Find eigenvectors of the $(n+1) \times (n+1)$-matrix Find eigenvectors of the $(n+1) \times (n+1)$-matrix: 
 $$\left(\begin {array}{cccccccc} 0&0&0&0&0&0&-1&0\\ 0&0&0&0&0&-2&0&n\\ 0&0&0&0&-3&0&n-1&0
\\ 0&0&0&\ldots&0&n-2&0&0\\ 0&0&-(n-2)&0&\ldots&0
&0&0\\ 0&-(n-1)&0&3&0&0&0&0\\ -n&0&2&0
&0&0&0&0\\ 0&1&0&0&0&0&0&0\end {array}  \right)$$
The eigenvalues of the matrix are 
$ n, n-2, n-4, \ldots, -(n-2), -n$  see the question 
Let $e_{\lambda}$ denotes the eigenvector corresponding to the eigenvalues $\lambda$. So far I have found
$$
e_0=[a_0, a_1, \ldots, a_n], a_i=\begin{cases} 0, \text{ $i$ odd}\\ \displaystyle\binom{\frac{n}{2}}{ \frac{i}{2}}, \text{$i$ even} \end{cases}
$$
For example for $n=8$ we have $e_0=[1, 0, 4, 0, 6, 0, 4, 0, 1]$.
Also
$$
e_1=[a_0, a_2, \ldots, a_n], a_i=\begin{cases} 0, \text{ $i$ even}\\ \displaystyle\binom{\frac{n-1}{2}}{ \frac{i-1}{2}}, \text{$i$ odd} \end{cases}.
$$
for example for $n=7$ we have $e_1=[0, 1, 0, 3, 0, 3, 0, 1]$, 
and 
$$
e_2=[a_0, a_2, \ldots, a_n], a_i=\begin{cases} \displaystyle 
       \binom{\frac{n}{2}-1}{\frac{i}{2}-1}-\binom{\frac{n}{2}-1}{\frac{i}{2}},  i\text{ even} \\ \\
      \displaystyle 2(-1)^{i-1}\binom{\frac{n}{2}-1}{\frac{i-1}{2}} , i\text{  odd}
   \end{cases}.
$$
for example for $n=8$ we have 
$e_2=[-1, 2, -2, 6, 0, 6, 2, 2, 1]$.
Question:
What is  an eigenvector $e_{\lambda}$ (with integer coordinates) for arbitrary $\lambda \in \{ n, n-2, n-4, \ldots, -(n-2), -n\}$?
I hope the problem was  already solved in 19th century.
 A: J.M. has commented in a Kac matrix question that the eigenvector matrix for the Kac matrix are given by the (zero-indexed) $(n+1)\times(n+1)$ "Krawtchouk matrix" $K$ defined by
$$
K_{ij}=\sum_{k=\max(0,\,i+j-n)}^{\min(i,j)} (-1)^k\binom{j}{k}\binom{n-j}{i-k},\quad (i,j\in\{0,1,\ldots,n\}).
$$
Since your matrix is similar to the Kac matrix, we expect that the coefficients of its eigenvectors are closely related to $K_{ij}$. If you try a few $n$s, you will see that some patterns do emerge. Apparently, an eigenvector matrix $V$ is given by
$$
V_{ij}=
\begin{cases}
(-1)^\left\lfloor(i+j+\color{red}{2})/2\right\rfloor\left(\frac{1\color{red}{-}(-1)^{i+j}}2\right)K_{ij}
&\text{ when }n\equiv1 (\!\!\!\mod 4),\\
(-1)^\left\lfloor(i+j+\color{red}{1})/2\right\rfloor K_{ij}
&\text{ when }n\equiv2 (\!\!\!\mod 4),\\
(-1)^\left\lfloor(i+j+\color{red}{1})/2\right\rfloor\left(\frac{1\color{red}{+}(-1)^{i+j}}2\right)K_{ij}
&\text{ when }n\equiv3 (\!\!\!\mod 4),\\
(-1)^\left\lfloor(i+j+\color{red}{2})/2\right\rfloor K_{ij}
&\text{ when }n\equiv0 (\!\!\!\mod 4).
\end{cases}
$$
To prove the correctness of this formula, one may verify that (if we call your matrix $B$) $BV=V\operatorname{diag}(n,n-2,\ldots,2-n,-n)$. This amounts to proving some identities that involve binomial coefficients. I haven't mathematically proved it, but I have verified that the formula is indeed correct for $1\le n\le 20$ using the Matlab program below:
function [B,V,D,diff]=kac(n)
% we should have B*V=V*D

B=fliplr(diag(n:-1:1,-1)-diag(1:n,1));
D=diag(n:-2:-n);

K=zeros(n+1,n+1);
for i=0:n
  for j=0:n
    for k=max(0,i+j-n):min(i,j)
      K(i+1,j+1)=K(i+1,j+1)+(-1)^k*nchoosek(j,k)*nchoosek(n-j,i-k);
    end
  end
end

r=rem(n,4);
if r==1
  S=(-1).^floor((2+repmat((0:n)',1,n+1)+repmat(0:n,n+1,1))./2);
  Z=(1-(-1).^(repmat((0:n)',1,n+1)+repmat(0:n,n+1,1)))./2;
  V=K.*S.*Z;
elseif r==2
  S=(-1).^floor((1+repmat((0:n)',1,n+1)+repmat(0:n,n+1,1))./2);
  V=K.*S;
elseif r==3
  S=(-1).^floor((1+repmat((0:n)',1,n+1)+repmat(0:n,n+1,1))./2);
  Z=(1+(-1).^(repmat((0:n)',1,n+1)+repmat(0:n,n+1,1)))./2;
  V=K.*S.*Z;
else
  S=(-1).^floor((2+repmat((0:n)',1,n+1)+repmat(0:n,n+1,1))./2);
  V=K.*S;
end

diff=max(max(abs(B*V-V*D)))  % should be zero

return

Matlab doesn't seem to use big integers to calculate binomial coefficients. So, the above program may return a false negative result if $n$ is large.
