Matrix Proofs

  • Thread starter jdstokes
  • Start date
  • #1
523
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Hi,

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

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.

James
 

Answers and Replies

  • #2
matt grime
Science Advisor
Homework Helper
9,420
4
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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