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Matrix Proofs

  1. May 22, 2005 #1

    I need help with this proof relating to strictly upper-triangular

    Let A be an n x n strictly upper triangular matrix. Then the (i,j-th
    entry of AA = A^2 is 0 if i >= j - 1.

    Here's what I have.

    Pf: Let B = A^2. The (i,j)-th entry of B is given by

    b_{ij} = \sum_{k=1}^{n} a_{ik} a_{kj}.

    If k >= j, a_{kj} = 0. If i >= k, a_{ik} = 0. If i >= j - 1, then there
    is no k s.t. i < k < j. Therefore

    b_ij = \sum_{i<k, k<j} a_{ik} b_{kj} ==> b_{ij} = 0.

    Also, if anyone has any clues on these related problems, it would be
    greatly appreciated.

    Suppose p is a given integer satisfying 1 <= p <= n -1 and that the
    entries b_{kj} of an n x n matrix B satisfy b_{kj} = 0 for k >= j - p.
    Show that the (i,j)-th entry of the product AB is zero if i >= j -
    (p+1). Deduce from the previous result that A^n = 0.

  2. jcsd
  3. May 23, 2005 #2

    matt grime

    User Avatar
    Science Advisor
    Homework Helper

    Just look at, say a 3x3 strictly uppwer triangular matirx and square it and then cube it, then raise it to the 4'th power. what happens to the entries? prove it happens in general.
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