# Does anyone know this kind of matrix decomposition?

Discussion in 'Numerical Analysis' started by seagle.99, Jul 9, 2008.

1. ### seagle.99Guest

Given a real matrix X, is there a unique way to decomp. X in to
D_1RD_2, where D_1, D_2 are two diagonal matrices, R is a rotation
matrix: RR^t=identity.

unique? If not, is there a way to achive this decomposition?

Thanks!

seagle.99, Jul 9, 2008

2. ### Ken PledgerGuest

Do you want all three matrices on the right-hand side to be real?
If so, then a little calculation in the 2 x 2 case shows that it's not
always possible. If your real matrix X is

(a b)
(c d)

then your decomposition requires the product abcd to be negative; so,
for example, you won't be able to do it with

(1 1)
(1 1).

Ken Pledger.

Ken Pledger, Jul 9, 2008

3. ### Helmut JarauschGuest

There is something similar, it's called the Singular Value Decomposition
(SVD for short).

The matrix X is decomposed into U S V' (' denotes transposition)
The matrices U and V, being so-called orthogonal matrices, are rotations
or reflections. The matrix S is diagonal with non-negative entries
in decreasing order of magnitude.
So a matrix can be imagined as a first rotation (V') then a scaling (S)
and finally one more rotation (U).

--
Helmut Jarausch

Lehrstuhl fuer Numerische Mathematik
RWTH - Aachen University
D 52056 Aachen, Germany

Helmut Jarausch, Jul 10, 2008
4. ### Peter SpellucciGuest

this decomposition is impossible for almost all quadratic matrices:
let us stick on invertible X. then of course also D1 and D2 must be invertible.
but this would mean that every regular n times n matrix X could be
transformed into a unitary one by diagonal scaling from the left and the
right. but now take the subset of triangular matrices (and any matrix can
be made triangular by an unitary transform) : the triangular structure is
unaltered by diagonal transforms, but a unitary triangular matrix is diagonal.
(easy exercise), hence also X would be diagonal, and the decomposition
isn't of interest any more.
maybe you confused it with the svd, which Helmut
hth
peter

Peter Spellucci, Jul 10, 2008
5. ### RRogersGuest

Are you trying for a projection type of decomposition? In any case it
would seem you are trying for something like
http://en.wikipedia.org/wiki/Polar_decomposition
I would like to help you out more but my memory is failing me right
now. I know there is a decomposition theorem somewhere; not identical
to your request but similar. Note that you can not just scale and
rotate; there are more degrees of freedom; for instance skew. I seem
to remember that skew (upper triangular matrices with 1's on the
diagonal), scale, and rotation are sufficient for decomposition; I am

Ray

RRogers, Jul 10, 2008
6. ### seagle.99Guest

Thank you all very much!

Actually, what I want to know is whether a GL group action can convert
a diagonal matrix to still a diagonal matrix,
that is, XDX^t=S, where D and S are diagonal matrices, X is
nonsingular square matrix. To solve this equation,
we have XD^(1/2)(XD^(1/2))^t=(S^(1/2)R)(S^(1/2)R)^t, where R is a
rotational matrix. So there exist an R such that
XD^(1/2)=S^(1/2)R, which means X=S^(1/2)RD^(1/2).

Thank you for noticing me this decomp. can not be made for all
nonsingular matrices! So are there some if and only if rule to decide
which matrix can and which can't? Thank you!

seagle.99, Jul 12, 2008
7. ### Peter SpellucciGuest

within some limits: yes.
remember Sylvesters law of inertia: from the nonsingularity of X
(.^t means transposition , I assume this!) it follows that D and S
have the same inertia. Now consider two symmetric real matrices with the same
inertia, A and B.
then, using Gaussian elimination with diagonal pivoting you may write

A= P_a* L_a* D_a * L_a^t *P_a^t
B= P_b* L_b* D_b * L_b^t *P_b^t

where the L's are invertible and P's are permutation matrices and
D_a and D_b diagonal. hence for
D_a and D_b you finally can write down your desired relation with X implicitly
defined by these decompositions

hth
peter

Peter Spellucci, Jul 14, 2008