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Suppose $c$ is a scalar, $\mathbf{A}$ is a symmetric positive definite matrix and $g(.)$ is some real-valued function. Define $\mathbf{B} = c \, \mathbf{A}$.

In this integral, $$ \int_{-\infty}^{\infty} g(c) \; det(c \, \mathbf{A}) \; dc = \int ? d\mathbf{B}, $$

how do I change the integration variable to $\mathbf{B}$?

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Assume $A\in{\mathbb R}^{n\times n},\,$ then $$\eqalign{ \det(cA) &= c^n\det(A) \\ B &= Ac \implies AB = BA\\ A^{-1}B &= Ic \;\implies A^{-1}dB = I\,dc \\ {\rm Tr}(I) &= n \\ f(x) &= x^ng(x) \\ }$$ Using this, the integration variable can be changed. $$\eqalign{ {\cal J} &= \int \det(A)\,f(c)\,dc \\ {\cal J}I &= \int \det(A)\,f(A^{-1}B)\,A^{-1}dB \\ {\cal J}I &= \int \det(A)\,A^{-1}f(A^{-1}B)\,dB \\ {\cal J}\;{\rm Tr}(I) &= {\rm Tr}\left(\int\det(A)\,A^{-1}f(A^{-1}B)\,dB\right) \\ {\cal J} &= \frac{1}{n}\;{\rm Tr}\left(\int\det(A)\,A^{-1}f(A^{-1}B)\,dB\right) \\ &= \int G:dB \\ }$$ where $G$ is the gradient of ${\cal J}$ with respect to $B$ $$\eqalign{ G &= \frac{\partial{\cal J}}{\partial B} \;=\; \frac{1}{n}\,\det(A)\,A^{-1}f(A^{-1}B) \\ }$$ and the colon denotes the trace/Frobenius product, i.e. $$A:B = {\rm Tr}(A^TB) = {\rm Tr}(AB^T)$$ When both matrices are symmetric, this can be simplified to $$A:B = {\rm Tr}(AB)$$

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  • $\begingroup$ Can you help me understand the $f(A^{-1} B)$ part. $g(x)$ is a scalar valued function, so how do I think about $f(x) = x^n g(x)$, with $x=A^{-1} B=Ic$? $\endgroup$
    – stollenm
    Oct 14, 2020 at 11:46
  • $\begingroup$ For any scalar valued function like f$(x)$, substituting $Ix$ as the argument yields $f(Ix) = I\,f(x)$. This seperabilty property does not hold for other matrices, i.e. $\,f(Ax) \ne A\,f(x)$ $\endgroup$
    – greg
    Oct 14, 2020 at 14:15

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