Jordan canonical form of a specific $n\times n$ matrix Let $A=(a_{ij})$ be a $n\times n$ matrix with entries $a_{ij}$ satisfy $a_{ij}=1$ only if $i+j=n$ or $n+1$; for other $i,j$, $a_{ij}=0$. How can we compute its Jordan canonical form? Thank you.
 A: Because $A$ is real and symmetric, its Jordan canonical form is a diagonal matrix whose diagonal entries are real numbers—$A$'s eigenvalues—which by the Gershgorin circle theorem are in the closed interval $[-2, 2].$ Those eigenvalues are the zeros of $A$'s characteristic polynomial, which I give below. If there is no convention for ordering Jordan blocks by eigenvalue, then there are $n!$ forms because the eigenvalues are distinct, which I prove next.
Eigenvalues distinct. I will use a theorem from graph theory: the number of distinct eigenvalues of a symmetric matrix that represents a graph is at least the number of vertices in a path that passes through a vertex no more than once. $A$ is such a matrix: The graph has $n$ vertices labeled $1, 2, \dots, n.$ If the $i, j$ and $j, i$ entries of $A$ are $1,$ then an edge connects vertices $i$ and $j;$ if those two entries are $0,$ then an edge does not connect vertices $i$ and $j.$ Due to the stairstep pattern of $1$s in $A,$ it is easy to construct a path that contains each vertex exactly once. For example, starting at the upper-right entry and working down and to the left we have the path $n$ → $1$ → $n - 1$ → $2$ → $n - 2$ → $\cdots$ → $\lceil n/2 \rceil$ where $\lceil \cdot \rceil$ is the ceiling function. There are $n$ vertices in the path, so the theorem implies that the eigenvalues are all distinct. (See "On the Minimum Number of Distinct Eigenvalues for a Symmetric Matrix Whose Graph is a Given Tree.")
Characteristic polynomial. The characteristic polynomials det$(\lambda I - A)$ for $n = 1, \dots, 8$ are—
\begin{array}{}
\lambda   & - 1,\\
\lambda^2 & - \lambda   & - 1,\\
\lambda^3 & - \lambda^2 & - 2\lambda   & + 1,\\
\lambda^4 & - \lambda^3 & - 3\lambda^2 & + 2\lambda   & +\;\,1,\\
\lambda^5 & - \lambda^4 & - 4\lambda^3 & + 3\lambda^2 & +\;\,3\lambda   & -\;\,1,\\
\lambda^6 & - \lambda^5 & - 5\lambda^4 & + 4\lambda^3 & +\;\,6\lambda^2 & -\;\,3\lambda   & -\;\,1,\\
\lambda^7 & - \lambda^6 & - 6\lambda^5 & + 5\lambda^4 & +   10\lambda^3 & -\;\,6\lambda^2 & -\;\,4\lambda   & + 1,\\
\lambda^8 & - \lambda^7 & - 7\lambda^6 & + 6\lambda^5 & +   15\lambda^4 & -   10\lambda^3 & -   10\lambda^2 & + 4\lambda & + 1,
\end{array}
so it appears that the characteristic polynomial is 
$$\sum_{i=0}^n (-1)^{\lceil i/2 \rceil}{n - \lceil i/2 \rceil \choose \lfloor i/2 \rfloor} \lambda^{n-i}$$
where $\lfloor \cdot \rfloor$ is the floor function. Here is a proof by induction:
Rename $A_n$ the matrix you define, and let $B_n$ be the $n \times n$ matrix with the $i, j$ entry equal to $1$ if $i + j$ equals $n + 1$ or $n + 2;$ otherwise, that entry equals zero. The characteristic polynomials of $B_n$ for $n = 1, \dots, 8$ are the same as those of $A_n$. So, assume that the characteristic polynomials of $A_n$ and $B_n$ are given by the polynomial above for $n = m.$ The characteristic polynomial of $A_{m + 1}$ is
$$\det \left( \begin{array}{r|rcrrr|r}
\lambda & 0       &      0 & \cdots &      0 &      -1 &      -1\\ \hline
      0 & \lambda &      0 & \cdots &     -1 &      -1 &       0\\
 \vdots &  \vdots & \vdots & \ddots & \vdots &  \vdots &  \vdots\\
     -1 &      -1 &      0 & \cdots &      0 & \lambda &       0\\ \hline
     -1 &       0 &      0 & \cdots &      0 &       0 & \lambda
\end{array}\right)$$
where the matrix has size $(m + 1) \times (m + 1),$ and I have added some lines to clarify Laplace expansion: Expand the determinant along the last row so that we have only two nonzero terms. The minor associated with $\lambda$ is $\lambda I - B_m,$ and the minor associated with $-1$ is the upper-right $n \times n$ matrix, say $C.$ To calculate det $C,$ expand it along the last column so that we have only one nonzero term; the minor associated with the first entry of that column, $-1,$ is $\lambda I - A_{m - 1}.$ Hence,
$$\begin{align}
\det(\lambda I - A_{m + 1}) & = (-1)^{m+1+m+1}\lambda \det(\lambda I - B_m) + (-1)^{m+1+1}(-1)(-1)^{m+1}(-1) \det(\lambda I - A_{m - 1})\\
& = \lambda \sum_{i=0}^m (-1)^{\lceil i/2 \rceil}{m - \lceil i/2 \rceil \choose \lfloor i/2 \rfloor} \lambda^{m-i} - \sum_{i=0}^{m-1} (-1)^{\lceil i/2 \rceil}{m - 1 - \lceil i/2 \rceil \choose \lfloor i/2 \rfloor} \lambda^{m-1-i}\\
& = \lambda^{m+1} - \lambda^m + \sum_{i=2}^m (-1)^{\lceil i/2 \rceil}{m - \lceil i/2 \rceil \choose \lfloor i/2 \rfloor} \lambda^{m+1-i}\\
& \quad\, - \sum_{i=0}^{m-2} (-1)^{\lceil i/2 \rceil}{m - 1 - \lceil i/2 \rceil \choose \lfloor i/2 \rfloor} \lambda^{m-1-i} - (-1)^{\lceil (m - 1)/2 \rceil}\\
& = \lambda^{m+1} - \lambda^m + \sum_{i=2}^m (-1)^{\lceil i/2 \rceil}\left({m - \lceil i/2 \rceil \choose \lfloor i/2 \rfloor} + {m - \lceil i/2 \rceil \choose \lfloor i/2 \rfloor - 1} \right) \lambda^{m+1-i}\\
& \quad\, + (-1)^{\lceil (m + 1)/2 \rceil}\\
& = \lambda^{m+1} - \lambda^m + \sum_{i=2}^m (-1)^{\lceil i/2 \rceil}{m + 1 - \lceil i/2 \rceil \choose \lfloor i/2 \rfloor} \lambda^{m+1-i} + (-1)^{\lceil (m + 1)/2 \rceil}\\
& = \sum_{i=0}^{m+1} (-1)^{\lceil i/2 \rceil}{m + 1 - \lceil i/2 \rceil \choose \lfloor i/2 \rfloor} \lambda^{m+1-i}.
\end{align}$$
From the third equation to the fourth, I combined the two sums into one by shifting the index of the second sum up by $2,$ which requires replacement of $i$ by $i - 2$ in the summand. From the fourth equation to the fifth, I used Pascal's identity. In some instances, I moved an integer into or out of the floor/ceiling functions or changed a sign by adding $1$ to the exponent of $-1.$
Hence, our claim about the characteristic polynomial holds for $n = m + 1$ whenever it holds for $n = m,$ and so the induction principle guarantees that it holds for every integer $n = 1, 2, \dots.$
It is easy to find the eigenvalues for $n = 1$ and $2,$ but the closed-form expressions for $n = 3$ and $4$ are "messy," so I do not think there is much value trying to find a general solution to the characteristic equation—just use your favorite root-finding algorithm.
