Peculiar Matrix I came up with this idea recently and I want to go deeper in this, but it has been difficult to me. Hope someone can help me on this.
Suppose I have a matrix of order $(n^2-1)\times (n^2-1)$ with eigenvalues $\lambda_1,\ldots, \lambda_{n^2-1}$. I can write its characteristic polynomial as $$p(z) = a_{n^2-1}z^{n^2-1}+a_{n^2-2}z^{n^2-2}+\ldots+a_1z + a_0.$$
After this, we can construct the $n\times n$ matrix given by
$$M = \begin{pmatrix}
a_0 & a_1 & \ldots & a_{n-1}\\
a_n & a_{n+1} & \ldots & a_{2n-1}\\
\vdots & \vdots & \ddots & \vdots\\
a_{n^2-n} & a_{n^2-n+1} & \ldots & a_{n^2-1}
 \end{pmatrix}$$
My question is: what can we say about the matrix $M$?
I know this question is a little vague, I'm just starting to explore this matrix. Thus, any interesting result is very welcome.
 A: Regarding the first part, since you have distinct $\lambda_1,\ldots,\lambda_{n^2-1}$ (was this the intention?) you know that your matrix is diagonalizable. 
From the characteristic polynomial you know that the trace of your matrix is $-a_{n^2-2}/a_{n^2-1}$ (and $a_{n^2-1}$ must be nonzero, else you don't have an $n^2-1 \times n^2-1$ matrix). 
Also from the characteristic polynomial the determinant of your matrix is $(-1)^{n^2-1}a_0/a_{n^2-1}$.

ok, I added this part later, once I understood that your main interest is in the matrix $M$ of coefficients of the characteristic polynomial, as defined by you above. Well from the above the only thing you can really say is that $a_{n^2-1}$ is nonzero. So really you have the set \begin{equation} \left \{ (m_{ij}) \in \mathbb{F}_{n \times n}:m_{nn}\neq0\right \}.\end{equation} In fact we could restrict it a little further by noticing that we always have $a_{n^2-1}=1$ since the characteristic polynomial is always monic. So then we have the set \begin{equation} \left \{ (m_{ij}) \in \mathbb{F}_{n \times n}:m_{nn}=1\right \}.\end{equation}
