Integral involving Hadamard product with volume element

Question

Consider the following integral: $$\int_{\mathbb{R}^{n\times m}}f(X)(H^T(A\odot dX)V)^{\wedge},$$ where $f:\mathbb{R}^{n\times m}\to \mathbb{R},$ $H\in O(n)$, $V\in O(m),$ $A$ is an $n\times m$ matrix with entries 0,1, and $\odot$ denotes the Hadamard product of two matrices. Is the above integral same as/different from the following integral? $$\int_{\mathbb{R}^{n\times m}}f(X)(A\odot (H^TdXV))^{\wedge}.$$

Answer 1

$ \def\o{\otimes} \def\J{{\cal J}} \def\LR#1{\left(#1\right)} \def\op#1{\operatorname{#1}} \def\vecc#1{\op{vec}\LR{#1}} \def\Diag#1{\op{Diag}\LR{#1}} \def\trace#1{\op{Tr}\LR{#1}} \def\frob#1{\left\| #1 \right\|_F} \def\qiq{\quad\implies\quad} \def\p{\partial} \def\grad#1#2{\frac{\p #1}{\p #2}} \def\c#1{\color{red}{#1}} $For typing convenience, define the variables... $$\eqalign{ a &= \vecc A,\qquad &\;\,x = \vecc X \\ D &= \Diag a,\qquad &M = \LR{V\o H}^T \\ }$$ The matrices in the first integral can be vectorized $$\eqalign{ J_1 &= \int f(X) \;{H^T(A\odot dX)V} \\ \vecc{J_1} &= \int f(x)\, \LR{V^T\o H^T}\vecc{A\odot dX} \\ &= \int f(x)\;MD\;dx \\ &= MD\int f(x)\;dx \\ }$$ Similarly, the second integral becomes $$\eqalign{ J_2 &= \int f(X)\; A\odot\LR{H^TdX\:V} \\ \vecc{J_2} &= \int f(x)\; D\,\vecc{H^TdX\:V} \qquad\quad \\ &= \int f(x)\;DM\:dx \\ &= DM\int f(x)\;dx \\ }$$ These integrals will be equal iff $M$ commutes with $D$.

Since $D$ is a diagonal matrix, this means that $M$ must also be diagonal. This implies that both $H{\rm\;and\;}V$ must be diagonal as well.