Question

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`.

Answer 1

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`