Re rational canonical form: Most of you probably know this already, but I looked at perhaps 25 presentations, and *no one* made it this easy. M is an arbitrary matrix (over the rationals, for example). C is the companion matrix for the characteristic polynomial of M. We here assume that the characteristic polynomial is *irreducible*. We need to find an invertible T such that (T^^-1).M.T = C. Choose T such that the first column T1 is random, the second column is T2=M.T1, and the third column is T3=(M^^2).T1=M.M.T1=M.T2, etc. If this matrix T1:T2:T3 is invertible, it will work. Unless you're exceedingly unlucky, it will work with the first random choice of T1. So, once you've got the characteristic polynomial and know that it doesn't factor, the rest is trivial. BTW, Cayley, Sylvester, etc. would have understood this argument; they would have found all that stuff about "modules" to be completely inscrutable and extremely unhelpful -- i.e., "abstract nonsense". (%i2) M; [ a b c ] [ ] (%o2) [ d e f ] [ ] [ g h i ] (%i3) C:polytocompanion(charpoly(M,x),x),expand; [ 0 0 a e i - b d i - a f h + c d h + b f g - c e g ] [ ] (%o3) [ 1 0 - e i - a i + f h + c g - a e + b d ] [ ] [ 0 1 i + e + a ] (%i4) T; [ x ] [ c z + b y + a x ] [ ] [ ] (%o4) Col 1 = [ y ] Col 2 = [ f z + e y + d x ] [ ] [ ] [ z ] [ i z + h y + g x ] [ c (i z + h y + g x) + b (f z + e y + d x) + a (c z + b y + a x) ] [ ] Col 3 = [ f (i z + h y + g x) + e (f z + e y + d x) + d (c z + b y + a x) ] [ ] [ i (i z + h y + g x) + h (f z + e y + d x) + g (c z + b y + a x) ] (%i5) M.T-T.C,expand; [ 0 0 0 ] [ ] (%o5) [ 0 0 0 ] [ ] [ 0 0 0 ]