## Monday, 21 September 2009

### BETA FUNCTION MATRIX DETERMINANT If $latex A_n$ is an square matrix of order $latex n$, whose elements are defined as:

$latex \displaystyle a_{i,j}=\frac{1}{\beta(i,j)}$, where $latex \beta$ is the beta function. Then:

$latex |A_{n}|=n!$

Proof:

If we use Cholesky method, we can decompose this matrix as:

$latex A_{n}=U_{n}^{T}*U_n$

Where $latex U_n$ is an upper triangular matrix.

But instead of applying the algorithm to a generic case, we are going to propose a factorization, and after we will check that this decomposition generates the appropriate matrix. This mean to use software to speed up the proof, like:

$latex \displaystyle A_{5}=\left(\begin{array}{ccccc} 1 & 2 & 3 & 4 & 5 \\ 2 & 6 & 12 & 20 & 30 \\ 3 & 12 & 30 & 60 & 105 \\ 4 & 20 & 60 & 140 & 280 \\ 5 & 30 & 105 & 280 & 630 \end{array} \right)= U_{n}^{T}* \left( \begin{array}{ccccc} 1 & 2 & 3 & 4 & 5 \\ 0 & \sqrt{2} & 3 \sqrt{2} & 6 \sqrt{2} & 10 \sqrt{2} \\ 0 & 0 & \sqrt{3} & 4 \sqrt{3} & 10 \sqrt{3} \\ 0 & 0 & 0 & 2 & 10 \\ 0 & 0 & 0 & 0 & \sqrt{5} \end{array} \right)$

The elements of $latex U_n$, seems to be:

$latex \displaystyle u_{i,j}=\sqrt{i}\cdot \binom{j}{i}$

$latex U_{n}^T$ is the transpose of $latex U_n$, so:

$latex \displaystyle u_{i,j}^{*}=u_{j,i}=\sqrt{j} \cdot \binom{i}{j}$

If we multiply both triangular matrices:

$latex \displaystyle a_{i,j}=\sum_{r=1}^n{u_{i,r}^{*} \cdot u_{r,j}} =\sum_{r=1}^n{r \cdot \binom{i}{r} \binom{j}{r}}=\sum_{r=1}^{min(i,j)}{r \cdot \binom{i}{r} \binom{j}{r}}$

This last binomial expression, can be added to a closed form, equal to the reciprocal of beta function, (See proof on reference )

$latex \displaystyle a_{i,j}=i \cdot \binom{i+j-1}{j-1}=\frac{1}{\beta(i,j)}$

Now, that we have found this decomposition, for $latex A_n$, then it is unique and $latex A_n$ is Hermitian and positive definite.

$latex \displaystyle |A_{n}|=|U_{n}^{T}*U_{n}|=|U_{n}|^{2}=\left(\prod_{i=1}^{n}{u_{i,i}}\right)^{2}=\prod_{i=1}^{n}{u_{i,i}^2}=\prod_{i=1}^{n}{\left(\sqrt{i}\cdot\binom{i}{i}\right)^2}=\prod_{i=1}^{n}{i}=n!$

Archives:

References:
-Ronald L. Graham, Donald E. Knuth, and Oren Patashnik (Reading, Massachusetts: Addison-Wesley, 1994 - Concrete Mathematics - page 181 Problem 5
-A000142-Factorial numbers: n! The On-Line Encyclopedia of Integer Sequences!