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 .
This
M^3+M+I_3 = 0 and this is its minimum polynomial.
Hence the identity matrix
1 Answer
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
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
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