Correspondencia entre dos matrices

Question

$B$ Es una matriz definida positiva y determinante $1 $ $$ A = \frac{1}{2} \int_0^\infty \frac{(B+sI)^{-1}}{\sqrt{\mbox{det}(B+sI)}} ds $ $

Entonces, cómo uno demostrar que esto proporciona un uno a uno a correspondencia entre matrices definidas positivas $B$ % determinante $1$y positiva definida matrices $A$ % rastro $1$.

Muchas gracias.

Answer 1

Sin pérdida de generalidad, que $B$ sea la matriz diagonal con elementos diagonales $b_i$. Considerar $$\begin{eqnarray} A_{11} &=& \frac{1}{2} \int_0^\infty \frac{1}{(s+b_1)^{3/2}} \prod_{k=2}^n \frac{1}{(s+b_k)^{1/2}} \mathrm{d} s \\ &=& \frac{1}{ \pi^{n/2}} \int_0^\infty \mathrm{d} s \int_0^\infty \mathrm{d} x_1 \cdots \int_0^\infty \mathrm{d} x_n x_1^{1/2} x_2^{-1/2}\cdots x_n^{-1/2} \mathrm{e}^{-(b_1+s)x_1} \cdots \mathrm{e}^{-(b_n+s)x_n} \\ &=& \frac{1}{ \pi^{n/2}} \int_0^\infty \mathrm{d} x_1 \cdots \int_0^\infty \mathrm{d} x_n \frac{x_1^{-1/2} x_2^{-1/2}\cdots x_n^{-1/2}}{x_1 + x_2 + \ldots + x_n} x_1 \mathrm{e}^{-\sum_i b_i x_i} \end{eqnarray} $$ Aviso esa representación integral será la suma de $\sum_i A_{ii}$: $$\begin{eqnarray} \operatorname{Tr}(A) &=& \frac{1}{ \pi^{n/2}} \int_0^\infty \mathrm{d} x_1 \cdots \int_0^\infty \mathrm{d} x_n \frac{x_1^{-1/2} x_2^{-1/2}\cdots x_n^{-1/2}}{x_1 + x_2 + \ldots + x_n} \left( \sum_i x_i \right) \mathrm{e}^{-\sum_i b_i x_i} \\ &=& \frac{1}{ \pi^{n/2}} \int_0^\infty \mathrm{d} x_1 \cdots \int_0^\infty \mathrm{d} x_n \left(x_1^{-1/2} x_2^{-1/2}\cdots x_n^{-1/2} \right) \mathrm{e}^{-\sum_i b_i x_i} = \frac{1}{\sqrt{\det B}} = 1 \end{eqnarray} $$