Show $U_A$ is a unitary operator. Let $A$ be an invertible $n \times n$ matrix with real entries.
Show that $(U_A f)(x) = f(A^{-1}x) \vert \det(A) \vert^{-1/2}$ defines a unitary operator in $L^2(\mathbb{R}^n,d\lambda)$.
I have some idea of what the questions asks but I have no idea where to begin.
Any help will be greatly appreciated
 A: Using that $A$ is injective (therefore bijective), and the change of variables formula we have that 
\begin{align}
 \langle f, g\rangle_{L^2} &= \int_{\mathbb R^n}fg \, d\lambda 
\\ &=
\int_{A^{-1}(\mathbb R^n)}fg \, d\lambda \\&= \int_{\mathbb R^n}f\circ A^{-1}\cdot g\circ A^{-1} \cdot  |\det A^{-1}| \, d\lambda 
\\ &=
\int_{\mathbb R^n}\color{green}{f\circ A^{-1}\cdot |\det A|^{-\frac 12}}\cdot {g\circ A^{-1} \cdot |\det A|^{-\frac 12}} \, d\lambda
\\ &= \int_{\mathbb R^n}\color{green}{U_Af}\cdot {U_Ag} \, d\lambda
\\ &= \langle U_Af, U_Ag\rangle_{L^2}.
\end{align}
Thus $U_A$ preserves the inner product (so it's bounded). It is also surjective, since if $g \in L^2(\mathbb{R}^n,\lambda)$, then $|\det A|^{\frac 12} g \circ A\in L^2(\mathbb{R}^n,\lambda)$, and$$ U_A\left(|\det A|^{\frac 12} g \circ A\right) =  g.$$
We can conclude that $U_A$ is a surjective, inner-product preserving operator i.e. it's unitary, and as we've seen $${U_A}^*(g) = {U_A}^{-1}(g) = g \circ A \cdot |\det A|^{\frac 12}.$$
