# How to get the characteristic polynomial of this matrix?

Consider a $$n\times n$$ matrix:

$$M_n = \begin{pmatrix} a_1 & 1 & 0 & 0 & 0 & \cdots & 1 \\ 1 & a_2 & 1 & 0 & 0 & \cdots & 0 \\ 0 & 1 & a_3 & 1 & 0 & \cdots & 0 \\ \vdots & \vdots& \vdots& \vdots & \vdots& \vdots & \vdots \\ 0 & \cdots & \cdots & 0 & 1 & a_{n-1} & 1 \\ 1 & \cdots & \cdots & \cdots & \cdots & 1 & a_n \end{pmatrix}$$

where $$a_k=2\cos(k\phi)+2\mathrm{i}\gamma\sin(k\phi)$$, with $$\phi=2\pi/n$$ and $$0<\gamma<1$$. $$~n\ge5$$, and can be assumed to be prime numbers if necessary.

How to get its characteristic polynomial $$P_n(x)=\det(M_n-xI)$$?

$$x^n$$, $$x^{n-2}$$ and $$x^0$$ terms are easy to get, can you get other terms?

• It does not look nice. For example for $n=5$: $$x^5+5 \left(g^2-2\right) x^3+\frac{5}{2} \left(2 g^4+\sqrt{5} g^2-7 g^2-\sqrt{5}+7\right) x-2 \left(5 g^4+10 g^2+2\right).$$
– user
May 13 '20 at 13:57
• I wonder whether it is not possible to find the traces of the powers of your matrix through induction. In that case, you could find the characteristic polynomial using the Newton-Girard identities. Just a suggestion... May 16 '20 at 17:49
• Last row entries are 1 ... 1 1 or 1 0 ... 0 1? May 19 '20 at 10:37
• @C.F.G see the third row. Basically it is a tridiagonal matrix with nonzero corners May 19 '20 at 10:55
• I am not brave enough to write explicitly the recurrence relations. It is more a suggestion. Because $M_n - x I_n$ has the same structure as $M_n$, we can compute the determinant. By doing cofactor expansion along the first line, and doing other cofactor expansions, we can obtain recurrence relation between $P_n$ and determinants of the same type but in inferior dimension. It will appear tridiagonal determinant which can be computed (math.stackexchange.com/questions/1318017/…).
– jvc
May 22 '20 at 6:55