Does anyone know this kind of matrix decomposition?

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

  1. seagle.99

    seagle.99 Guest

    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.

    Does anyone have any idea about this ? If so do you think it is
    unique? If not, is there a way to achive this decomposition?

    Thanks!
     
    seagle.99, Jul 9, 2008
    #1
    1. Advertisements

  2. seagle.99

    Ken Pledger Guest


    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
    #2
    1. Advertisements

  3. 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
    #3
  4. 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
    Jarausch did mention already
    hth
    peter
     
    Peter Spellucci, Jul 10, 2008
    #4
  5. seagle.99

    RRogers Guest

    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
    not sure about uniqueness.

    Ray
     
    RRogers, Jul 10, 2008
    #5
  6. seagle.99

    seagle.99 Guest

    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
    #6
  7. 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
    #7
    1. Advertisements

Ask a Question

Want to reply to this thread or ask your own question?

You'll need to choose a username for the site, which only take a couple of moments (here). After that, you can post your question and our members will help you out.