Let $A=(a_{ij})$ be a real symmetric, positive definite $n \times n$ matrix and set
$$ F(x_1,x_2,\ldots, ,x_n)= \sum_{i,j}a_{ij}x_ix_j.$$
I am trying to show that for any non-negative measurable function $\alpha$ on the real line
$$ \int_{\mathbb{R}^n} \alpha(F(x_1,x_2,\ldots, ,x_n))\,dm=
\frac{1}{\sqrt{\det A}}\int_{\mathbb{R}^n} \alpha(x_1^2+x_2^2+\ldots +x_n^2)\,dm,$$
where $dm$ is the Lebesgue measure on $\mathbb{R}^n$.
I thought in the following sense: Somehow, change of variables theorem for $\mathbb{R}^n$ should be used and here we need the Jacobian matrix for the transformation $F$. Is $A$ the matrix representation of $F$? How to write Jacobian for $F$?
Moreover, $A$ is diagonalizable and all of its eigenvalues, $\lambda_1\lambda_2\ldots\lambda_n$, are positive. Therefore, $D= P^{-1}A P$, where $D$ is the diagonal $n \times n$ matrix with eigenvalues on the diagonal and $P$ is the orthogonal $n \times n$ matrix consisting of eigenfunctions as column vectors. Then, it follows that
$$\det A=\det D=\lambda_1\lambda_2\ldots\lambda_n.$$
How can we combine all these data? Can you please help me?
Thank you!
Best Answer
As you notice, we can write $A=P^tDP$, where $D$ is diagonal, with real entries and $P$ orthogonal. The LHs is $$I:=\int_{\Bbb R^n}F(x^tP^tDPx)dx_1\dots dx_n.$$ Do the change $y=Px$; since $P$ is orthogonal the absolute value of the Jacobian is $1$, hence $$I=\int_{\Bbb R^n}F(y^tDy)dy_1\dots dy_n=\int_{\Bbb R^n}F\left(\sum_{j=1}^n\lambda_jy_j^2\right)dy_1\dots dy_n.$$ Since $\lambda_j>0$, let $t_j:=\sqrt{\lambda_j}y_j$ for $1\leq j\leq n$. The inverse of the Jacobian of this transformation is $\frac 1{\sqrt{\det D}}$, which is what we want since $\det D=\det A$.