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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.

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    $\begingroup$ By the way, computing the trace of this matrix and the sum of eigenvalues we get a famous identity $\sum \binom{2i}{i}\binom{2(n-i)}{n-i}=4^n$. $\endgroup$
    – Fedor Petrov
    Commented Aug 10, 2017 at 10:13
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    $\begingroup$ There is a paper by Amdeberhan that proves the determinant of this matrix is the product of those eigenvalues, but that does not seem to help in finding the individual eigenvalues. @FedorPetrov $\endgroup$
    – Alexander Burstein
    Commented Aug 10, 2017 at 18:32
  • $\begingroup$ To see where this result might have been useful (and also, speaking of traces of matrices), take a look at Section 4 of this paper. @FedorPetrov $\endgroup$
    – Alexander Burstein
    Commented Aug 10, 2017 at 18:34
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    $\begingroup$ @AlexanderBurstein btw the positive definiteness of the matrix $A_m$ in Sec. 4 of your paper can follow "more directly" by recalling that the Pascal matrix is positive definite (proof: $\binom{p+q}{p} = c\int_0^{2\pi} (1+e^{i\theta})^p(1+e^{-i\theta})^qd\theta$) and invoking Schur's theorem on psdness of the elementwise product of psd matrices. $\endgroup$
    – Suvrit
    Commented Aug 10, 2017 at 19:23
  • $\begingroup$ The eigenvectors for $n$ at most 5 form a pretty nice orthogonal basis, but I do not see a general formula from them. $\endgroup$
    – Fedor Petrov
    Commented Aug 10, 2017 at 19:30

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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 ${2n+1 \choose n-k}$.

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  • $\begingroup$ Thanks! At least this yields the eigenvalues. However, I saw this result referred to in passing in some paper as if it were known for some time, and when I needed to use it a few years later, I could neither find where it was proven nor the paper where I saw a reference to it. $\endgroup$
    – Alexander Burstein
    Commented Aug 11, 2017 at 20:34
  • $\begingroup$ oh, this gives a concrete basis in which the matrix is triangular, that in theory allows to get formulae for eigenvectors as well $\endgroup$
    – Fedor Petrov
    Commented Aug 12, 2017 at 9:09
  • $\begingroup$ Just saw the added part today. Thanks! $\endgroup$
    – Alexander Burstein
    Commented Oct 1, 2017 at 4:43

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