# 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 ${x}^{5} + 4 x + 2$. It satisfies: ${M}^{5} + 4 M + 2 = 0$ and since ${x}^{5} + 4 x + 2$ is irreducible (over $\mathbb{Q}$), this is its minimum polynomial. Hence the identity matrix ${I}_{5}$ with $M$ generates a field of matrices all expressible in the form $a {M}^{4} + b {M}^{3} + c {M}^{2} + \mathrm{dM} + e$.

Mar 10, 2018

Yes

#### Explanation:

Note that the powers of $M$ are:

${M}^{0} = \left(\begin{matrix}1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1\end{matrix}\right)$

${M}^{1} = \left(\begin{matrix}0 & 0 & 0 & 0 & - 2 \\ 1 & 0 & 0 & 0 & - 4 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0\end{matrix}\right)$

${M}^{2} = \left(\begin{matrix}0 & 0 & 0 & - 2 & 0 \\ 0 & 0 & 0 & - 4 & - 2 \\ 1 & 0 & 0 & 0 & - 4 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0\end{matrix}\right)$

${M}^{3} = \left(\begin{matrix}0 & 0 & - 2 & 0 & 0 \\ 0 & 0 & - 4 & - 2 & 0 \\ 0 & 0 & 0 & - 4 & - 2 \\ 1 & 0 & 0 & 0 & - 4 \\ 0 & 1 & 0 & 0 & 0\end{matrix}\right)$

${M}^{4} = \left(\begin{matrix}0 & - 2 & 0 & 0 & 0 \\ 0 & - 4 & - 2 & 0 & 0 \\ 0 & 0 & - 4 & - 2 & 0 \\ 0 & 0 & 0 & - 4 & - 2 \\ 1 & 0 & 0 & 0 & - 4\end{matrix}\right)$

Notice in particular that the left hand column of:

$a {M}^{4} + b {M}^{3} + c {M}^{2} + \mathrm{dM} + e I$

is:

$\left(\begin{matrix}e \\ d \\ c \\ b \\ a\end{matrix}\right)$

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

$\left(\begin{matrix}{a}_{11} \\ {a}_{21} \\ {a}_{31} \\ {a}_{41} \\ {a}_{51}\end{matrix}\right)$

we find that the matrix:

$A - \left({a}_{51} {M}^{4} + {a}_{41} {M}_{3} + {a}_{31} {M}_{2} + {a}_{21} M + {a}_{11} I\right)$

has left hand column:

$\left(\begin{matrix}0 \\ 0 \\ 0 \\ 0 \\ 0\end{matrix}\right)$

...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:

$A = {a}_{51} {M}^{4} + {a}_{41} {M}^{3} + {a}_{31} {M}^{2} + {a}_{21} M + {a}_{11} I$