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find a recursive relation for the characteristic polynomial of the $k \times k $ matrix $$\begin{pmatrix} 0 & 1 \\ 1 & 0 & 1 \\ \mbox{ } & 1 & . & . \\ \mbox{ } &\mbox{ } & . & . &. & \mbox{ } \\ \mbox{ } &\mbox{ } &\mbox{ } & . & . & 1 \\ \mbox{ } & \mbox{ } & \mbox{ } & \mbox{ } & 1 & 0\end{pmatrix}$$

and compute the polynomial for $k\le 5$

My attempt : Let $M_k$ be the $k\times k$ matrix and $P_k(x)=\det(M_k-xI_k)$ be its characteristic polynomial. We have

$$P_k(x)=\det\begin{pmatrix} -x & 1 \\ 1 & -x & 1 \\ \mbox{ } & 1 & \ddots & \ddots \\ \mbox{ } &\mbox{ } & \ddots & . &. & \mbox{ } \\ \mbox{ } &\mbox{ } &\mbox{ } & . & . & 1 \\ \mbox{ } & \mbox{ } & \mbox{ } & \mbox{ } & 1 & -x\end{pmatrix}$$

after that im not able proceed further

Any hints/solution will be appreciated

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    $\begingroup$ do it first for $k=2$ and $k=3. \;$ I found a quote "There is nothing impossible to him who will try." $\endgroup$ – Will Jagy Jan 4 '19 at 22:15
  • $\begingroup$ ya ,,,im trying @WillJagy.....haa $\endgroup$ – jasmine Jan 4 '19 at 22:17
  • $\begingroup$ what do you get for $k=2?$ $\endgroup$ – Will Jagy Jan 4 '19 at 22:19
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    $\begingroup$ now do $k=3.$ By the way, there is a pretty good recipe for writing the char poly in dimension 3, where $\sigma_1$ is the trace and $\sigma_3$ is the determinant, we also need $\sigma_2$ which is the sum of the little 2 by 2 prinipal minor determintns, Then char poly is $x^3 - \sigma_1 x^2 + \sigma_2 x - \sigma_3.$ I think there is also a recipe for $\sigma_2$ involving traces of the matrix squared, I will see if I can find that. Worth getting right and memorizing for dimension 2 and 3 $\endgroup$ – Will Jagy Jan 4 '19 at 22:24
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    $\begingroup$ Alright, worked it out, for matrix $M,$ $$\sigma_2 = \frac{1}{2} \left( \sigma_1^2 - \operatorname{trace} M^2 \right)$$ where $\sigma_1 = \operatorname{trace} M$ as i indicated $\endgroup$ – Will Jagy Jan 4 '19 at 22:35
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Hint: Look at the formula for the determinant of using the first row. Then you get a recursive definition for the determinant.

Solution:

$P_k(x)=-xP_{k-1}(x)-P_{k-1}(-1)$

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