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Is there an existing or elementary proof of the determinant identity $ \det_{1\le i,j\le n}\left( \binom{i}{2j}+ \binom{-i}{2j}\right)=1 $?

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    $\begingroup$ Did you have a look in Krattenthaler's "advanced determinant calculus"? arxiv.org/pdf/math/9902004v3.pdf $\endgroup$ – Wolfgang Jan 11 '14 at 9:51
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    $\begingroup$ "Is there an existing proof" implies that the identity is new/unproved. Can you, please, clarify how do you know that the identity holds and give some context? $\endgroup$ – Victor Protsak Jan 11 '14 at 15:36
  • $\begingroup$ Many thanks to all for the comments. I have a proof based on an LU decomposition as suggested by Douglas (@Douglas-Zare) with \begin{eqnarray*} L_{ij}=\left\{ \begin{array}{cc} \binom{2i-j-1}{j-1}& \mbox{for } i\le j ,\\ 0 & \mbox{for } i>j, \end{array} \right. \qquad U_{jk}= \left\{ \begin{array}{cc} \frac{k}{j}\binom{j+k-1}{2j-1} &\mbox{for } j \le k,\\ 0 & \mbox{for }j>k. \end{array} \right. \end{eqnarray*} and using induction in k. Gjergji's (@Gjergji-Zaimi) general formula is very nice! $\endgroup$ – MPTuite Jan 17 '14 at 10:56
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This is true, and in fact you can show a slightly more general fact:

$$\det_{1\le i,j\le n}\left( \binom{x_i}{2j}+ \binom{-x_i}{2j}\right)=\prod_{i=1}^n x_i^2 \prod_{i<j} (x_j^2-x_i^2) \prod_{j=1}^n \frac{2}{(2j)!}.$$

It is easy to show that this evaluates to $1$ when you set $x_i=i$. To prove this, notice that every column is an even polynomial in the $x_i$ of degree $2j$ with leading coefficient $\frac{2}{(2j)!}$. So elementary column operations will bring the matrix in a familiar Vandermonde form.

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I don't have a proof yet, but it's likely that this matrix has a simple LU decomposition which makes the determinant obviously $1$.

\begin{multline} \scriptsize\begin{pmatrix} 1 & 1 & 1 & 1 & 1 & 1 \\\ 4 & 5 & 7 & 9 & 11 & 13 \\\ 9 & 15 & 28 & 45 & 66 & 91 \\\ 16 & 35 & 84 & 165 & 286 & 455 \\\ 25 & 75 & 210 & 495 & 1001 & 1820 \\\ 36 & 141 & 463 & 1287 & 3003 & 6188 \end{pmatrix} \\ \scriptsize\hskip1cm= \begin{pmatrix} 1 & 0 & 0 & 0 & 0 & 0 \\\ 1 & 1 & 0 & 0 & 0 & 0 \\\ 1 & 3 & 1 & 0 & 0 & 0 \\\ 1 & 5 & 6 & 1 & 0 & 0 \\\ 1 & 7 & 15 & 10 & 1 & 0 \\\ 1 & 9 & 28 & 35 & 15 & 1 \end{pmatrix} \begin{pmatrix}1 & 4 & 9 & 16 & 25 & 36 \\\ 0 & 1 & 6 & 20 & 50 & 105 \\\ 0 & 0 & 1 & 8 & 35 & 112 \\\ 0 & 0 & 0 & 1 & 10 & 54 \\\ 0 & 0 & 0 & 0 & 1 & 12 \\\ 0 & 0 & 0 & 0 & 0 & 1 \end{pmatrix} \end{multline}

The first factor appears to be A054142 as a lower diagonal matrix. The second factor appears to be A156308 as an upper diagonal matrix.

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  • $\begingroup$ @Mariano Suárez-Alvarez: I can't read the matrices now... I think it's ok to have the equation split between lines. $\endgroup$ – Douglas Zare Jan 12 '14 at 19:55
  • $\begingroup$ Better? ${}{}{}$ $\endgroup$ – Mariano Suárez-Álvarez Jan 12 '14 at 19:57
  • $\begingroup$ @Mariano Suárez-Alvarez: I think it might depend on the settings each person uses. The original still looks much better to me. What improvement do you see on your screen? $\endgroup$ – Douglas Zare Jan 12 '14 at 19:59
  • $\begingroup$ Well, for me the original did not fit in the column: none of the individual matrices did! :-( $\endgroup$ – Mariano Suárez-Álvarez Jan 12 '14 at 20:10
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    $\begingroup$ @Mariano Suárez-Alvarez: I'll leave your formatting even though I have to squint to see the numbers unless I use the zoom. I think there is something wrong with your display if you can't see a $6\times 6$ matrix. Next time please wait at least a few minutes since I had an incorrect row of the second factor, and I don't want to change the content while you are changing the formatting. $\endgroup$ – Douglas Zare Jan 12 '14 at 20:19

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