Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Singular Values & Linear Transformations

  1. Apr 8, 2009 #1
    I'm struggling to grasp what should be a trivial property of singular value decomposition. Say that I have a linear transformation T that is non-singular (i.e. [tex]T^{-1}[/tex] exists) and relates matrices A and B:

    [tex]B = T A[/tex]

    or

    [tex]A = T^{-1} B[/tex]

    What I would like to know is how the singular values of A and B are related?

    A and B are both dimension (m x n), i.e. T is square (m x m). My thoughts are:

    [tex]A = U_A \Lambda_A V_A^T[/tex]

    [tex]B = U_B \Lambda_B V_B^T[/tex]

    where U is (m x m), [tex]\Lambda[/tex] is (m x n) and V is (n x n). Substitution gives

    [tex]T A = U_B \Lambda_B V_B^T[/tex]

    [tex]T U_A \Lambda_A V_A^T = U_B \Lambda_B V_B^T[/tex]

    but I'm not sure where to go next. From the above, can I simply write:

    [tex]T U_A = U_B[/tex]

    [tex]\Lambda_A = \Lambda_B[/tex]

    [tex]V_A = V_B[/tex]

    This implies the singular values are identical (with the caveat that T is non-singular) and that only the basis vectors of A are affected by the transformation... this is what I want to show but I'm unsure if I've actually succeeded with the above.
     
  2. jcsd
  3. Apr 9, 2009 #2
    After a bit more thought... the above is only true if T is orthogonal/unitary. I still think it is possible to relate the singular values of A and TA for the slightly more general case where T is full-rank (i.e. non-singular) but I'm not entirely sure how. Obviously, if T is not of full rank then all bets are off.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Singular Values & Linear Transformations
Loading...