I agree with the first part "it is necessary and sufficient that M(B,x) divide C(A,x)", but not the second part "the multiplicity in B cannot exceed the multiplicity in A". See this Maxima script: Maxima 5.28.0-2 http://maxima.sourceforge.net using Lisp GNU Common Lisp (GCL) GCL 2.6.8 (a.k.a. GCL) Distributed under the GNU Public License. See the file COPYING. Dedicated to the memory of William Schelter. The function bug_report() provides bug reporting information. (%i1) A:diag_matrix(a,b,c); [ a 0 0 ] [ ] (%o1) [ 0 b 0 ] [ ] [ 0 0 c ] (%i2) B:diag_matrix(a,b,b); [ a 0 0 ] [ ] (%o2) [ 0 b 0 ] [ ] [ 0 0 b ] (%i3) Ap:charpoly(A,z),expand; 3 2 2 2 (%o3) - z + c z + b z + a z - b c z - a c z - a b z + a b c (%i4) Bp:charpoly(B,z),expand; 3 2 2 2 2 (%o4) - z + 2 b z + a z - b z - 2 a b z + a b (%i5) -B.B.B+(a+b+c)*B.B-(a*b+a*c+b*c)*B+a*b*c*ident(3),expand; [ 0 0 0 ] [ ] (%o5) [ 0 0 0 ] [ ] [ 0 0 0 ] (%i6) At 08:15 AM 5/20/2014, Eugene Salamin wrote:
Let C(A,x) and M(A,x) be the characteristic and minimal polynomials of A. Matrix B satisfies M(B,x) = 0. In order that B also satisfy C(A,x) = 0, it is necessary and sufficient that M(B,x) divide C(A,x). This requires that each eigenvalue of B must be an eigenvalue of A, and the multiplicity in B cannot exceed the multiplicity in A.
-- Gene
________________________________ From: Henry Baker <hbaker1@pipeline.com> To: math-fun@mailman.xmission.com Sent: Tuesday, May 20, 2014 7:46 AM Subject: [math-fun] Stupid Cayley-Hamilton question
Cayley-Hamilton tells us that a square matrix A satisfies its own characteristic polynomial.
Any square B with a subset of A's eigenvalues should also satisfy A's characteristic polynomial. (Note that B could be larger or smaller than A, and may have repeated eigenvalues.)
Are there any other B's that would satisfy A's characteristic polynomial ?