The first step of the end game is to run NCProcess2 on the last spreadsheet which was produced in §14.3. The aim of this run of NCProcess2 is to shrink the spreadsheet as aggressively as possible without destroying important information. The spreadsheet produced by NCProcess2 is the same as the last spreadsheet which was produced 4 in §14.3.
Note that it is necessary that all of the equations in the spreadsheet have solutions, since they are implied by the original equations. The equations involving only knowns play a key role. In particular, they say precisely that, there must exist a projection P1 such that
![]() | (14.4) |
are satisfied.
The converse is also true and can be verified with the assistance of the above spreadsheet. To do this, we assume that the matrices A, B, C and P1 are given and that (14.4) holds, and wish to define m1, m2, n1, n2, a, b, c, e, f and g such that each of the equations in the above spreadsheet hold. If we can do this, then each of the equations from the starting polynomial equations (FAC) given in §14.2 will hold and we will have shown that a minimal factorization of the [A,B,C, 1] system exists.
Here we have used the fact that we are working with matrices and not elements of an abstract algebra.
With the assignments made above, every equation in the spreadsheet above holds. Thus, by backsolving through the spreadsheet, we have constructed the factors of the original system [A,B,C, 1]. This proves
Theorem ([BGKvD]) The system [A,B,C, 1] can be factored if and only if there exists a projection P1 such that P1 AP1 = P1 A and P1 B C P1 = P1 A - AP1 + B C P1.
Note that the known equations can be neatly expressed in terms of P1 and P2 = 1 - P1. Indeed, it is easy to check with a little algebra that these are equivalent to (14.1). It is a question of taste, not algebra, as to which form one chooses.
For a more complicated example of an end game, see §17.4.
We saw that this factorization problem could be solved with a 2-prestrategy. It was not a 1-prestrategy because there was at least at one point in the session where the user had to make a decision about an equation in two unknowns. On the other hand, the assignment (14.3) was a motivated unknown. We will see in §16.3.1 that this is a 1-prestrategy. For example, the equation
![]() | (14.5) |
in the two unknowns m1 and n1 can be transformed into an equation in the one unknown m1n1 by multiplying by n1 on the right:
![]() | (14.6) |
If we do not restrict ourselves to the original variables but allow constructions of new variables (according to certain very rigid rules), then the factorization problem is solvable using a generalization of a 1-prestrategy, called a 1-strategy. Section 5 of [HS] describes 1-strategies.
The brevity of this presentation suppresses some of the advantages and some of the difficulties. For example, one might not instantly have all of the insight which leads to the second spreadsheet. In practice, a session in which someone “discovers” this theorem might use many spreadsheets.