By Susan Norman
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Example text
C0, • • • , c n ) is not identically 0. Since the rank of y is 1, its second row (ci, c2, • • • , cn+\] must be a multiple of the first. ) The third row of y, namely (c2, c3, • • • , c n + 2 ) must also be a multiple of the first, and since we already know that c2 = g2c0, the third row is g2 times the first. It follows therefore that cn+2 = g2cn = gn+2c0. Continuing the argument we see that c, = g'c0 for all / = 0, 1, • • • , m + n. So in this case we have (CQ, c\, • • •, cm+n) = c0(i,g, g2, • • • ,g1m+n), CO^Q.
There exists a nonsingular matrix C~l (with entries in G) such that for every row c 0/C~V(dW(jt)y) = i. For every C~l there exists an algorithm A = A(C~l) computing M(x)y, satisfying n(A) = /tt(M(jt)y) such that its /cth m/d step mk is given by mk = (£/=i «ifc*«)(Z/=i j3/fcy/), where the a,fc's and /S/fc's are in G. (M(x)y) there exists a matrix C~ 1 suchthatA=A(C" 1 ). IVb. Classification of the algorithms. We will use Theorem 1 of the last subsection to exhibit all the algorithms for computing z = x * y using m+n + 1 m/d steps.
For every C~l there exists an algorithm A = A(C~l) computing M(x)y, satisfying n(A) = /tt(M(jt)y) such that its /cth m/d step mk is given by mk = (£/=i «ifc*«)(Z/=i j3/fcy/), where the a,fc's and /S/fc's are in G. (M(x)y) there exists a matrix C~ 1 suchthatA=A(C" 1 ). IVb. Classification of the algorithms. We will use Theorem 1 of the last subsection to exhibit all the algorithms for computing z = x * y using m+n + 1 m/d steps. We first write this system of bilinear forms as M(x)y where M(x) is PRODUCT OF POLYNOMIALS 29 the (m + n +1) x n matrix whose (/, /)th entry is *,•_/ whenever 0 ^ / —/ ^ m, and 0 otherwise.
We mean business - an elementary course in business English by Susan Norman
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