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Cochran's Theorem (algebraic version)

  1. Sep 29, 2010 #1
    1. The problem statement, all variables and given/known data
    Trying to show that if a matrix A=A_1+...+A_n is idempotent and that if the sum of the ranks of A_i is equal to the rank of A then each A_i is idempotent. The A_i's are symmetric and real, of course.

    This isn't really a homework question per se, I am writing notes for a reading class and I am trying to show that a particular distribution is chi^2 and I decided to add an appendix with cochrans theorem.

    2. Relevant equations


    I found this but the extension from beyond n=2 is not that obvious.

    3. The attempt at a solution
    It's obvious that A has an orthogonal decomposition of the form A=V^T diag(I_r,0)V and each of the A_i's have a similar decomposition by the spectral theorem. I realize that the rank will be unchanged by such an orthogonal decomposition. Each A_i will decompose into V_i^T diag(eigen1,eigen2,...,eigen(rank A_i),0,...) V_i.

    Any nudge in the right direction would be appreciated.
  2. jcsd
  3. Sep 29, 2010 #2


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    If you can prove the result for A = A1 + B, then consider the case where B= A2+A3. The general result will follow by induction.
  4. Sep 29, 2010 #3
    The size of the matrix and the number of A_i's are related.

    If you're talking about that particular link, if you do n=2 you're limited to looking at the sums A_1+A_2 since the rank of a nonzero quadratic form will never be 0.

    So for an inductive argument you would be going from n=2 to n=3 the size of the matrix increases from 2x2 to 3x3, so you would have to break up this argument somehow to get your result in the 2x2 and show that it works for the 3x3 while holding the assumptions. Which to me, is not all that obvious.
  5. Sep 29, 2010 #4


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  6. Sep 29, 2010 #5
    No, that doesn't work. B_i's are assumed to be positive definite.

    Oh wait! Aren't any two symmetric matrices of the same rank similar?
  7. Sep 29, 2010 #6


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    Similarity would imply that they have the same eigenvalues, so not in general.
  8. Sep 29, 2010 #7
    I am just drawing a blank.

    The real importance here is how A_i becomes an orthogonal family of matrices, ie [tex]A_iA_j=0[/tex] for [tex]i \not= j[/tex]. I think the key to this is looking at the nullspace of I-A_i...
    Last edited: Sep 30, 2010
  9. Sep 29, 2010 #8
    Idempotent matrices with the same rank are similar.

    Which doesn't help.
  10. Sep 30, 2010 #9
    I am still having trouble with this...

    I'll restate the problem to make it easier for anyone who could help me:

    Suppose [tex]A_1,\dots,A_k[/tex] are real symmetric matrices of order n with rank [tex] A_i = r_i[/tex].

    If [tex]A_1+\dotsb+A_k=A[/tex] is idempotent and rank [tex]A=r_1+\dots+r_k[/tex] then each [tex]A_i[/tex] must be idempotent.
  11. Sep 30, 2010 #10
    Last edited by a moderator: Apr 25, 2017
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