A few years ago, I wanted to cite a result in a paper, for which I could not find a reference. I ended up not using the full strength of it, and the part that I needed could be easily proved. Still, I'd like to know where the full version appears. The result is as follows:



> 
> The eigenvalues of the matrix
>  $$\left[\binom{i+j}{i}\binom{2n-i-j}{n-i}\right]\_{0\le i,j\le n}$$ 
>  are $$\binom{2n+1}{k}, \quad 0\le k\le n.$$
> 
> 
> 


If a formula for the corresponding eigenvectors also appears somewhere, that would be helpful, too. (I only needed the fact that $\binom{2n+1}{n}$ has eigenvector $[1,1,\dots,1]^T$, and that's easy to see.) Thanks.



Result attribution for eigenvalues of a matrix of Pascal-type
I don't know a reference. One way to show the eigenvalues starts from the observation (which can be proved using 
generating functions) 
that $\sum\_{i=0}^n {i\choose k} A\_{i,j}={2n+1 \choose n-k} {j+k \choose k}={2n+1 \choose n-k}\,\sum\_{\ell=0}^k {k\choose \ell} { j \choose \ell}$ (where $A$ is the matrix above).
With the row vectors $\mathbf{v}\_k$ with coordinates $\mathbf{v}\_k(i)={i \choose k}$ that is 
$$\mathbf{v}\_k A={2n+1 \choose n-k}\left(\sum\_{\ell=0}^k {k \choose \ell} \mathbf{v}\_\ell\right).$$
The rest is routine.


ADDED:
 (for the record)  

 With some patience one finally finds that
$$\mathbf{e}\_k=\sum\_{j=0}^k (-1)^j{n-j \choose k-j}{k+j \choose j}\mathbf{v}\_j$$
is an eigenvector to the eigenvalue ${2n+1 \choose n-k}$. 


