1 by only utilization of knowledge of linear algebras. To demonstrate an interesting proof of some proposition in linear algebras and its applications, we present two proofs of (2) in Theorem 1 for infinite fields and arbi- trary fields, respectively, and then use the related results to prove the Cayley-Hamilton theorem.
The following lemma provide an interesting proof of an proposition in linear algebras that a vector space over
Copyright © 2012 SciRes. ALAMT
D. B. ZHENG, H. G. LIU 49
an infinite field can not be an union of a finite number of its proper subspaces by Vandermonde determinants.
Lemma 1. Let F be an infinite field, and be a vector space over
V F with dimension , and V
ibe nontrivial subspaces of for i . Then there exists infinite many bases of such that any element of them is not in each for . Therefore, if
then
Vfor some i . n
, , V
V
iVi
, n
1, 2,
1, 2, i
s V
s
1V
i
Proof: Let
s
V
i 1, 2,
n
be a F-base of . For any we set
V
b F
2 1
1 2 3
n
b b b b
. Let
b b1, 2,,bndistinct elements in
F. We have
1, 2, ,
1, 2, ,
1, 2, ,b b bn n Van b b bn
where is a Vandemonde matrix. So
1 2
b b
1, 2, , n Van b b b, , ,
bn
b b1, 2,,bnis a base of because the determinant of is nonzero. Let be the follow- ing set with an infinite number of vectors:
VVan S
b
S
bF s.
Since
iwith is a nontrivial subspace of one can verify that
V i1, 2,,
V SVi n 1
. And so
1 1
1
s s
i i
i i
S V S V s n
.
Therefore, is infinite, and any distinct
1
\
s i i
S S V
n vectors in the set constitute a base of
V.
Proposition 1. Let
Fbe an infinite field. Let be a
V F
-vector space with dimension n , and be a lin- ear transformation of
V. Then there exists a vector
such that
vV v
m
x , 2, m x .
Proof: It is clear that
1,
,
n2are linearly de- pendent over
F. So the degree deg
m
x
n2. For any , the minimal polynomial of is a monic factor of . So there exist finite number of
vectors such that
vV v ii
xv
mv
m x
2,
,
s,
1,
v1
v2
vs
m x m x m x m x
, where
vi
are mutually coprime irreducible poly- nomials. Set
m x s
i 0
i v
V V m . One can verify that
1 2 s
V V
V
V .
By Lemma 1, there exists with such that . Which shows that
k 1 k s V V
k 0, for all
vk
m V ,
and so m
vk is a zero linear transformation. Hence we have m x m
x .
vk
In fact, Proposition 1 holds for arbitrary fields from
the introduction. To obtain a general proof we first give the following lemma.
Lemma 2. Let
Fbe a field, be a -dimensional linear space over
V n
F
, and be a linear transformation of V . For any
0
, V, there exists V such that
,
m
x lcm m
x m
x ,
here and the following stand for the least common multiple and greatest common divisor of two polynomials, respectively.
lcm gcd
Proof: By properly arrangement, the minimal poly- nomials of
,with respective to have the fol- lowing irreducible factorization respectively,
1 1
1 2
1 r r1 s
k
k k k
r r s
u x u x
m
x
p
p
p
p ,
1 1
1 2
1 1
s
r r l
l l l
r r s
v x v x
m
x p
p p
p
.
Moreover, k
i l
ifor i 1, 2,
, r , and k
i l
ifor 1, ,
i r
s . So, we have
,
1
2lcm m
x m
x u x v x ,
1 2
gcd u x v , x 1 .
One can verify that the minimal polynomials of
2
u and v
1 are
2 1
,
1 2u v
m
x u x m
x v x , respectively. Set u
2 v
1 , then
1 2 0
u
v
. Which implies that
1 2m
x u x v x . (3) Conversely, from u
2 v
1 it follows that
v1
2
0 m
m
u
. Which shows that
2 1 , . . 1 2
u v
m x m x m x i e u x m x v x
So,
u x m1
xsince
gcd
u x v1
, 2 x
1Simi- larly,
v2
x m
x. By
gcd
u x v1
, 2 x
1again, we have
1 2
u x v x m
x . (4) Equations (3) and (4) imply that
1
2
m x u x v x lcm m x m x
. Proposition 2. Let
Fbe a field. Let
Vbe a
Copyright © 2012 SciRes. ALAMT
D. B. ZHENG, H. G. LIU
Copyright © 2012 SciRes. ALAMT 50
F
-vector space with dimension n and be a linear transform of
V. Then there exists a vector
vVsuch that
and
Cis an n m square matrix, and
Xis an
m
m n
matrix. So the characteristic polynomial of
is
n 0 m
B X mv x m x
.
x xI xI B xIn m C
C
,
Proof: Let
1, 2,,
nbe a
F-base of
V. One
can verify that and
1
2m x lcm m x m,
x ,
x
mv
,mn
x
V lcm
1
1 1
m m
m m 0
xI B x b x b x b
. .
Hence,
m
x x, and
0. By repeatedly utilization of Lemma 2, we can find a
vector
vsuch that
1
, 2
, ,
m x m x m x mn
x .Actually, the Cayley-Hamilton theorem can be ob- tained by only using the minimal polynomial of a vector.
Another Proof of Cayley-Hamilton Theorem: Let
x
be the characteristic polynomial of . For any
vVlet
mv
xbe the minimal polynomial of the vector v with respective to . To prove the Cayley- Hamilton theorem, it is enough to show that
According to Proposition 2, we can easily deduce the Cayley-Hamilton theorem.
Proof of Cayley-Hamilton Theorem: Let
xbe the characteristic polynomial of . We show
m
v x
x for any v V .
m
x
x . By Proposition 2 there exists
vVsuch that
m
x. Let
m
v m
x
v
mm x x xmb 1xm1 b1
x b0
This statement can be verified by the same arguments as that in above proof.
. So, one can verify that vectors
v,
v ,,
m1
vare linearly independent over
F. We extend them to a basis of
Vas follows:
3. Acknowledgements
1
, , , m , ,
v
v
v ,
1, ,
n m
n m
0 0 0 1
n mB X C
. We have
1 1
1 1
, , , ,
0
m
m
v v
v v
, ,
, ,
The authors would like to thank the anonymous referees for helpful comments. The work of both authors was supported by the Fund of Linear Algebras Quality Course of Hubei Province of China. The work of D.
Zheng was supported by the National Natural since Foundation of China (NSFC) under Grant 11101131.
REFERENCES
where the m square matrix
Bhas the form
0 1 2
1
0 0
1 0
0 1
0 0 m
b b
B b
b
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