Path: utzoo!utgpu!jarvis.csri.toronto.edu!cs.utexas.edu!samsung!usc!jarthur!uci-ics!ucla-cs!math.ucla.edu!sonia!pmontgom From: pmontgom@sonia.math.ucla.edu (Peter Montgomery) Newsgroups: comp.theory Subject: Re: Matrix Properties Message-ID: <2061@sunset.MATH.UCLA.EDU> Date: 30 Nov 89 03:29:27 GMT References: <1989Nov29.155406.23647@ee.rochester.edu> Sender: news@MATH.UCLA.EDU Reply-To: pmontgom@math.ucla.edu (Peter Montgomery) Organization: UCLA Mathematics Department Lines: 32 In article ben@nsf1.mth.msu.edu writes: >>>>>> On 29 Nov 89 15:54:06 GMT, bobm@ee.rochester.edu (Bob Molyneaux) said: > >Bob> Is there a classification for the square matrix A for which A^k >Bob> = A for not all but some value(s) of k? > >Let k be the smallest integer greater than 1 such that A^k = A. The >minimal polynomial of A must then divide x^k - x. The roots of the >minimal polynomial are either 0 or a (k-1)st root of unity. Since any >root has multiplicity 1, A is diagonalizable. > The claim that the roots of x^k - x have multiplicity 1 assumes A is defined over a field of characteristic 0, such as the real or complex numbers. It is not true in general. For example, x^4 - x = x(x-1)^3 in a field of characteristic 3 (one in which 1 + 1 + 1 = 0). If A = ( 1 1 ) , A^k = ( 1 k ), ( 0 1 ) ( 0 1 ) then A^k = A if the characteristic of the base field divides k-1, yet A is never diagonalizable (its minimal polynomial is (X-1)^2). As regards the original (Bob's) question, a necessary condition for A^k = A is that A^(k-1) be idempotent, but this condition is not sufficient for k > 2 (let A be a matrix whose square is 0). I don't know a name for the condition. -------- Peter Montgomery pmontgom@MATH.UCLA.EDU Department of Mathematics, UCLA, Los Angeles, CA 90024 Brought to you by Super Global Mega Corp .com