Let be a field, be a vector space over with dimension, and be a linear transformation of. It is known that becomes a -module according to the following definition:

.

For a fixed linear transformation and a vector, the annihilator of with respective to is defined to be

.

Similarly, the annihilator of with respective to is defined to be

.

Since 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 in abstract algebras, respectively. They are also called the minimal polynomials of and with respective to in linear algebras, respectively. It is 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:

and are vectors in such that

where. Let be the characteristic polynomial of. By (1) and (2) one has

• ;

• .

Furthermore, these results straightforwardly imply the following theorem:

Theorem 1. [3,4] With the notations as above, we have 1) [Cayley-Hamilton Theorem]

, and so.

2) There exists a vector such that

.

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 arbitrary 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 an infinite field can not be an union of a finite number of its proper subspaces by Vandermonde determinants.

Lemma 1. Let be an infinite field, and be a vector space over with dimension, and be nontrivial subspaces of for. Then there exists infinite many bases of such that any element of them is not in each for. Therefore, if then for some i .

Proof: Let be a F-base of. For any we set

.

Let distinct elements in. We have

where is a Vandemonde matrix. So is a base of because the determinant of is nonzero. Let be the following set with an infinite number of vectors:

.

Since with is a nontrivial subspace of one can verify that. And so

.

Therefore, is infinite, and any distinct

vectors in the set constitute a base of.

Proposition 1. Let be an infinite field. Let be a -vector space with dimension, and be a linear transformation of. Then there exists a vector such that.

Proof: It is clear that are linearly dependent over. So the degree. For any, the minimal polynomial of is a monic factor of. So there exist finite number of vectors such that

where are mutually coprime irreducible polynomials. Set. One can verify that

.

By Lemma 1, there exists with such that. Which shows that

and so is a zero linear transformation. Hence we have.

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 be a field, be a -dimensional linear space over, and be a linear transformation of. For any, there exists such that

here and the following stand for the least common multiple and greatest common divisor of two polynomials, respectively.

Proof: By properly arrangement, the minimal polynomials of with respective to have the following irreducible factorization respectively,

,

.

Moreover, for, and for. So, we have

,

.

One can verify that the minimal polynomials of and are

respectively. Set, then

.

Which implies that

Conversely, from it follows that

.

Which shows that

So, since Similarly,. By again, we have

Equations (3) and (4) imply that

.

Proposition 2. Let be a field. Let be a -vector space with dimension and be a linear transform of. Then there exists a vector such that

.

Proof: Let be a -base of. One can verify that

.

By repeatedly utilization of Lemma 2, we can find a vector such that

According to Proposition 2, we can easily deduce the Cayley-Hamilton theorem.

Proof of Cayley-Hamilton Theorem: Let be the characteristic polynomial of. We show . By Proposition 2 there exists such that. Let

.

So, one can verify that vectors are linearly independent over. We extend them to a basis of as follows:

.

We have

where the square matrix has the form

and is an square matrix, and is an matrix. So the characteristic polynomial of is

and

.

Hence, , and.

Actually, the Cayley-Hamilton theorem can be obtained by only using the minimal polynomial of a vector.

Another Proof of Cayley-Hamilton Theorem: Let be the characteristic polynomial of. For any let be the minimal polynomial of the vector with respective to. To prove the CayleyHamilton theorem, it is enough to show that

.

This statement can be verified by the same arguments as that in above proof.

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.