Adjoints of square matrices

I was wondering if anyone could point me to an explanation as to why
taking the adjoint of three commas produces a temperament (plus
generators to primes) or the adjoint of three temperaments gives a
comma.

I understand the geometry of it, but don't really get the linear math
involved. I am reading through some of Graham's website, perhaps the
answers lie therein. I also know that the adjoint relates very closely
to the wedge product. (Cofactors of a matrix * determinant because
inverse is cofactors /determinant, etc). Sorry if I am being lazy!

Paul

--- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
<paul.hjelmstad@m...> wrote:
> I was wondering if anyone could point me to an explanation as to why
> taking the adjoint of three commas produces a temperament (plus
> generators to primes) or the adjoint of three temperaments gives a
> comma.
>
> I understand the geometry of it, but don't really get the linear math
> involved. I am reading through some of Graham's website, perhaps the
> answers lie therein. I also know that the adjoint relates very
closely
> to the wedge product. (Cofactors of a matrix * determinant because
> inverse is cofactors /determinant, etc). Sorry if I am being lazy!

Oops meant Inverse * determinant, not that it matters...
>
> Paul

--- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
<paul.hjelmstad@m...> wrote:
> I was wondering if anyone could point me to an explanation as to why
> taking the adjoint of three commas produces a temperament (plus
> generators to primes) or the adjoint of three temperaments gives a
> comma.

How are you getting on? I think the key is first to understand how the equal temperament mapping comes out from the solution of a matrix equation. Then you can think of the other mappings as bizarre examples of equal temperaments with 0 notes to the octave.

> I understand the geometry of it, but don't really get the linear math
> involved. I am reading through some of Graham's website, perhaps the
> answers lie therein. I also know that the adjoint relates very
closely
> to the wedge product. (Cofactors of a matrix * determinant because
> inverse is cofactors /determinant, etc). Sorry if I am being lazy!
> Oops meant Inverse * determinant, not that it matters...

I don't think my website's going to give you the answers, because I rarely understand it myself :-S Well, at least the old idea of only equal temperaments is explained there. I discovered by accident that the generator mappings were also being produced, so I had algorithms that worked without knowing why.

Linear independence might be something to think about. It's the common thread to both the matrix and wedge product approaches. With wedge products you can split a matrix into complementary parts. So instead of linearly independent commas, you have some commas and some mappings (vals, ets, whatever). If they match, the wedge product should be zero. Matrices should give identical results if you use parts of the identity matrix for missing vectors, and the adjoint instead of the complement operation. So understanding one approach might help you with the other.

Graham

--- In tuning-math@yahoogroups.com, Graham Breed <gbreed@g...> wrote:
> --- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
> <paul.hjelmstad@m...> wrote:
> > I was wondering if anyone could point me to an explanation as
to why
> > taking the adjoint of three commas produces a temperament (plus
> > generators to primes) or the adjoint of three temperaments
gives a
> > comma.
>
> How are you getting on? I think the key is first to understand
how the
> equal temperament mapping comes out from the solution of a matrix
> equation. Then you can think of the other mappings as bizarre
examples
> of equal temperaments with 0 notes to the octave.
>
> > I understand the geometry of it, but don't really get the
linear math
> > involved. I am reading through some of Graham's website,
perhaps the
> > answers lie therein. I also know that the adjoint relates very
> closely
> > to the wedge product. (Cofactors of a matrix * determinant
because
> > inverse is cofactors /determinant, etc). Sorry if I am being
lazy!
> > Oops meant Inverse * determinant, not that it matters...
>
> I don't think my website's going to give you the answers, because
I
> rarely understand it myself :-S Well, at least the old idea of
only
> equal temperaments is explained there. I discovered by accident
that
> the generator mappings were also being produced, so I had
algorithms
> that worked without knowing why.
>
> Linear independence might be something to think about. It's the
common
> thread to both the matrix and wedge product approaches. With
wedge
> products you can split a matrix into complementary parts. So
instead of
> linearly independent commas, you have some commas and some
mappings
> (vals, ets, whatever). If they match, the wedge product should be
zero.
> Matrices should give identical results if you use parts of the
> identity matrix for missing vectors, and the adjoint instead of
the
> complement operation. So understanding one approach might help
you with
> the other.
>
>
> Graham

I'm finally making progress, after being on this group for over a
year. I am happy to say I have mastered your pages on Linear
Temperaments and the like. I am glad to find out that the adjoint
method is not easy (and actually is adjoint the proper term to use
for inverse*determinant of a matrix? Mathworld defines adjoint as
the transpose of the complex conjugate of matrix) but I am having
fun with Excel - taking three commas in the 7-limit and finding a
temperament, or taking three temperaments and finding a comma, or
most impressively, taking two commas and getting two temperaments
(linear temperament). Thanks for your response, I'll mull it over

Paul

--- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
<paul.hjelmstad@m...> wrote:

> I am glad to find out that the adjoint
> method is not easy (and actually is adjoint the proper term to use
> for inverse*determinant of a matrix? Mathworld defines adjoint as
> the transpose of the complex conjugate of matrix)

Unfortunately, it means both. A word sometimes used just to mean
"adjoint" in the classical, old-fashioned sense of M^(-1) * det(M) is
"adjugate", but I think "adjoint" is more common, even if ambiguous.
We could shift to "adjugate" if that would be less confusing.