Re: Nilpotent matrix

From: Stephen Bianchi (swbianchi_at_earthlink.net)
Date: 09/26/04

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    Date: Sun, 26 Sep 2004 00:23:57 +0000 (UTC)

    On 27 Jan 2000 21:34:34 GMT, Robert Israel wrote:
    >In article <01bf68f6$ea6fc420$dda0a98e@default>,
    > "Serge Garneau" <garnote@globetrotter.qc.ca> writes:
    >
    >> Question 1 : Is it possible to generate a nilpotent randmatrix of
    index p
    >> ?
    >
    >I don't know how to find one that's uniformly distributed over the
    manifold
    >of all such matrices. But you might do the following:
    >
    >1) Choose n random vectors v1,...,vn, forming a random basis of R^n.

    >2) Choose integers p = p1 >= p2 >= ... >= pr where p1 + p2 + ... + pr
    = n.
    >
    >Construct a matrix N which is all 0 except for square blocks of sizes
    p1, p2, ...,
    >pr on the diagonal, each block having 1's on the first superdiagonal
    and 0's
    >elsewhere. Then do a change of basis, i.e. A = B N B^(-1) where B is
    >the matrix whose columns are v1,...,vn.
    >
    >> Question 2 : If I have a square matrix A of order n, is it
    possible to
    >> find
    >> his index of nipotence without try A , AA , AAA
    , AAAA
    >> ... ?
    >
    >Take a random vector v and compute v1 = A v, v2 = A v1, ... until vp
    = 0.
    >Then with probability 1, p is the index of nilpotence.
    >
    >> Question 3 : I suppose that the index of nilpotence of a matrix of
    order n
    >> is less than n+1. It's correct ?
    >
    >Yes, by the Cayley-Hamilton Theorem.
    >
    >Robert Israel israel@math.ubc.ca
    >Department of Mathematics <a
    href="http://www.math.ubc.ca/~israel">http://www.math.ubc.ca/~israel>

    >University of British Columbia
    >Vancouver, BC, Canada V6T 1Z2

    Is there a way to prove Question 3, without using Cayley-Hamilton or
    eigenvalues?

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