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Let $\phi: \mathbb{R}^n\to \mathbb{R}^n; x\mapsto Ax+b$, where $A$ is non-singular and $b\in \mathbb{R}^n$.

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$$\mathcal{F}^\pm(f\circ \phi)(y) = \frac{1}{\det{A}}e^{\mp i \langle A^{-1}b,y\rangle} \mathcal{F}^\pm f( (A^T)^{-1}y)$$

I've tried:

$$\mathcal{F}^\pm (f\circ \phi)(y) = \frac{1}{(2\pi)^{\frac{n}{2}}} \int_{\mathbb{R}^n}e^{\pm i \langle Ax+b, y\rangle} f(Ax+b) \operatorname d x$$ Then let $Ax+b = z\Rightarrow x = A^{-1}(z-b) = A^{-1}z -A^{-1}b$, and then: $$= \frac{1}{(2\pi)^{\frac{n}{2}}} e^{\mp i \langle A^{-1}b, y\rangle}\int_{\mathbb{R}^n}e^{\pm i \langle A^{-1}z, y\rangle} f(z) \cdot J(A^{-1})\operatorname d z$$ Where $J(A^{-1})$ is the Jacobian of $A^{-1}$. How should I continue?

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dietervdf
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1 Answers1

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You're practically done. What is the Jacobian of $A^{-1}$ and how can you get the $A^{-1}$ on the other side of the inner product within the integral?

Dominik
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  • I guess $\langle A^{-1} x, y\rangle = \langle x, A^{-1^} y\rangle$ where $A^{-1^}$ is the adjoint operator of $A^{-1}$, and since $A\in \mathbb{R}^{n\times n}, A^{-1^*} = A^{T^{-1}}$, but I don't see how I could simplify the Jacobian of $A^{-1}$ – dietervdf Jan 11 '16 at 16:25
  • $J(A^{-1})=det(A^{-1})=\ldots$ – Dominik Jan 11 '16 at 16:42
  • I see, I guess I should have written $J(A^{-1}(z-b) ) = \det A^{-1}$, and this is valid because $A^{-1}(z-b)$ is a linear map right? – dietervdf Jan 11 '16 at 16:50
  • Yeah that would be more detailed. (The map is not linear in the usual sense, but I get what you mean.) – Dominik Jan 11 '16 at 16:56