Page 241 - DMTH502_LINEAR_ALGEBRA

Unit 22: Computation of Invariant Factors

We now define elementary column operations and column-equivalence in a manner analogous Notes
to row operations and row-equivalence. We do not need a new concept of elementary matrix
because the class of matrices which can be obtained by performing one elementary column
operation on the identity matrix is the same as the class obtained by using a single elementary
row operation.
Definition: The matrix N is equivalent to the matrix M if we can pass from M to N by means of
a sequence of operations

M = M M ... M = N
0 1 k
each of which is an elementary row operation or an elementary column operation.
Theorem 2: Let M and N be m n matrices with entries in the polynomial algebra F[x ]. Then N
is equivalent to M if and only if
N = PMQ
where P is an invertible matrix in F[x] m m and Q is an invertible matrix in F[x] n n .
Theorem 3: Let A be an n n matrix with entries in the field F, and let p , ... , p be the invariant
1 r
factors for A. The matrix x I – A is equivalent to the n n diagonal matrix with diagonal entries
p , ... , p , 1, 1, ... , 1.
1 r
–1
Proof: There exists an invertible n n matrix P, with entries in F, such that PAP is in rational
form, that is, has the block form
A 1 0  0
0 A 2  0
PAP –1 =
   
0 0  A
r
where A is the companion matrix of the polynomial p . According to Theorem 2, the matrix
i i
P(xI – A)P –1 = xI – PAP –1 ...(6)
is equivalent to xI – A. Now

xI A 1 0  0
0 xI A 2  0
xI – PAP –1 = ...(7)
  
0 0  xI A
r
where the various I's we have used are identity matrices of appropriate sizes. At the beginning
of this section, we showed that xl – A, is equivalent to the matrix
p i 0  0
0 1  0
.
  
0 0  1

From (6) and (7) it is then clear that xl – A is equivalent to a diagonal matrix which has the
polynomials p , and (n – r) 1's on its main diagonal. By a succession of row and column
i
interchanges, we can arrange those diagonal entries in any order we choose. For example: p , ...,
1
p , 1, ... ,1.
r
Theorem 3 does not give us an effective way of calculating the elementary divisors p , ... , p
1 r
because our proof depends upon the cyclic decomposition theorem. We shall now give an

LOVELY PROFESSIONAL UNIVERSITY 235