How to invert the matrix [n choose 2j - i] ? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T17:15:20Z http://mathoverflow.net/feeds/question/68875 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/68875/how-to-invert-the-matrix-n-choose-2j-i How to invert the matrix [n choose 2j - i] ? Leonid Petrov 2011-06-26T20:15:42Z 2011-06-28T11:15:10Z <p>In a certain model of a stat-physics type, one encounters a matrix $$A_n:=\left[\binom{n}{2j-i}\right]_{i,j=1}^{n-1}.$$ The determinant of this matrix (equal to $2^{\binom n2}$) counts the number of all possible configurations, and our understanding of the model would greatly increase if we would know the inverse of this matrix. So the question is,</p> <blockquote> <p>is there a closed-form expression for the inverse of the matrix $A_n$?</p> </blockquote> <p>A little more information about the matrix: its eigenvalues are $2^i\colon i=1,\dots,n-1$. The eigenvectors are not orthogonal to each other (and $A_n$ is not a symmetric matrix, for sure), but the vectors corresponding to even $i$'s are orthogonal to the ones which correspond to the odd $i$'s (i.e., the whole space orthogonally decomposes into the "even" and the "odd" parts).</p> <p>PS I would appreciate any help or reference. I've looked through Krattenthaler's seminal "Advanced determinant calculus", but didn't find such a matrix there.</p> http://mathoverflow.net/questions/68875/how-to-invert-the-matrix-n-choose-2j-i/68880#68880 Answer by Emmanuel Briand for How to invert the matrix [n choose 2j - i] ? Emmanuel Briand 2011-06-26T21:52:06Z 2011-06-28T11:15:10Z <p>This is more an idea to explore than a complete answer. </p> <p>You may interpret the binomial coefficient $\binom{n}{k}$ as the elementary symmetric function $e_k$ of $1,1,\ldots,1$ ($n$ variables evaluated at $1$). The coefficients of the adjoint matrix of $A_n$ become skew Schur functions of $1,1,\ldots,1$. Then there may be some further simplifications. </p> <p>(By the way, this approach gives a nice proof for the value of the determinant of $A_n$: it is the value of the staircase Schur function $s_{(n-1,n-2,\ldots,1,0)}$ evaluated at $1,1,\ldots,1$. Note that the staircase Schur function at $x_1,x_2,\ldots,x_n$ is equal to $\prod_{i \lt j} (x_i+x_j)$).</p> <p>EDIT: I find that the coefficient $(i,j)$ of the inverse is $(-1)^{i+j} s_{[j]'/(n-i)}(1,1,\ldots,1)/2^{\binom{n}{2}}$, where $[j]$ stands for the partition obtained from $(n-1,n-2,\ldots,1)$ by removing $j$, and $[j]'$ is its conjugate. At this point there is some hope to find a nice formula. First by expressing the skew Schur function as a sum a Schur functions by means of dual Pieri rule: $$s_{\lambda/(k)}=\sum s_{\nu}$$ where the sum is carried over all partitions $\nu$ obtained from $\lambda$ by removing a horizontal strip with $k$ boxes. After that by using the following formula found in Macdonald, I.3. Ex. 4: $$s_{\lambda'}(1,1,\ldots,1)=\prod_{x \in \lambda} \frac{n-c(x)}{h(x)}$$ where the evaluation is at $(1,1,\ldots,1)$ with $n$ ones, $\lambda'$ is the conjugate of $\lambda$ and $h(x)$ and $c(x)$ are the <i>hook length</i> and <i>content</i> respectively of the box $x$ in the diagram of $\lambda$. </p> <p>Hopefully the formulas simplify.</p>