The projection operator of a least-squares tuning

The exact solution of a least-squares tuning for a regular temperament
will involve fractional exponents of log2(p) for the various primes p.
As such, they can be taken as examples of Joe's beloved fractional
monzos, and a square matrix of them formed; call it P. P is an example
of a projection operator:

http://en.wikipedia.org/wiki/Projection_%28linear_algebra%29

http://en.wikipedia.org/wiki/Projection_operator

P has eigenvalues of 0 and 1. The zero eigenvectors of P, on the left,
are the kernel elements of the temperament. The eigenvectors for 1, on
the right, are the vals associated to the temperament, including the
equal temperament vals and the generator mapping vals. The
eigenvectors for 1 on the left are the monzo for 2 and elements
orthogonal, in some sense, to the temperament. For unweighted least
squares in the 5 or 7 limit, that sense is just the symmetric lattice
distance sense.

Here's an example, unwighted least squares for 7-limit meantone.

Solving for the exact solution of the least squares, and writing the
result as a matrix of fractional monzos, gives:

P = [|1 0 0 0>, |329/243 -11/243 1/243 25/243>,
|344/243 -44/243 4/243 100/243>, |131/243 -110/243 10/243 250/243>]

You can check that P^2 = P (P is idempotent) and that P sends monzos
for 81/80 and 126/125 to zero, and vals for 12, 19, 31 etc, as well
as <0 1 4 10|, to themselves. It also sends the monzo for 2 and
|0 -11 1 25> to themselves; the latter is the note class perpendicular
under the symmetric lattice metric to meantone.

--- In tuning-math@yahoogroups.com, "Gene Ward Smith" <gwsmith@s...>
wrote:

> P = [|1 0 0 0>, |329/243 -11/243 1/243 25/243>,
> |344/243 -44/243 4/243 100/243>, |131/243 -110/243 10/243 250/243>]

By way of comparison, here is the projection operator for 1/4 comma,
which is the minimax tuning:

[|1 0 0 0>, |1 0 1/4 0>, |0 0 1 0>, |-3 0 5/2 0>]

The same arguments apply: if r is the rank, and n = phi(p) the number
of primes, there must be a dimension r space of right eigenvectors for
1, for the vals belonging to the temperament, and a dimension n-r
space of monzo eigenvectors for 0. These will be the same for any
tuning of thre temperament; the other eigenvectors will differ, but we
will alwasy have a projection operator with r eigenvalues of 1 and n-r
of 0.

--- In tuning-math@yahoogroups.com, "Gene Ward Smith" <gwsmith@s...>
wrote:
> --- In tuning-math@yahoogroups.com, "Gene Ward Smith" <gwsmith@s...>
> wrote:
>
> > P = [|1 0 0 0>, |329/243 -11/243 1/243 25/243>,
> > |344/243 -44/243 4/243 100/243>, |131/243 -110/243 10/243 250/243>]
>
> By way of comparison, here is the projection operator for 1/4 comma,
> which is the minimax tuning:
>
> [|1 0 0 0>, |1 0 1/4 0>, |0 0 1 0>, |-3 0 5/2 0>]

Given such a projection, the notes of the tuning can be defined in a
way independent of any choice of generators, as the image under the
projection map of the p-limit intervals. This lies in an invariant
subspace,the left eigenspace corresponding to the eigenvalue 1. Of
course given a tuning map we can in any case define the notes of the
tuning as the image, in R, of the p-limit intervals.

--- In tuning-math@yahoogroups.com, "Gene Ward Smith" <gwsmith@s...>
wrote:

I forgot to mention a key point about the notes of the tuning in the
left eigenspace with eigenvalue 1: they consitute a lattice.