RMS projection map

I've already called the unweighted version of this the Frobenius projection map, so I'll call this rms unless someone has a better idea.

In a prime limit with n primes, take k independent vals, and form M, the kxn matrix whose row are the vals. Now take N = pseudoinverse(M). If we take the product M.N, we get a kxk identity matrix, not very exciting. If we take the product N.M, we get an nxn square matrix, the RMS projection map P. This has k eigenvalues of 1, corresponding to the vals which are left untouched, and n-k eigenvalues of 0, corresponding to the kernel of commas, which are collapsed to 0. The JIP times N (on either side) morphs the JIP into a tuning map which just happens to be TOP-rms tuning.

<x|P|x> defines an inner product on the subspace of tuning space (weighted vals or monzos) where P projects things, which is the subspace which defines the temperament, spanned by vals of the temperament.

On 08/07/2010, genewardsmith <genewardsmith@sbcglobal.net> wrote:
> I've already called the unweighted version of this the Frobenius projection map, so I'll call this rms unless someone has a better idea.
>
> In a prime limit with n primes, take k independent vals, and form M, the kxn matrix whose row are the vals. Now take N = pseudoinverse(M). If we take the product M.N, we get a kxk identity matrix, not very exciting. If we take the product N.M, we get an nxn square matrix, the RMS projection map P. This has k eigenvalues of 1, corresponding to the vals which are left untouched, and n-k eigenvalues of 0, corresponding to the kernel of commas, which are collapsed to 0. The JIP times N (on either side) morphs the JIP into a tuning map which just happens to be TOP-rms tuning.

Is this like eigenmonzos? I remember them. I thought you should use
rational weights so that real intervals come out.

I didn't bring anything to make notes today, so I won't be able to
give a fully informed reply tomorrow.

Graham

--- In tuning-math@yahoogroups.com, Graham Breed <gbreed@...> wrote:

>
> Is this like eigenmonzos? I remember them.

The eigenmonzos were a spinoff idea.

On 08/07/2010, genewardsmith <genewardsmith@sbcglobal.net> wrote:
> I've already called the unweighted version of this the Frobenius projection map, so I'll call this rms unless someone has a better idea.
>
> In a prime limit with n primes, take k independent vals, and form M, the kxn matrix whose row are the vals. Now take N = pseudoinverse(M). If we take the product M.N, we get a kxk identity matrix, not very exciting. If we take the product N.M, we get an nxn square matrix, the RMS projection map P. This has k eigenvalues of 1, corresponding to the vals which are left untouched, and n-k eigenvalues of 0, corresponding to the kernel of commas, which are collapsed to 0. The JIP times N (on either side) morphs the JIP into a tuning map which just happens to be TOP-rms tuning.

Right, I remembered or found out that, except for the TOP-RMS bit, and
the kernel of commas. That makes some of what I'll say redundant.

> <x|P|x> defines an inner product on the subspace of tuning space
> (weighted vals or monzos) where P projects things, which is the
> subspace which defines the temperament, spanned by vals of
> the temperament.

What you can also do is take P-I, which is still a projection, and
transforms any val in the temperament class to a zero vector. That
is, it projects the temperament class to the origin. This is similar
to what the wedgie does -- wedge with a val belonging to the class and
you get zero.

With unweighted vals, P-I is rational. So you can scale it up (by
scalar complexity squared) to get an integer matrix that isn't a
projection, but still sends the temperament class to the origin. That
means it's a list of things like unison vectors. Unfortunately, they
have a horrible amount of torsion. But I think it's an interesting
result.

Graham

--- In tuning-math@yahoogroups.com, Graham Breed <gbreed@...> wrote:

> What you can also do is take P-I, which is still a projection, and
> transforms any val in the temperament class to a zero vector.

Taking I-P is equivalent, and brings out the duality better: it also has eigenvalues of 0 and 1, only role of vals and monzos is exchanged.

> With unweighted vals, P-I is rational. So you can scale it up (by
> scalar complexity squared) to get an integer matrix that isn't a
> projection, but still sends the temperament class to the origin. That
> means it's a list of things like unison vectors. Unfortunately, they
> have a horrible amount of torsion. But I think it's an interesting
> result.

The horrible torsion is removable, since you can Hermite reduce and divide it out. Hence you could use this as a way of finding commas for the temperament, starting from vals.

You can also use the same idea to start from monzos rather than vals. Take a matrix whose rows are independent monzos, and do the same pseudoinverse construction. Now I-P will be just what you would have gotten starting from vals, and you can in fact extract the vals out in the same way as with the monzos--clear denominators, Hermite reduce, take out the GCDs, and Hermite reduce again.

--- In tuning-math@yahoogroups.com, "genewardsmith" <genewardsmith@...> wrote:

> The horrible torsion is removable, since you can Hermite reduce and divide it out. Hence you could use this as a way of finding commas for the temperament, starting from vals.

Sadly, this works a lot of the time but not always.

On 10 July 2010 09:35, genewardsmith <genewardsmith@sbcglobal.net> wrote:
>
>
> --- In tuning-math@yahoogroups.com, "genewardsmith" <genewardsmith@...> wrote:
>
>> The horrible torsion is removable, since you can Hermite reduce and divide it out. Hence you could use this as a way of finding commas for the temperament, starting from vals.
>
> Sadly, this works a lot of the time but not always.

That's about where I got to. There are easier ways of getting the
original commas, although this method's still nice to know about. I
still don't know how to be torsion-free for more than two things (rank
2 vals or 2 unison vectors).

Graham

--- In tuning-math@yahoogroups.com, Graham Breed <gbreed@...> wrote:
I
> still don't know how to be torsion-free for more than two things (rank
> 2 vals or 2 unison vectors).

Have you tried first finding the (r+1)-prime commas, where r is the rank of the temperament?

On 11 July 2010 02:28, genewardsmith <genewardsmith@sbcglobal.net> wrote:
>
>
> --- In tuning-math@yahoogroups.com, Graham Breed <gbreed@...> wrote:
>  I
>> still don't know how to be torsion-free for more than two things (rank
>> 2 vals or 2 unison vectors).
>
> Have you tried first finding the (r+1)-prime commas, where r is the rank of the temperament?

What does r+1 have to do with it? And if r>2, how are these commas defined?

Graham

--- In tuning-math@yahoogroups.com, Graham Breed <gbreed@...> wrote:

> What does r+1 have to do with it? And if r>2, how are these commas defined?

Let me see if I can get this other thing to work; I was thinking that an analogy to rank two could work but it needs some work, as you have noticed.