Re: SVD and repeat singular values




"john2" <john2@xxxxxxxxxxxxxxxx> wrote in message
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Steve Hawkwind wrote:
"john2" <john2@xxxxxxxxxxxxxxxx> wrote in message
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Steve Hawkwind wrote:

"john2" <john2@xxxxxxxxxxxxxxxx> wrote in message
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Steve Hawkwind wrote:


Hi,

I am following a college text book algorithm that finds the SVD of a
rectangular matrix, but it in fact seems to fail if the matrix
involved

has


repeat singular values.

What can you do in this case? How do you form the 'V' matrix? Is it
possible to use the Jordan decomposition here to find the V matrix?



AIUI the decomposition is degenerate if there are repeated eigenvalues
and there is genuinely no 'unique' solution. The associated U and V
vectors can be selected arbitrarily in the left and right subspaces
and
any such selection will give a useful solution for practical purposes.


sorry ive just woken up... AIUI? Yes i have tried introducing
randomly
selected unit vectors to replace the 'repeat eigenvectors'
of the repeat eigenvalues. This will work but in some cases only in

certain

selected permutations, no idea why.

i wondered if the geometric multiplicity is equal to the algebraic
mutiplicity then i could fill all the remaining repeat eigenvectors
with

the

remaining vectors of the eigencector basis.


AIUI = "As I understand it". Meaning I take no responsibility for my
major errors, basic lack of understanding of the problem and/or the
solution.

In the decomposition A = UDV' where D is positive diagonal, if D has m
repeated values these define orthogonal subspaces of dimension m in U
and V and orthogonal eigenvectors can be arbitrarily selected in one or
other of these subspaces. Each vector selection U1 in, say, U is
attached to a fixed vector V1 in V and vice versa so they can't be
arbitrary on both the left and right sides. i.e. U1'A = D1*V1 is
unique. but U1 is arbitrary within its subspace.
Hope this makes sense.


Are you talking here of the form of the SVD where if our original matrix
is
m x n, then the U matrix will be m x n, the diagonal D will be n x n, as
will the V matrix? If so then yes there seems to be no problem here
with
repeat singular values and random vectors can be selected to 'fill in
the
gaps'.

But if the other convention is adopted where U is m x m, D is m x n and
V is
n x n, then there seems to be problems here with using random repeat
eigenvectors.


How can D be diagonal if it's m x n. The only form I know is U is m x
r, D is r x r, and V' is r x n
where r <= m, r <=n
is the rank of the matrix. This is the way that Matlab does it and other
forms must be equivalent by column, row exchanges of U and V'.


john2

there are apparently 'two conflicting notations' as according to mathworld
:-
http://mathworld.wolfram.com/SingularValueDecomposition.html

i have myself used several CAS's in the past and most seem to use the
convention U as m x m, D as m x n, and V as n x n. When i called D
diagonal I meant it loosely in the sense that the matrix could be
rectangular but have non-zero entries going across its 'diagonal'




.