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Commuting Invertible Matrices

  1. Oct 16, 2004 #1
    Hey all,
    I'm having trouble showing the following point:
    if A is an invertible (n×n, real) matrix that commutes with ALL other invertible (n×n, real) matrices, then A is of the form cI, where c is any real number not equal to 0.

    Anyone know how to show this?
     
    Last edited: Oct 16, 2004
  2. jcsd
  3. Oct 16, 2004 #2

    Hurkyl

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    Try picking some particularly simple matrices B (such as having only a single nonzero entry!) and use AB = BA to get equations for the entries of A.
     
  4. Oct 16, 2004 #3
    Thanks for the suggestion. So I had to use matrices B of the form I with an additional 1 at one other entry. By brute force, calculating the corresponding entries of AB would eventually give that all diagonal entries of A were the same, and that all other non diagonal entries must be zero. Sweet.
     
  5. Oct 16, 2004 #4

    Hurkyl

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    Actually, I would pick my B's so they have only one nonzero entry, rather than being the identity with an additional nonzero entry.

    I think this is the easiest approach to understand, though not the shortest.
     
  6. Oct 16, 2004 #5
    Yes, I was considering that, but B must also be invertible, so I had to complicate B a little bit by adding the diagonal 1's.
     
  7. Oct 16, 2004 #6

    Hurkyl

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    Ah, good point, I had missed that.

    However, note this (reversible) deduction:

    A(I+B) = (I+B)A
    A + AB = A + BA
    AB = BA


    B commutes with A iff (I+B) commutes with A, so you could still work with my B's, if you choose.
     
  8. Oct 16, 2004 #7
    That's wonderful! Thanks a lot for the help :smile:
     
  9. Oct 17, 2004 #8

    mathwonk

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    Here's a conceptual suggestion. Think of A as a transformation and let v be any non zero vector. We want to show first that Av = cv for some constant c.

    Choose a basis involving v as first vector and define another transformation B that takes v to v and the other basis vectors to 0. Then applying AB = BA to v shows that ABv = Av = BAv, so B acts on Av != 0 as the identity. Since A is invertible, Av is not zero, so then Av = cv for some non zero constant c, since multiples of v are the only non zero vectors B takes to themselves.

    Thus every vector v is an "eigenvector" for A, i.e. Av always equals cv for some c possibly depending on v. Now if every vector is an eigen - vector, we claim all the eigenvalues must be equal.

    To see that, assume that Av = cv and Aw = dw, where v and w are in different directions, and look at A(v+w) = cv + dw = e(v+w), and check that we must have c=d=e.

    Now why is that? well then cv + dw -ev -ew = 0 = (c-e)v + (d-e)w, hence
    (c-e)v = (e-d)w, so these multiples of v and w are equal.

    But since v, w are in different directions, their only equal multiples are zero, so d=e, c=e. QED.
     
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