Eigenvalues of certain Hankel matrix For each $n\in\mathbb N$, let $A_n$ be the $n\times n$ matrix with entries
$$
A_n(k,j)=\begin{cases} 1,&\ k+j=n+1\\ 
1,&\ k+j=n+2,\\
0,&\ \text{ otherwise}
\end{cases}
$$
So, for instance, 
$$
A_4=\begin{bmatrix} 0&0&0&1\\ 0&0&1&1\\ 0&1&1&0\\ 1&1&0&0\end{bmatrix}.
$$
I'm looking for a proof of the following:

  
*
  
*If $n=2m$, then $A_n$ has $m$ positive and $m$ negative eigenvalues.
  
*if $n=2m+1$, then $A_n$ has $m+1$ positive and $m$ negative eigenvalues. 

Finding the eigenvalues explicitly is definitely not an option, judging by the formulas already in the case $n=3$. 
If $p_n$ denotes the characteristic polynomial of $A_n$, then we have the recursion
$$
p_{n+1}(t)=-t\,p_n(t)-p_{n-1}(t),
$$
but I don't know if one can obtain meaningful information from this. 
 A: With a bit of help from a friend, I found this way to answer the question. 
Consider first the case $n$ even, say $n=2m$. Then $A_n$ has an $n\times n$ block of zeroes in the upper left corner. Cauchy's Interlacing then gives (note that $A_n$ is invertible for all $n$)
$$
\lambda_k(A)<0<\lambda_{2m-m+k}=\lambda_{m+k}.
$$
So the first $m$ eigenvalues of $A_{2m}$ are positive, and the other $m$ are negative. 
When $n$ is odd, $n=2m+1$, we have 
$$
A_{2m+1}=\begin{bmatrix} 
0_{m\times m} & E & Y\\
E^*&\begin{bmatrix} 1&1\\1&0\end{bmatrix}&0_{2\times (m-1)}\\
X&0_{(m-1)\times 2}& 0_{(m-1)\times (m-1)}
\end{bmatrix},
$$
where $E$ is the $m\times2$ matrix with $1$ in the $m,2$ entry and zeroes elsewhere; $Y$ is the $m\times(m-1)$ matrix (using matrix unit notation)
$$
Y=E_{1,m-1}+E_{m,1}+\sum_{j=2}^{m-1}E_{j,m-j+1}+E_{j,m-j},
$$
and
$$
X=\sum_{j=1}^{m-1}E_{j,m-j+1}+E_{j,m-j}
$$
For an example, consider $m=3$:
$$
\left[
\begin{array}{ccc|cc|cc}
0&0&0&0&0&0&1\\
0&0&0&0&0&1&1\\
0&0&0&0&1&1&0\\
\hline0&0&0&1&1&0&0\\
0&0&1&1&0&0&0\\
\hline0&1&1&0&0&0&0\\
1&1&0&0&0&0&0
\end{array}
\right].
$$
Now define
$$
V=\begin{bmatrix}
I_{m }&0_{m\times 2}&0_{m\times(m-1)}\\
0_{2\times m}&\begin{bmatrix} 1&0\\-1&1\end{bmatrix}& 0_{2\times(m-1)}\\
0_{(m-1)\times m}&0_{(m-1)\times 2}& I_{m-1}
\end{bmatrix}.
$$
One then checks
$$\tag1
VA_{2m+1}V^*=\begin{bmatrix}
0_{m\times m}&E&Y\\
E^*&\begin{bmatrix}1&0\\0&-1\end{bmatrix}& 0_{2\times(m-1)}\\
X&0_{(m-1)\times2}& 0_{(m-1)\times(m-1)}
\end{bmatrix}
$$
By Sylvester's Law of Inertia, the matrix in $(1)$ has the same number of positive and negative eigenvalues as $A_{2m+1}$. 
If we now look at row and column $m+1$, we see that both are zero with the exception of the $1$ at the $m+1,m+1$ entry. So, by conjugating with the right permutation (the one that flips rows $1$ and $m+1$), we obtain 
$$
\left[\begin{array}{c|ccc}
1&0_{m\times m}&0_{1\times m}&0_{1\times(m-1)}\\
\hline 0_{m\times1}&0_{m\times m}&e_m&Y\\
0_{1\times m}&e_m^*&-1&0_{2\times(m-1)}\\
0_{(m-1)\times1}&X&0_{(m-1)\times1}&0_{(m-1)\times(m-1)}.
\end{array}\right]
$$
This matrix has $1$ has an eigenvalue, and the remaining eigenvalues are those of the $2m\times2m$ block down right; this block is selfadjoint, and has an $m\times m$ block of zeroes, so reasoning as in the even case we conclude that it has $m$ positive and $m$ negative eigenvalues. In conclusion, $A_{2m+1}$ has $m+1$ positive eigenvalues and $m$ negative eigenvalues.
A: Cauchy's interlacing inequality says that if $Y$ is an $n\times n$ Hermitian matrix and $X$ is the matrix obtained by deleting the last row and the last column of $Y$, and the eigenvalues of $X$ $X$ are interlaced between the eigenvalues of $Y$ if they are arranged in ascending order:
$$
\lambda_1(Y)\le\lambda_1(X)\le\lambda_2(Y)\le\lambda_2(X)\le\cdots\le\lambda_{n-1}(Y)\le\lambda_{n-1}(X)\le\lambda_n(Y).
$$
Now, suppose the dimension $n$ of your $A$ is even, so that $n=2m$. Denote  by $B_k$ the leading principal $k\times k$ submatrix of $A$ (so that $B_n=A$)of $A$. Clearly, all eigenvalues of $B_m=0$ are zero. So, by Cauchy's interlacing inequality, $B_{m+1}$ has a non-positive eigenvalue (that can be zero), $m-1$ eigenvalues that are surely zero, and a non-negative eigenvalue (that can be zero). However, since $B_{m+1}$ has rank $2$, that non-positive eigenvalue must be negative and that non-negative eigenvalue must be positive. Thus $B_{m+1}$ has one negative eigenvalue, $m-1$ zero eigenvalues and one positive eigenvalues. Proceed inductively using Cauchy's interlacing inequality and the rank of $B_k$, the conclusion follows.
When $n=2m+1$ is odd, start with $B_{m+1}$ instead.
