6
$\begingroup$

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?

$\endgroup$
10
  • 1
    $\begingroup$ It does not look nice. For example for $n=5$: $$x^5+5 \left(g^2-2\right) x^3+\frac{5}{2} \left(2 g^4+\sqrt{5} g^2-7 g^2-\sqrt{5}+7\right) x-2 \left(5 g^4+10 g^2+2\right).$$ $\endgroup$
    – user
    May 13 '20 at 13:57
  • 3
    $\begingroup$ 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... $\endgroup$ May 16 '20 at 17:49
  • 1
    $\begingroup$ Last row entries are 1 ... 1 1 or 1 0 ... 0 1? $\endgroup$
    – C.F.G
    May 19 '20 at 10:37
  • 1
    $\begingroup$ @C.F.G see the third row. Basically it is a tridiagonal matrix with nonzero corners $\endgroup$ May 19 '20 at 10:55
  • 1
    $\begingroup$ 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/…). $\endgroup$
    – jvc
    May 22 '20 at 6:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.