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高等代数英文版Chapter 6.1

Chapter 6

Jordan canonical form of matrices In the previous chapter we have discussed the problem of reducing matrixes to diagonal matrices. In general, matrices that cannot be reduced to diagonal matrices can be reduced to block diagonal matrices, or the Jordan canonical form. In this chapter we shall consider this problem, specifically the following:

1.the necessary and sufficient condition for λ-matrices to be

equivalent.

2.methods of finding Jordan canonical forms.

This chapter is divided into four sections. The second and third sections deal with the first problem, the other sections with the second problem.

6.1. Necessary and sufficient condition for two

matrices to be similar

For a given matrix A, how does one find a simple matrix B such that A~B?

We proceed as follows. First find the necessary and sufficient condition for two matrices to be similar, and then using the sufficient condition find the matrix B required.

Suppose two matrices A and B are similar, i.e., A~B. Then there exists matrix P such that AP

=. Thus

B1-

P

P A E P B E )(1-=--λλ.

We easily prove A E -λ is equivalent to B E -λ, i.e.,B E A E -?-λλ. Conversely, if B E A E -?-λλ, we will prove that there exist constant matrices R and S such that

R A E S B E )(-=-λλ.

Consequently, SAR B SR E ==,, and hence A ~B . Thus we have:

Theorem 1. two matrices A and B are similar if and only if their characteristic matrices are equivalent:

B E A E -?-λλ.

This theorem is very important. It not only gives the necessary and sufficient condition for two matrices to be similar, but, more importantly, it changes the similarity relation into an equivalence relation.

It is difficult for us to deal with a similarity relation, but, using elementary operations, we can make the equivalence relation more specific. However we have given only the conclusion of the theorem, no its proof. Here the characteristic matrices are λ-matrices. However, the equivalence concept of λ-matrices has not been given either. Therefore before a proof of the above theorem we first have to define some concepts, such as the concepts of elementary operations on λ-matrices, the concept of equivalence of two λ-matrices.

The concept of the rank of a λ-matrices )(λA is the same as that of a constant matrix. Thus, the highest order of the minors of )(λA not

being identically zero is called the rank of a λ-matrices)

A. It is to be

noted that minors of )

A are polynomials inλ. By a polynomial in λbeing identically zero we mean that any number can be its root. Hence its coefficients are all zero, and we often say that the polynomial is identically equal to zero. Because any polynomial of degree n only can have n roots, it is not a null polynomial, i.e., not a polynomial vanishing identically.

Suppose the matrix A is of order n. Since A

λis a polynomial

E-

of degree n in λ, and not identically vanishing, the characteristic matrix A

λof A is of rank n. In other words, whether A is

E-

nonsingular or not the characteristic matrix of A is always nonsingular.

We know that the elements in a λ-matrices are polynomials in λ, and the power exponents of λin a λ-matrices can only be positive integer or zero. Owing to this constraint, elementary operations of λ-matrices are not quite the same as those of matrices with scalar elements as in Sec.2.5. The concrete definition is as follows.

Definition.The following operations are called elementary operations on a λ-matrices )

A:

(1). Interchanging two rows or two columns of )

A.

(2). Multiplying a row or a column by a nonzero number.

(3). Adding a row(column) multiplied by an arbitrary polynomial in λto another row(column).

Note that the operation of type (2) here is the same as in the case of constant matrices in Sec.2.5. They are multiplied by a nonzero constant. If we should be allowed to multiply )(λA by polynomial in λ, then since the elements in )(λA are polynomials in λ, the symmetric law of the equivalence relation might be destroyed. For instance, if we multiply

???

? ??000λ by λ we obtain ???? ??0002λ. But in order to get ???? ??000λ from ???

? ??0002

λ,we must divide the latter by λ. Divided by a polynomial in λ, which is obviously not allowed. The definition such as given above is formulated so that the equivalence relation of λ-matrices can satisfy the three equivalence laws as for constant matrices.

The elementary operations of type (1) follows from that of types (2) and (3), as in Se.2.5. We list them separately λ-matrices. It corresponds to elementary operations.

As in a matrices with scalar elements, two λ-matrices )(λA and )(λB are said to be equivalent, denoted by )()(λλB A ?, if there exists a finite sequence of elementary operations that can carry )(λA into )(λB . The equivalent relation of λ-matrices satisfies also three laws of equivalence. The ranks of two equivalent λ-matrices are equal.

By analogy with Theorem 5 in Sec. 3.3, we can at once give the following theorem from definition:

Theorem 2. Two λ-matrices

)(λA and )(λB of the same size

are equivalent if and only if there are two invertible λ-matrices )(λP and )(λQ such that

)()()()(λλλλQ A P B =.

The reason is that every elementary operation corresponds to a multiplication on the left side or the right side of )(λA by an elementary λ-matrix and any nonsingular λ-matrices can be expressed as a product of elementary λ-matrices.

If a matrix A has an inverse matrix, we say that A is invertible (or nonsingular). For matrices with scalar elements, a nonsingular matrix has an inverse matrix and is thus an inventible matrix. But a nonsingular λ-matrices is not necessarily an invertible matrix, which follows directly the following theorem.

T heorem 3. A λ-matrices )(λA is invertible if and only if )(λA is a nonzero constant.

Proof. If 0)(≠=c A λ, then the elements of the matrix *)(λA are minors of order 1-n of )(λA . After dividing by 0≠c we obtain

*)()

(1λλA A , Whose elements are polynomials in λ. Thus from Theorem 2 in Sec.3.3

*1)()

(1)(λλλA A A =-. Again 1)(-λA is a λ-matrices. Therefore )(λA is an invertible

λ-matrices.

Conversely, if )(λA is invertible, then its inverse matrix 1)(-λA is a λ-matrices. From the equality E A A =-)()(1λλ, we obtain

1)()(1==-E A A λλ, And so )(λA is a nonzero constant. The proof is complete.

Now we verify Theorem 1.

From Theorem 2, we at once obtain the necessary condition for Theorem 1. The sufficient condition for the theorem is proved as follow.

Before going on to prove the sufficient condition, we first present a basic property of λ-matrices which plays an important role in the proof.

In high school we learnt division algorithm: Dividing a polynomial )(x p by a x - we obtain a quotient )(x q and a remainder r :

()r x q a x x p +-=)()(,

Where r is a constant, and both )(x q and r are unique.

For any λ-matrices )(λA , by matrix addition and scalar multiplication, it can be written as a polynomial in λ with matric coefficients:

m m m A A A A +++=-10)( λλ,

Where m A A A ,,,10 are constant matrices. When 00≠A ,we say that the matrix )(λA is of degree m . Thus a λ-matrix of degree zero is simply a constant matrix.

The above result just given about division carries over for matric

polynomials, and the proof is analogous.

Theorem 4. Suppose )(λP is a λ-matrix, A is a constant matrix. Then there exist a λ-matrices )(λQ and a constant matrix R such that

R Q A E P +-=)()()(λλλ,

and )(λQ and R are unique.

Proof. Supposing

m m m P P P P +++=- 110)(λ

λλ, where m P P P ,,),0(10 ≠ are all constant matrices, we have

m m m m P P AP P P A E P ++++=----- 22

10110)()()(λλλλλ. The degree of λ-matrix on the righ-hand side does not exceed 1-m . If we can prove that its degree is zero, then the theorem is established. This can be done by an absurdity argument. If its degree is more that zero, we continue in the same way to decrease the degree of λ-matrix on the right-hand side, until it becomes zero. Combining the terms which contain A E -λ, we obtain

R Q A E P +-=)()()(λλλ,

where )(λQ is a λ-matrix and R a constant matrix. This proves the

existence of )(λQ and R , As for uniqueness, from

00)()()()(R Q A E R Q A E +-=+-λλλλ,

we obtain

R R Q Q A E -=--00)]()()[(λλλ.

Since the right-hand side of the above expression is a constant

matrix, we have )()(0λλQ Q =. Therefore 0R R =. Thus the theorem is established.

Similarly we have

11)()()(R Q A E P +-=λλλ,

where )(1λQ is a λ-matrix, 1R is a constant matrix and )(1λQ and 1R are both unique.

Note that since matrix multiplication is not generally commutative, the above is a constant matrix and )(λQ and )(1λQ are not necessarily identical. Similarly R and 1R are also not necessarily identical theorem, which can be used to prove readily the above formula (1).

Theorem 5. Suppose matrices A and C are nonsingular. If B A +λ is equivalent to D C +λ, i.e.,

)())((λλλλQ B A P D C +=+,

where )(λP and )(λQ are invertible, then there exist nonsingular constant matrices S and R such that

R B A S D C )(+=+λλ.

Proof. Suppose

S M D C P ++=)()()(λλλ,

R D C N Q ++=))(()(λλλ,

Where S and R are constant matrices. From the assumption we have

)()())((1λλλλQ B A D C P +=+-.

Substitution of the expression of )(λQ gives

}))((){())((1R D C N B A D C P +++=+-λλλλλ,

Or

R B A D C N B A P )())}(()()({1+=++--λλλλλ (2)

Comparing the degree of the two sides of the above identity we obtain a constant matrix

)()()(1λλλN B A P T +-=-,

Or

)()()()())(()(1λλλλλλλN Q D C E N B A P E T P -+-=+-=.

Thus

}.

)()()(){()()()(})(){()

()()()(111ST N Q T M D C N Q D C T S M D C N Q D C T P E +++=++++=++=---λλλλλλλλλλλλλ

Again comparing the degree of the two sides of the above identity we get

ST E =,or S T =-1.

Therefore, from (2) we obtain

R B A S D C )(+=+λλ,

where S is nonsingular. Similarly, R is also nonsingular. Thus the above theorem is completely proved.

Exercises

1. find the Jordan canonical forms of the following matrices

(1) ????? ??---112020021, (2) ????

? ??-----3104252373 (3) ??????? ??----01617121700140013, (4) ???????

? ??0101000 2.Find the rational canonical form of the matrix

????

? ??=011101110A .