If M = ((0,0,-1),(1,0,-1),(0,1,0)) and A is an invertible rational 3xx3 matrix which commutes with M, then is A necessarily expressible as A = aM^2+bM+cI_3 for some scalar factors a, b, c?

This M is the companion matrix of the polynomial x^3+x+1. It satisfies:

M^3+M+I_3 = 0 and this is its minimum polynomial.

Hence the identity matrix I_3 with M generates a field of matrices all expressible in the form aM^2+bM+cI_3.

1 Answer
Feb 22, 2017

Yes...

Explanation:

Given:

M = ((0,0,-1),(1,0,-1),(0,1,0))

Then:

M^2 = ((0,0,-1),(1,0,-1),(0,1,0))((0,0,-1),(1,0,-1),(0,1,0)) = ((0, -1, 0),(0, -1, -1),(1, 0, -1))

Suppose A = ((c, u, v), (b, w, x), (a, y, z)) is a 3xx3 matrix that commutes with M.

We have:

AM = ((c, u, v), (b, w, x), (a, y, z))((0,0,-1),(1,0,-1),(0,1,0)) = ((u, v, -c-u), (w, x, -b-w), (y, z, -a-y))

MA = ((0,0,-1),(1,0,-1),(0,1,0))((c, u, v), (b, w, x), (a, y, z)) = ((-a, -y, -z), (c-a, u-y, v-z), (b, w, x))

Equating the elements of AM and MA, we find:

u = -a

w = c-a

y = b

v = -y = -b

x = -a-y = -a-b

z = w = c-a

So:

A = ((c, -a, -b),(b, c-a, -a-b),(a, b, c-a))

color(white)(A) = a((0, -1, 0),(0, -1, -1),(1, 0, -1))+b((0, 0, -1),(1, 0, -1),(0, 1, 0))+c((1, 0, 0),(0, 1, 0),(0, 0, 1))

color(white)(A) = aM^2+bM+cI_3