bobm@ee.rochester.edu (Bob Molyneaux) (11/29/89)
Hi All, A square matrix A is said to be nilpotent if A^k = 0 for some value of k. A square matrix A is said to be idempotent if A^2 = A ==> A^k = A for all values of k. Is there a classification for the square matrix A for which A^k = A for not all but some value(s) of k? Please email resonses along with references. (Subject references not yours!) Thanks bobm.ee.rochester.edu
ben@nsf1.mth.msu.edu (Ben Lotto) (11/30/89)
>>>>> 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 characterization: A is such a matrix if and only if A is
diagonalizable and every eigenvalue of A either equals 0 or an integral
root of unity. The smallest k that works is 1 greater than the least
common multiple of the order of the root of unity eigenvalues.
A reference: Herstein's *Topics in Algebra*, chapter on linear
transformations. In particular, the section on Jordan canonical form.
Any good linear algebra book that covers Jordan canonical form should
cover this material as well.
--
-B. A. Lotto (ben@nsf1.mth.msu.edu)
Department of Mathematics/Michigan State University/East Lansing, MI 48824pmontgom@sonia.math.ucla.edu (Peter Montgomery) (11/30/89)
In article <BEN.89Nov29121420@lamm.nsf1.mth.msu.edu> 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