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From "Tyler Ward" <ward.ty...@gmail.com>
Subject Description of SVD
Date Sat, 26 Aug 2006 23:19:38 GMT
I'm going to send in basically the textbook description of SVD, in case
someone wants to implement it. If you guys get your householder reductions,
then that's the most difficult part. I have code for all of this, but it is
held up by legal, so I'll just give you guys the basic ideas instead.

Forgive me if this seems like condescension, you can find this in any decent
math book, and it is about one of the most well understood algorithms out
there. However, since you're so concerned with clean room implementations,
here I'll describe one from first principals, so you can be confident that
its clean if someone implements it.

Here's the idea. Lets assume the matrix A is n by m, and A` would be
A-transpose. n >= m, in this case, so A`A (m by m) is smaller than AA` (n by
n).

For a matrix A, it can be broken down as A = USV`, where U and V are
orthogonal, and S is diagonal.

Let look at eigenvectors for a moment. Eigenvector is a vector of the form
Ab = ub, for some scalar (eigenvalue) u. This means that given an
Eigenvector matrix D (which is orthogonal), where the eigenvectors make up
the columns of D, this eigenvector equation can be written as AD = ZD, where
Z is the diagonal matrix of the eigenvalues. Rearranging, D`AD = Z, the
eigenvector matrices will diagonalize the matrix A. Much talk about the
number of eigenvectors and their orthogonality can be found almost anywhere.


Now imagine A`A (A-transpose multiplied by A, a symmetrical matrix). Given
that A = USV`, A`A = VSU`USV` = VSSV`. SS, is really just a diagonal matrix
of values, lets call it (suggestively) Z.

A`A = VZV`, so if we can find the eigenvectors of A`A (which is symmetric),
and the eigenvalues Z, then we can compute the elemenst of S (taking the
square root of each element of Z).

Given that we have V, and S, we can compute AV(S-inverse) = U, and get the
other part of the decomposition.

So, how do we find the eigenvectors? Easily enough.

1) Start by householder transforming a symmetric matrix into tridiagonal
form by multiplying by householder matrixes H, like so A2 = H1A1H1`. This
will reduce to tridiagonal form after N-2 such multiplications, producing
Q1` = H1`H2`H3`....

2) Reduce to diagonal form using jacoby reductions with Givens plane
rotations. Because of the simple form, literally mulitply it out by hand,
and you'll see that only a few elements of A change wiith each reduction,
and the formulas for the new values there are quite easy. Use this
simplified code to reduce A. Each step here produces a Givens Plane Rotation
matrix G, such that A2 = G1A1G1`. Again, Q2 = G1`G2`G3`.... These
multiplications are easy, as only two rows and two columns of Q2 change with
each new plane rotation. This means that Z = Q2Q1AQ1`Q2`, or that V = Q2Q1.
A has been diagonalized when this process is finished.

3) Now that we have Z and V, the rest is just two matrix multiplications
away.


Once again, the givens reductions are really trivial. remember, a givens
matrix is like this.....

1 0 0 0
0 r  s 0
0 -s r 0
0 0 0 1

Make sure that the box of values ends up at the correct position to wipe out
one of the off diagonal elements (the one in the position of -s, here
element 3, 2). It's easy to compute which s and r are needed to eliminate
the element, do it by hand and put the equation into the code. Do the
multiplication by hand, and put those equations into the code as well.
G1A1G1` can be multiplied out by hand, and only a half a dozen elemnts of A
change, A is symmetrical, so you only have to consider half of those,
something like 4 elements. Really easy. The off-diagonal elements will come
back each time. compute a new matrix and wipe them out again. They will be
smaller each time, and soon you'll reduce them down to the level where you
can just throw them away. Perhaps when the off diagonal element divided by
the diagonal is less than 10^-14 or so. Then make a new matrix and get the
next elements. The next matrix after the one above would be this one.

1 0 0 0
0 1 0 0
0 0 r  s
0 0 -s r

And so on, get them one at a time.

Ok, there you go. The complete algorithm. Efficient householder reductions,
and this algorithm will be efficient.

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