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I'm stuck on a matrix multiplication problem

  1. Oct 7, 2009 #1
    1. The problem statement, all variables and given/known data

    If AB = BA and p is a nonnegative integer, show that (AB)^p = (A^p)(B^p)

    2. Relevant equations

    N/A

    3. The attempt at a solution

    I would love to attempt a solution, but I don't even know where to begin. I guess my first question is, how can I generalize matrices so that AB = BA? And why is it important that AB = BA?

    Thanks in advance.
     
  2. jcsd
  3. Oct 7, 2009 #2

    Avodyne

    User Avatar
    Science Advisor

    You're not generalizing matrices. The problem means, suppose you have two particular matrices, called A and B, and for these two particular matrices, it happens to be the case that AB = BA. Now, you are supposed to show that (AB)^p = A^p B^p.

    It might help to first try a specific example, p=2.
     
  4. Oct 7, 2009 #3
    thanks for the reply. i find this kind of weird...so is it...

    (AB)^p
    = ABABAB...AB
    = A(BA)(BA)(BA)...(BA)B
    = A(AB)(AB)(AB)...(AB)B
    = AA(BA)(BA)...(BA)BB
    = AAA(BA)(BA)...(BA)BBB

    and so on? it makes sense to me, but i'm not sure if this is the correct way to do it.
     
  5. Oct 7, 2009 #4

    Mark44

    Staff: Mentor

    Rather than a proof with "..." in the middle, I think you want to do this by mathematical induction. Show that the statement is true for some base case, such as p = 2. Assume that it is true for p = k. Finally show that if it is true for p = k, it must also be true for p = k + 1.
     
  6. Oct 7, 2009 #5
    how about an inductive proof?

    (AB)^2 = (AB)(AB) = A(BA)B = A(AB)B = (A^2)(B^2)

    assume that (AB)^p = (A^p)(B^p).

    (AB)^(p+1) = (AB)^p (AB)
    = (A^p)(B^p)(AB)
    = (A^p)(B^p)(BA)
    = (A^p)(B^p+1)(A)
    ...
    not sure where to go from here. am i allowed to just move that last (AB) into the middle?

    thanks mark, i got your reply while i was typing this.
     
  7. Oct 7, 2009 #6
    i'm really confused now. the next couple of homework questions i have are:

    If p is a nonnegative integer and c is a scalar, show that (cA)^p = c^p A^p.

    For a square matrix A and a nonnegative integer p, show that (A^T)^p = (A^p)^T.

    For a nonsingular matrix A and a nonnegative integer p, show that (A^p)^-1 = (A^-1)^p.

    For a nonsingular matrix A and nonzero scalar k, show that (kA)^-1 = (1/k)(A^-1).

    i'm frustrated because my textbook tells me practically nothing about A^p and A^-1. my instructor didn't go over it much either. we basically just did problems with concrete numbers in class, no proofs.

    well, if anyone has any hints, i'd really appreciate it. otherwise, i'm just going to keep trying to figure it out.
     
  8. Oct 7, 2009 #7

    Mark44

    Staff: Mentor

    (AB)p + 1 = (AB)pAB = ApBpAB
    = ApBp - 1BAB
    = ApBp - 1ABB
    = ApBp - 1AB2

    If I continue this process p - 1 more times, I'll end up with all the A's on one end and all the B's on the other end. At each step I'm using the facts that AB = BA and that matrix multiplication is associative.

    That's the best I can come up with at the moment. Maybe someone else in this forum has a better idea.
     
  9. Oct 7, 2009 #8
    Thank you.
     
  10. Oct 7, 2009 #9

    Mark44

    Staff: Mentor

    For a nonnegative integer p, Ap = AA...A where there are p factors of A.
    A-1 is read as "A inverse" and that's the reason for the requirement that A be nonsingular, which is the same as saying that A has an inverse. It is implicit that A is a square matrix. For a nonsingular or invertible matrix A, AA-1 = A-1A = I, the identity matrix, a matrix with 1s down the main diagonal and 0s everywhere else. If A is n x n, I has to be the n x n identity matrix.

    You didn't ask, but just to be complete, AT is the transpose of A, a matrix obtained by switching the rows and the columns.

    Being that all these problems have some matrix raised to the power p, induction proofs are called for.
     
  11. Oct 7, 2009 #10
    tyvm, that helps a lot.
     
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