Page 224 - DMTH502_LINEAR_ALGEBRA

Linear Algebra

Notes Proof: Let T be the linear operator on F which is represented by B in the standard ordered basis.
n
As we have just observed, there is some ordered basis for F in which T is represented by a
n
matrix A in rational form. Then B is similar to this matrix A. Suppose B is similar over F to
another matrix C which is in rational form. This means simply that there is some ordered basis
n
for F in which the operator T is represented by the matrix C. If C is the direct sum of companion
matrices C of monic polynomials g , ...., g such that g divides g for i = 1, ..., s 1, then it is
i 1 s i + 1 i
apparent that we shall have non-zero vectors , ....., in V with T-annihilators g , ...., g such that
1 s 1 s
T
T
V Z ( ; ) ... Z ( ; ).
s
1
But then by the uniqueness statement in the cyclic decomposition theorem, the polynomials g ,
i
are identical with the polynomials p which define the matrix A. Thus C = A.
i
The polynomials p , ...., p are called the invariant factors for the matrix B. We shall describe an
1 r
algorithm for calculating the invariant factors of a given matrix B. The fact that it is possible to
compute these polynomials by means of a finite number of rational operations on the entries of
B is what gives the rational form its name.
Example 1: Suppose that V is a two-dimensional vector space over the field F and T is a
linear operator on V. The possibilities for the cyclic subspace decomposition for T are very
limited. For, if the minimal polynomial for T has degree 2, it is equal to the characteristic
polynomial for T and T has a cyclic vector. Thus there is some ordered basis for V in which T is
represented by the companion matrix of its characteristic polynomial. If, on the other hand, the
minimal polynomial for T has degree 1, then T is a scalar multiple of the identity operator.
If T = cI, then for any two linear independent vectors and in V we have
1 2
T
T
V ( Z 1 ; ) ( Z 2 ; )
p 1 p 2 x . c
For matrices, this analysis says that every 2 × 2 matrix over the field F is similar over F to exactly
one matrix of the types

c 0 0 c 0
,
0 c 1 c 1

Example 2: Let T be the linear operator on R which is represented by the matrix.
3
5 6 6
A 1 4 2
3 6 4

in the standard ordered basis. We have computed earlier that the characteristic polynomial for
2
T is f (x 1)(x 2) and minimal polynomial for T is p (x 1)(x 2). Thus we know that in
the cyclic decomposition for T the first vector will have p as its T-annihilator.
1
Since we are operating in a three-dimensional space, there can be only one further vector, .
2
It must generate a cyclic subspace of dimension I, i.e., it must be a characteristic vector for T.
T-annihilator p must be (x 2), because we must have pp = f. Notice that this tells us immediately
2 2
that the matrix A is similar to the matrix
0 2 0
B 1 3 0
0 0 2

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