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In a paper by Mathai, he uses the following integral representation of a determinant, (or, really, what I give is a simple special case of what he gives), without any explication. All matrices are real $p\times p$ symmetric positive definite.

\begin{equation} | I-U |^{-a} = \frac{1}{\Gamma_p(a)} \int_{T>0} |T|^{a-(p+1)/2} \exp(-\text{Tr}(I-U)T) \;dT \end{equation}

where $U$ satisfies $0\lt U \lt I$ (in the cone order in the cone of positive definte matrices), the integral is over the cone of positive definite matrices and $\Gamma_p(a)$ is the generalized gamma function in dimension $p$, and $\Re(a) > (p-1)/2$.

Any references for this?

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Alexander: I do not understand your simpler example, I get $\frac{1}{\text{Tr}M}$. The role of the integer $p$ is as the dimension of the positive-definite matrix$U$ and $T$. –  kjetil b halvorsen Jun 21 '12 at 21:45
    
@Kjetil B Halvorsen Sorry I was in hurry... –  Alexander Chervov Jun 24 '12 at 7:26

2 Answers 2

up vote 11 down vote accepted

OLDER EDIT. (Elementary derivation) I realized that my original answer was actually overkill for the question. The said integral in question follows from the definition of the multivariate Gamma function

\begin{equation*} \Gamma_p(a) := \int_{A > 0} \exp(-\mbox{tr}(A))\det(A)^{a-(p+1)/2}(dA), \end{equation*} where $\Re(a)>(p-1)/2$.

From this it follows (by a change of variables) that for a positive definite matrix $S$, \begin{equation*} \int_{A > 0} \exp(-\mbox{tr}(S^{-1}A))\det(A)^{a-(p+1)/2}(dA) = \Gamma_p(a)\det(S)^a, \end{equation*} so that with $S=(I-U)^{-1}$ we obtain the integral in question.

Of course, to complete the picture it may be helpful to express $\Gamma_p(a)$ in more elementary terms. Chapter 2 of Muirhead's book provides these details. I cite the result that provides this expression.

Theorem (Muirhead (1982), Thm 2.1.2) Let $\Re(a) > (p-1)/2$. Then, $$ \Gamma_p(a) = \pi^{p(p-1)/4}\prod_{j=1}^p \Gamma(a - (j-1)/2) $$

(Hint: To prove the above, write the Cholesky decomposition $A=T'T$, with these change of variables, the original Gamma integral factorizes into a product of Gaussian and Gamma integrals.)


The part that I recalled below provides yet another representation that expresses the multiplicative determinantal lhs in terms of an infinite sum.

OLDER STUFF

This is actually somewhat classical knowledge. Here are two related pointers.

A partition $\tau=(t_1,\ldots,t_m)$ is a vector of nonnegative integers listed in increasing order, and $|\tau|$ denotes $t_1+\cdots+t_m$. The generalized Pochhammer symbol $(a)_\tau$ is defined as \begin{equation*} \newcommand{\risingf}[2]{{{#1}}^{\overline{{#2}}}} (a)_\tau := \frac{\Gamma_d(a+\tau)}{\Gamma_d(a)} = \prod_{l=1}^m \risingf{\bigl(a - \tfrac{1}{2}(l-1)\bigr)}{t_l} \end{equation*}

Let $C_\tau(X)$ be the Zonal Polynomial with signature partition $\tau$. Then, the following representation exists

For a matrix $U$ satisfying $\|U\| < 1$, we have the following "binomial-theorem"

\begin{equation} \frac{1}{|I-U|^a} = \sum_{k\ge 0}\sum_{|\tau| = k} \frac{(a)_\tau C_\tau(U)}{k!}. \end{equation}

Using representations for these Zonal polynomials, one can obtain the integral representation mentioned in the original post.

More directly, you can look at Chapter 7 of R. Muirhead, "Aspects of Multivariate Statistical Theory", where you'll see that actually, $|I-U|^{-a}={}_1F_0(a;U)$, a matrix argument hypergeometric function. I've to run now, if I get a chance I'll clean up my answer and fill in the details.

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A very similar (but different) expression for log of the determinant is given by Du and Ji in their paper "an integral representation of the determinant of a matrix and its applications". I am guessing a slight adaptation of their thing can get your formula.

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