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Matrix Operation

  1. Feb 4, 2009 #1
    1. The problem statement, all variables and given/known data

    If S is a lower triangular matrix (n by n), S^n is the zero matrix.



    2. Relevant equations



    3. The attempt at a solution

    S =

    0 0 0 0 ... 0
    1 0 0 0 ... 0
    0 1 0 0 ... 0
    0 0 1 0 ... 0
    ...
    0 0 0 0 .1.0


    S^2 =


    0 0 0 0 ... 0
    0 0 0 0 ... 0
    1 0 0 0 ... 0
    0 1 0 0 ... 0
    ...
    0 0 0 0 1.. 0


    How do I show this using induction?
     
  2. jcsd
  3. Feb 4, 2009 #2
    Are you sure it's true?
     
  4. Feb 4, 2009 #3
    No. So given S as I defined above then computing S^2, and S^3 so forth, the entries lower each row for each n. how do I generalize this?
     
  5. Feb 4, 2009 #4
    Use the definition of matrix multiplication to show that the value of Sk-1k,1 = 1 but Sk-1< k,* = 0 and Sk-1k, >1and<n = 0 => Sk*,* = 0.

    Where * means any row and any column.

    edit: If you are trying to prove that statement in general, you might want to consider the identity matrix which is both n x n and lower triangular.
     
  6. Feb 5, 2009 #5

    Mark44

    Staff: Mentor

    First, a lower triangular matrix is one where the entries above the diagonal are zero. The matrix you show has ones down the lower subdiagonal and zeroes everywhere else.
    Second, the statement isn't true.

    Here's a counterexample for n = 3.

    S =
    [1 0 0]
    [1 1 0]
    [1 1 1]

    S^2=
    [1 0 0]
    [2 1 0]
    [3 2 1]

    S^3=
    [1 0 0]
    [3 1 0]
    [6 3 1]

    The upshot is that S is a lower triangular 3 x 3 matrix, but S^3 != 0.
     
  7. Feb 5, 2009 #6

    gabbagabbahey

    User Avatar
    Homework Helper
    Gold Member

    Are you sure the question isn't in regards to a 'strictly lower triangular matrix'?
     
  8. Feb 5, 2009 #7
    The matrix is this:

    S =

    0 0 0 0 ... 0
    1 0 0 0 ... 0
    0 1 0 0 ... 0
    0 0 1 0 ... 0
    ...
    0 0 0 0 .1.0

    it's not a lower triangular matrix.
     
  9. Feb 5, 2009 #8

    Mark44

    Staff: Mentor

    It would be helpful if we knew exactly how the problem is stated. In your first post, here is what you had:
    The somewhat sparse n x n matrix with 1s on the subdiagonal appeared in your attempt at a solution.

    Is the problem statement to show that for that matrix S, S^n = 0?

    Inquiring minds would like to know...
     
  10. Feb 5, 2009 #9
    Okay.
    Given
    =

    0 0 0 0 ... 0
    1 0 0 0 ... 0
    0 1 0 0 ... 0
    0 0 1 0 ... 0
    ...
    0 0 0 0 .1.0

    Compute (S^n)(S^n*) and (S^*n)(S^n)

    where * is the complex conjugate transpose.
     
  11. Feb 6, 2009 #10

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Actually, that is a "lower triangular matrix" (it has only 0s above the main diagonal), just not a general lower triangular matrix as was intially implied.
     
  12. Feb 7, 2009 #11
    So in Compute (S^n)(S^n*) and (S^*n)(S^n)
    they're both zero correct?
     
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