It seems strange to organize a set of linear equations according to which unknown is somehow ‘principal’ for a given equation.

x+y = 5
x−y = −1

seems not to require or even admit such bias. In the systems we are looking at, however, this relation identifies the ‘diagonal’ of the matrix A in the matrix form of our problem:

Ax = B

Often A is near the identity is some appropriate sense. That sense is missing without this bias. For better or worse, it is near the center of my intuitions on how to solve these equations efficiently.

Symmetrizing A

Suppose we transform the original problem to make A symmetric. Do subsequent stages of the solution remain symmetric? I think I will play with this code to get an empirical answer.

In the context of these definitions the following refutes the above conjecture:

(define s (vg 10 10))
; (vector-ref (cdr s) 33) => (0 (23 . 1/4) (32 . 1/4) (34 . 1/4) (43 . 1/4) (100 . 0))
(retire s 33)
; (vector-ref (cdr s) 33) => ("d" 0 (23 . 1/4) (32 . 1/4) (34 . 1/4) (43 . 1/4) (100 . 0))
; (vector-ref (cdr s) 23) => (0 (13 . 4/15) (22 . 4/15) (24 . 4/15) (32 . 1/15) (34 . 1/15) (43 . 1/15) (100 . 0))
; (vector-ref (cdr s) 13) => (0 (3 . 1/4) (12 . 1/4) (14 . 1/4) (23 . 1/4) (100 . 0))
; (vector-ref (cdr s) 43) => (0 (23 . 1/15) (32 . 1/15) (34 . 1/15) (42 . 4/15) (44 . 4/15) (53 . 4/15) (100 . 0))

Note that after the unknown at point 33 has been retired (eliminated) the relation between points 23 and 13 is no longer symmetric!

I have forgotten what this is about!