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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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Trace and determinant are invariants under coordinate changes and positive definite matrices are diagonalizable, so try diagonalizing $B$. After shifting to diagonals, I think the sum of the integrals on the diagonal of $A$ should collapse down some and leave you with a manageable computation. Maybe someone else has a better (i.e. slick) way to proceed. – Bill Cook Nov 8 '11 at 22:39
Thanks for the hint. How does one show that this mapping is one to one and onto though. – Shibi Vasudevan Nov 8 '11 at 23:36
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. – Bill Cook Nov 9 '11 at 0:18

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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Thank you so much for the computation. How does one show that this map is one to one and onto. – Shibi Vasudevan Nov 8 '11 at 23:35
@ShibiVasudevan Notice that $A$ and $B$ are simultaneously. Is that true for every pair $(A,B)$ of positive definite matrices matrices subject to $\operatorname{Tr}(A)=1=\det(B)$ ? – Sasha Nov 9 '11 at 0:15
Thanks for the comment. Sorry, but I didn't fully understand the statement ....A and B are simultaneously. – Shibi Vasudevan Nov 9 '11 at 0:19
@ShibiVasudevan I mean that there exists a basis where both matrices $A$ and $B$ are diagonal. – Sasha Nov 9 '11 at 2:21
I am a bit lost though. How does this help us. Thanks. – Shibi Vasudevan Nov 9 '11 at 2:32

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