Precise Implementation of Cad Primitives Using Rational Parametrizations of Stan

Cover Precise Implementation of Cad Primitives Using Rational Parametrizations of Stan
Precise Implementation of Cad Primitives Using Rational Parametrizations of Stan
S Ocken
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, ej_i respectively. Then EBF. Omits one row and one column of B^, and hence its rank is either r-1 or r-2.
-17- Suppose first that the rank of EBE is r-1, i. E. Identical to the rank of EB, so that these two matrices have the same range. Then there exists a u = Eu such that EB(e^^^^-u) = 0. Plainly ei^+^'u = = (Cei^)«u since Ce^ = e^+[ . Define a linear transformation T on X by putting Tx = X - (e|^+px)u + (Cu«x)e{^ - 1 /2(e^+^' x)(Cu« u)e^.
Then if Tx = 0, we have e^^^p Tx = e^^+px, so Tx = x
... + (Cu'x)ei^. Hence = Cu« Tx = Cu« X + (Cu« x)(Cu« e^) = Cu* x, and therefore x = 0, showing that T is nonsingular. Plainly Te^ = e^^ for i = l... K. Moreover FBFej^ = 0, so that (Ty). FBF(Tx) = (y-(e^^^>. Y)u + ((Cu-y) - l/ZCe^^+l* y)(Gu. U) )e, J •FBF(x-(e^^-^px)u + ((Cu-x) - 1 /2(e, . +i. X) (Cu* u) )ei^) = ((e^. +l«y) (e^^+i-u) + Ey )• FBF( (e^+1' x) (e;^+^-u) + Ex) =yEBEx + (e^^py) (e^+i« x) (e^+i - u)« B(e|^+l-u) .
Thus in the basis Tej^, . . . Te^^, the form B has the representation A, ^o' where B^' has one less row and column than B^.


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