A look at self-referential matrices: conclusion

This is essentially the summing-up of the reasoning found in Part 1.

6g. A Necessary Digression regarding Scalar and Additive Constants for Systems of Three or More Variables

It is important to note at this point that I have proven that adding, subtracting, multiplying or dividing any constant (the k's) to a system of three or more n's also leads to the appropriate degenerate cases of all zeroes, all ones, both or neither.

For a system of only three equations with possibly different operators with scalars and/or additive constants put in:

n₁ = k₁n₂ [operator₁] k₂n₃ + k₃

n₂ = k₄n₁ [operator₂] k₅n₃ + k₆

n₃ = k₇n₁ [operator₃] k₈n₂ + k₉

the same reasoning applies. The only value for the n's that can be fitted in that will avoid the infinite regress is the identity value (0, 1 or none) that all the operators in the equation have in common - and all three variables have to be that value. There are two possible cases:

1. all n's equal zero, which is true for all combinations that have finite solutions except for the ones that have only multiplication or division:

   0 = k₁ * 0 [operator₁] k₂ * 0 + k₃

   0 = k₄ * 0 [operator₂] k₅ * 0 + k₆

   0 = k₇ * 0 [operator₃] k₈ * 0 + k₉

   or:

   0 = 0 [operator₁] 0 + k₃

   0 = 0 [operator₂] 0 + k₆

   0 = 0 [operator₃] 0 + k₉

   and this implies that k₃, k₆ and k₉ have to equal zero and that the value of the others are irrelevant because the answer ends up being all zeroes anyway.
2. all n's equal one, which is only true for ones that have multiplication or division operators:

   1 = k₁ * 1 [operator₁] k₂ * 1 + k₃

   1 = k₄ * 1 [operator₂] k₅ * 1 + k₆

   1 = k₇ * 1 [operator₃] k₈ * 1 + k₉

   or:

   1 = k₁ [operator₁] k₂ + k₃

   1 = k₄ [operator₂] k₅ + k₆

   1 = k₇ [operator₃] k₈ + k₉

   Note that if the operators are multiplication or division, any nine constants that fit the above template will be consistent with the above set.

   And so, any set of nine constants that fulfill this will not make the system spin into an infinite regress.

   But note that this is dependent upon the three n-variables themselves being one, and the operators being either multiplication or division, and this cannot be generalized to a self-referential matrix of four or more variables. This has been shown above. In otherwords, we're back to the degenerate case for the variables themselves, only in this particular case, the constants aren't limited to one value. [But note that the variables - the values you plug in - disappear.]

   The only exception I can think of [where the variables don't vanish completely] is if the nine constants are collapsed into six -three scalars (ks) and three additives (ka) - like so:

   n₁ = ks₁n₂ [operator₁] n₃ + ka₁

   n₂ = ks₂n₁ [operator₂] n₃ + ka₂

   n₃ = ks₃n₁ [operator₃] n₂ + ka₃

   with each operator (to remind you) being either multiplication or division. Set:
   • ks₁ equal to (1/ n₂ n₃) if operator_i is multiplicative, or (n₃ / n₂) if operator_i is divisive;
   • ks₂ equal to (1/ n₁ n₃) if operator_i is multiplicative, or (n₃ / n₁) if operator_i is divisive;
   • ks₁ equal to (1/ n₁ n₂) if operator_i is multiplicative, or (n₂ / n₁) if operator_i is divisive.
   Note that this is impossible if any of the divisors are equal to zero. This reduces to:

   n₁ = ka₁

   n₂ = ka₂

   n₃ = ka₃

   Note too that this is the only exception - and it only applies to the specific three-variable system described above. I must repeat that the three-variables all-multiciplative-or-divisive system, with those specifically-described constants, subject to the restriction that none of the scalar constants are zero, is the only exception - and what you end up with is a list of already-supplied constants.

   And if you're tempted to infer that this might be extended to a system of four or more variables, it is possible; you can generalize this to as many variables as you like. For this system,

   n₁ = ks₁n₂ [operator_1,1],…,[operator_1,x-1] n_x + ka₁

   n₂ = ks₂n₁ [operator₂],…,[operator_2,x-1] n_x + ka₂ [etc.]

   n₃ = ks₃n₁ [operator₃],…,[operator_x,x-1] n_x-1 + ka₃

   it'll work if, for each ks_i, i being one to x, you:
   • begin with ks_i = 1;
   • for each j, j being 1 to (x-1):
     • if operator_ij is multiplicative, divide ks_I by n_j;
     • if operator_ij is divisive, divide ks_I by n_j;
   • and, when ks_i is derived after working through each equation, stick it into the above system.

   Notice, though, that it collapses to the same list as above:

   n₁ = ka₁

   n₂ = ka₂ [etc]

   n_x = ka_x

   and, if one of the divisors of any ks_i (i being from 1 to x) turns out to be zero, the system is unsolvable.

So, while it is true that the constants can in some cases take on a range of possible values, as shown above, the variables themselves have to equal all zeroes, all ones, both or neither because you've stuck the constants onto a system for which that's already proven. (See 6a to 6f above.) Adding scalar and additive constants doesn't change that. And what does that imply for the value of the constants? In otherwords, the n's have already been proven to be either all zero or all one, or infinite.

Also note that the above system contradicts the definition of a self-referential matrix in 0 - so this is like division by zero. Adding the scalars using the above method leads to a self-referential matrix turning into something else - a simple list of the additive constants.

7. Conclusion You realize, of course, that putting 6a through 6e and 6g together is sufficient for the stepladder procedure. And this in turn, alongside with 1 through 5, proves that any self-referential matrix as defined in 0 above has only four possible solutions: all zeroes {(0,…,0)}; all ones {(1,…,1)}; both of the preceding cases {(0,…,0),(1,…1)}; or none {}. These are all trivial. Adding both scalar and additive constants doesn't change that.

Q.E.D.

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