I've been chewing on the following problem for some time and I just don't have any more ideas how to tackle it. I have matrices $A\_1,...,A\_k \in \mathbb R^{n\times n}$ and I'm observing that the spectrum of the product is the same whatever order of multiplication I choose, that is, for an arbitrary permutation $\pi$ on ${1,...,k}$, I have
$$\sigma(A\_1\cdot ...\cdot A\_k) = \sigma(A\_{\pi(1)}\cdot ...\cdot A\_{\pi(k)}).$$
Of course, the form of the matrices is important since the above equality doesn't hold in general. All are of the form:
$$ A\_j = \left[
 \array{0 & 1 & 0 & \cdots & 0 \\
 0 & \ddots & \ddots & \ddots & 0 \\
 0 & \cdots & 0 & 1 & 0 \\
 0 & \cdots & \cdots & 0 & 1 \\
 a\_j & \cdots & \cdots & \cdots & a\_j }
 \right],$$
with possibly all different $a\_j \in\mathbb{R},\ j=1,...,k$. I find this quite peculiar and I don't see the reason, although all tested examples confirm the statement... I would be very curious and grateful if someone has an idea how to solve that...
Thanks ahead for your comments! 



Why is the spectrum of this matrix product invariant with respect to order of the multiplicants?
It seems that the claim is false when $n=2$ and $k=4$, because
$${\rm Tr} (A\_1A\_2A\_3A\_4)=a\_1a\_3+a\_2a\_4+\sigma\_3({\bf a})+\sigma\_4({\bf a})$$
is not symmetric in ${\bf a}=(a\_1,a\_2,a\_3,a\_4)$.


**Edit**: just let me add a classical result. Let $C\_1,\ldots,C\_r$ be the companion matrices of unitary polynomials $P\_1,\ldots,P\_r$ of same degree $n$. Suppose that $\lambda$ is a common root to all the $P\_j's$. Then $\lambda^r$ is an eigenvalue of the product $C\_1\cdots C\_r$.


