Proof of Strictly Upper Triangular Matrix Property by James

[jdstokes]

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

 

[matt grime]

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.

 

[thepatientmental]

,

Your proof for the strictly upper triangular matrix property is correct. You have correctly shown that the (i,j)-th entry of AA is 0 if i >= j - 1. This is because for i >= j - 1, there is no k such that i < k < j, therefore the sum \sum_{i<k, k<j} a_{ik} b_{kj} is equal to 0.

For the related problem, we can use a similar proof technique. Let B = AB where A is a strictly upper triangular matrix and B is an n x n matrix with entries b_{kj}. Then, the (i,j)-th entry of B is given by b_{ij} = \sum_{k=1}^{n} a_{ik} b_{kj}. We can show that if i >= j - (p+1), then there is no k such that i < k < j - p. Therefore, the sum \sum_{i<k, k<j-p} a_{ik} b_{kj} is equal to 0. This means that the (i,j)-th entry of B is also equal to 0, and thus A^n = 0. This is because B = A^p * A^{n-p} = 0 * A^{n-p} = 0.

I hope this helps! Let me know if you have any further questions.