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 ?