# How to simplify the characteristic polynomial of a given matrix?

Reading through this paper I've come across a statement that I don't follow, could someone give some pointers/hints?

Let $A$ be the $2n\times 2n$ matrix given by

$$A=(I_n\otimes F)+(G\otimes H),\quad\quad(*)$$

where

$$F=\begin{bmatrix}0&1\\-c_2c_4&-(c_2+c_4)\end{bmatrix},\quad\quad H=\begin{bmatrix}0&0\\ c_1c_3\alpha&0\end{bmatrix},$$

with $\alpha,c_1,\dots,c_4\in\mathbb{R},$ and $G$ is the $n\times n$ circulant matrix

$$G=\begin{bmatrix}0&0&0&\dots&1\\1&0&0&\dots&0\\&&\vdots&&\\0&\dots&1&0&0\\0&\dots&0&1&0\end{bmatrix}.$$

The authors write: "Fourier-diagonalizing the circulant matrix $G$, we obtain the following complex quadratic equation for the eigenvalues $\lambda_l$ of the matrix $A$:

$$\lambda_l^2+(c_2+c_4)\lambda_l+c_2c_4\alpha\left(1-\frac{c_1c_3}{c_2c_4}\alpha e^{i\frac{2\pi}{n}l}\right),\quad\quad (**)$$

with $l=0,\dots,n-1$."

From the above I assume they showed that the characteristic polynomial of $A$ can be factored into $n$ quadratic polynomials in $(**)$. Fourier-diagonalizing $G$ gives

$$G=\frac{1}{n}U^*\text{diag}(u)U,\quad\quad U=\begin{bmatrix}\omega_n^{0\cdot0}&\omega_n^{0\cdot1}&\dots&\omega_n^{0\cdot(n-1)}\\ \omega_n^{1\cdot0}&\omega_n^{1\cdot1}&\dots&\omega_n^{1\cdot(n-1)}\\ \vdots&\vdots&\ddots&\vdots\\ \omega_n^{(n-1)\cdot0}&\omega_n^{(n-1)\cdot1}&\dots&\omega_n^{(n-1)\cdot(n-1)}\end{bmatrix},\quad\quad u =\begin{bmatrix}\omega_n^{0\cdot1}\\\omega_n^{1\cdot1}\\\vdots\\\omega_n^{(n-1)\cdot1}\end{bmatrix},$$

where $\omega_n=e^{-\frac{2\pi}{n}}$, and diag$(u)$ denotes the diagonal matrix with vector $u$ on its diagonal. This is as far as I get, any hints on how to proceed?

Clearly, the formula is false. Indeed, let $\alpha=0$ and assume $c_2c_4\not=0$.Then $A=I\otimes F$ is invertible and necessarily $\lambda_l\not= 0$, that is a contradiction. On the other hand, your formula concerning $\omega_n$ is false. Indeed $G^n=I$, then $w_n=exp(2i\pi/n)$. Here $U$ has nothing to do because $I$ is invariant by a change of basis. More precisely you can take $U=I$. The calculation is easy; we find (we note $\omega$ for $\omega_n$) a diagonal of $2\times 2$ matrices in the form $\begin{pmatrix} 0&1\\-c_2c_4+\omega^lc_1c_3\alpha&-(c_2+c_4)\end{pmatrix}$. Finally the eigenvalues are the roots of the polynomials $\lambda_l^2+(c_2+c_4)\lambda_l+c_2c_4-\omega^lc_1c_3\alpha$ and $\alpha$ is not in the factor.