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Suppose $B$ is a positive definite matrix with determinant $1 $ and $ A = \frac{1}{2} \int_0^\infty \frac{(B+sI)^{-1}}{\sqrt{\mbox{det}(B+sI)}} ds $

Then, how does one prove that this provides a one to one onto correspondence between positive definite matrices $B$ with determinant $1$ and positive definite matrices $A$ with trace $1$.

Thank you very much.

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    My comment just suggests how to compute with this formula. It certainly doesn't address whether the transformation is bijective. Where is this map from? I'm not familiar with it. Maybe some context will help.2011-11-09

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Without loss of generality, let $B$ be diagonal matrix with diagonal elements $b_i$. Consider $ \begin{eqnarray} A_{11} &=& \frac{1}{2} \int_0^\infty \frac{1}{(s+b_1)^{3/2}} \prod_{k=2}^n \frac{1}{(s+b_k)^{1/2}} \mathrm{d} s \\ &=& \frac{1}{ \pi^{n/2}} \int_0^\infty \mathrm{d} s \int_0^\infty \mathrm{d} x_1 \cdots \int_0^\infty \mathrm{d} x_n x_1^{1/2} x_2^{-1/2}\cdots x_n^{-1/2} \mathrm{e}^{-(b_1+s)x_1} \cdots \mathrm{e}^{-(b_n+s)x_n} \\ &=& \frac{1}{ \pi^{n/2}} \int_0^\infty \mathrm{d} x_1 \cdots \int_0^\infty \mathrm{d} x_n \frac{x_1^{-1/2} x_2^{-1/2}\cdots x_n^{-1/2}}{x_1 + x_2 + \ldots + x_n} x_1 \mathrm{e}^{-\sum_i b_i x_i} \end{eqnarray} $ Notice that integral representation for the sum $\sum_i A_{ii}$ will be: $ \begin{eqnarray} \operatorname{Tr}(A) &=& \frac{1}{ \pi^{n/2}} \int_0^\infty \mathrm{d} x_1 \cdots \int_0^\infty \mathrm{d} x_n \frac{x_1^{-1/2} x_2^{-1/2}\cdots x_n^{-1/2}}{x_1 + x_2 + \ldots + x_n} \left( \sum_i x_i \right) \mathrm{e}^{-\sum_i b_i x_i} \\ &=& \frac{1}{ \pi^{n/2}} \int_0^\infty \mathrm{d} x_1 \cdots \int_0^\infty \mathrm{d} x_n \left(x_1^{-1/2} x_2^{-1/2}\cdots x_n^{-1/2} \right) \mathrm{e}^{-\sum_i b_i x_i} = \frac{1}{\sqrt{\det B}} = 1 \end{eqnarray} $

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    I am a bit lost though. How does this help us. Thanks.2011-11-09