How to find the eigenvalues and Jordan canonical form of this matrix Question:
let $a_{i,j}\in R,A=(a_{i,j})_{n\times n} $,and 
$a_{i,j}=\begin{cases}
1&i+j\in\{n,n+1\}\\
0&i+j\notin\{n,n+1\}
\end{cases}$
that's meaning:
$$A=\begin{bmatrix}
0&0&0&\cdots&0&1&1\\
0&0&0&\cdots&1&1&0\\
0&0&\cdots&1&1&0&0\\
\vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\vdots\\
0&1&1&\cdots&0&0&0\\
1&1&0&\cdots&0&0&0\\
1&0&0&\cdots&0&0&0
\end{bmatrix}_{n\times n}$$
Problem (1): Find the Jordan canonical form of $A$.
I know this matrix Jordan is 
$$diag(\lambda_{1},\lambda_{2},\cdots,\lambda_{n})$$
where $\lambda_{i}$ is eigenvalue
But this problem key find the eigenvalue is hard,
Thank you.maybe this problem is not easy,But I hope see someone can solve it.Thank you very much!

Although Ewan Delanoy was more quick than I, :-)  I’ll continue user1551‘s answer.  
It seems the following. Denote the determinant of the $n\times n$ matrix $xI-A$ as $q_n(x)$. Then we can calculate the first polynomials: 
$q_1=x-1$ 
$q_2=x^2-x-1$
$q_3=x^3-x^2-2x+1$
$q_4=x^4-x^3-3x^2+2x+1$
$q_5=x^5-x^4-4x^3+3x^2+3x-1$
$q_6=x^6-x^5-5x^4+4x^3+6x^2-3x-1$
$q_7=x^7-x^6-6x^5+5x^4+10x^3-6x^2-4x+1$
Investigating decomposition of the determinant $|xI-A|$ by the first row, we obtain the following recurrence: 
$q_n=(x^2-1)q_{n-2}-xq_{n-3}$ for each $n\ge 4$.
Adding for the completeness the polynomial $q_0\equiv 1$, we can easily prove the recurrence 
$q_n=xq_{n-1}-q_{n-2}$ for each $n\ge 2$. 
The Google search by these polynomials suggests that roots of the polynomial $q_n$ are $x_{k}=2\cos\frac{2k+1}{2n+1}\pi $, where $k=0,\dots,n-1$ (but not always for $k=n$). 
Geometric interpretation (and Ewan Delanoy’s solution :-) ) suggest to try to take as $q_n$ a sum $s_n(x)=(-1)^n\sum_{j=-n}^n \cos jt,$ where $t=\arccos (-x/2)$. 
We can check that $s_n=q_n$ for all $n\le 3.$  Let $n\ge 2$. Then 
$$s_n(x)+s_{n-2}(x)= (-1)^n\sum_{j=-n}^n \cos jt+(-1)^{n-2}\sum_{j=-n+2}^{n-2} \cos jt=
(-1)^n\left(\sum_{j=-1}^1 \cos jt + 2\sum_{j=2}^n \left( \cos jt+\cos (j-2)t \right)-\sum_{j=0}^0 \cos jt \right)= (-1)^n\left(-x+2\sum_{j=2}^n 2\cos (j-1)t\cos t \right)=
(-1)^n\left(-x-2x\sum_{j=1}^{n-1} \cos jt \right)=s_{n-1}x.$$
That is $s_n(x)+s_{n-2}(x)=xs_{n-1}x$. Therefore $s_n=q_n$ for each $n$. At last, we show that $s=s_n(2\cos\frac{2k+1}{2n+1}\pi)=0$ for $k=0,\dots,n-1$. But in this case $$s=
(-1)^{n-1}\sum_{j=-n}^n \cos j\left(\frac{2k+1}{2n+1}+1\right)\pi$$ is a multiple of the sum of the abscissas of vertices of a right $(2n+1)/d$-gon $P$ centered at the zero of the complex plane, where $d$ is a greatest common divisor of the numbers $k+n+1$ and $2n+1$. The rotational symmetry of $P$ implies that $se^{2d\pi i/(2n+1)}=s$. Therefore $s=0$.
PS. The Google  search by “x^5-x^4-4x^3+3x^2+3x-1” yielded me only two links. One of then to *.jp site, and the other to this MSE question. The recurrent relations and the roots of the polynomials $q_n$ are similar to these of Chebyshev (and cyclotomic) polynomials, so I added relative tags to the question. But the general trigonometric formula for polynomials $q_n$ is much more complex than the trigonometric formula for Chebyshev polynomials.
