Question

If `M = ((0,0,0,0,-2),(1,0,0,0,-4),(0,1,0,0,0),(0,0,1,0,0),(0,0,0,1,0))` and `A` is an invertible rational `5xx5` matrix which commutes with `M`, then is `A` necessarily expressible as `A = aM^4+bM^3+cM^2+dM+e` for some scalar factors `a, b, c, d, e`?

This `M` is the companion matrix of the polynomial `x^5+4x+2`. It satisfies:

`M^5+4M+2 = 0` and since `x^5+4x+2` is irreducible (over `QQ`), this is its minimum polynomial.

Hence the identity matrix `I_5` with `M` generates a field of matrices all expressible in the form `aM^4+bM^3+cM^2+dM+e`.

Answer 1

Yes

Explanation:

Note that the powers of `M` are:

`M^0 = ((1, 0, 0, 0, 0),(0, 1, 0, 0, 0),(0, 0, 1, 0, 0),(0, 0, 0, 1, 0), (0, 0, 0, 0, 1))`

`M^1 = ((0, 0, 0, 0, -2), (1, 0, 0, 0, -4), (0, 1, 0, 0, 0), (0, 0, 1, 0, 0), (0, 0, 0, 1, 0))`

`M^2 = ((0, 0, 0, -2, 0), (0, 0, 0, -4, -2), (1, 0, 0, 0, -4), (0, 1, 0, 0, 0), (0, 0, 1, 0, 0))`

`M^3 = ((0, 0, -2, 0, 0), (0, 0, -4, -2, 0), (0, 0, 0, -4, -2), (1, 0, 0, 0, -4), (0, 1, 0, 0, 0))`

`M^4 = ((0, -2, 0, 0, 0), (0, -4, -2, 0, 0), (0, 0, -4, -2, 0), (0, 0, 0, -4, -2), (1, 0, 0, 0, -4))`

Notice in particular that the left hand column of:

`aM^4+bM^3+cM^2+dM+eI`

is:

`((e),(d),(c),(b),(a))`

So given an invertible matrix `A` with left hand column:

`((a_11),(a_21),(a_31),(a_41),(a_51))`

we find that the matrix:

`A-(a_51M^4+a_41M_3+a_31M_2+a_21M+a_11I)`

has left hand column:

`((0),(0),(0),(0),(0))`

...so is not invertible.

If `A` commutes with `M` then `A` and `M` generate a field of matrices, so the only non-invertible matrix in the field is `0` and we deduce that:

`A=a_51M^4+a_41M^3+a_31M^2+a_21M+a_11I`