By Hans Blomberg and Raimo Ylinen (Eds.)

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The first row multiplied by 2p is added to the second row. This results in 8 [f 2p2 i 3p 6 +1 -3p 6p2 -6p2 3 Step 7. The first column in ( 8 ) has attained a satisfactory form, and so has the entry in the position ( 2 , 2 ) . At this stage we therefore augment ( 8 ) with the previously deleted row and column to get 2p +1 0 0 9 16 p2+3 2p P3 0 2P2 6 -1 3p +1 6P2 -3p 1 -6p2 :]. 3 Step 8. The first column in (9) is satisfactory. The second column is also satisfactory, because the entry in the position (1,2) is of lower degree than the corresponding diagonal entry in the position (2,2).

It was already concluded that this problem has a solution. e. by 50 and the corresponding matrix having as its rows the leading coefficients of the rows of (50) is consequently (cf. e. (50) is indeed strictly proper and row proper. 3 (cf. 17)). 19) we already constructed a unimodular matrix P @ ) such that [ A @ )i - B @ ) ] P ( p ) = [ I i 01 with [ A @ )i -B@)] as in (50) (cf. 20), . . 22). To get a suitable matrix (36) as a first candidate Z @ ) for (29) we have accordingly first of all to compute P@)-'.

Such a matrix equation cannot generally be solved with respect to y for a given x in the usual vector space sense, because we cannot perform division by arbitrary nonzero scalars. Thus the question arises: Would it be possible 13 Algebraic Theory for Multivariable Linear Systems to strengthen the C[p]-module % to a vector space % over some suitable field which contains C[p] as a subring? Now it is known that we can form the field of quotients of Cb], denoted by C(p), in much the same way as the field of rational numbers is formed from the ring of integers.