# fourier transform rotation property

I need to show that $$\widehat{f(Ax)}(\xi)=\hat{f}(A\xi)$$ where $A$ is a orthogonal $d\times d$ matrix (thus $A^{-1}=A^T$). The definition of the Fourier transform is: $$\hat{f}(\xi):=\int_{\mathbb{R}^d}f(x)e^{-2\pi ix\cdot \xi}dx$$

Now I think this can be done with a change of variables $y=Ax$ and since we know that $A$ is orthogonal we have $det(A)=1$. And I also know from the hints to use $A^{-1}y\cdot \xi=y\cdot A\xi$.

But my math is a bit rusty and I am not sure how to express $\widehat{f(Ax)}(\xi)$ in integral form. It seems to simple just to say $$\widehat{f(Ax)}(\xi)=\int_{\mathbb{R}^d}f(Ax)e^{-2\pi iAx\cdot \xi}Adx=\int_{\mathbb{R}^d}f(y)e^{-2\pi iy\cdot \xi}AA^{-1}dy$$

Can someone help me with the first steps?

Starting with the definition of the Fourier transform, use the change of variable $z = Ax$; $dz = |\det A| dx = dx$ then in the shoulder of the exponential use \begin{equation*} x\cdot\xi = A^Tz\cdot \xi= z\cdot A\xi. \end{equation*} We want to compute the transform of the function $g(x) = f(Ax)$, so using the above we have \begin{eqnarray*} \widehat g(\xi) & = & \int_{\mathbb R^d}g(x)e^{-2\pi i x\cdot \xi}\; dx \\ & = & \int_{\mathbb R^d}f(Ax)e^{-2\pi i x\cdot \xi}\; dx \\ & = & \int_{\mathbb R^d}f(z)e^{-2\pi i A^Tz\cdot \xi}\; dz \\ & = & \int_{\mathbb R^d}f(z)e^{-2\pi i z\cdot A\xi}\; dz \\ & = & \widehat f(A\xi) \end{eqnarray*}
• @BindersFull Nice answer! Why exactly is dz = |det A| dx ? – Ethunxxx Jul 21 '17 at 11:08
• @Ethunxxx Thats the Jacobian determinant for the change of variable $z = Ax$. – BindersFull Jul 21 '17 at 11:26