Let $X: \Omega \rightarrow \mathbb{R}$ and $Y: \Omega \rightarrow \mathbb{R}$ be two real-valued, independent and identically distributed random variables. Now we additionally assume that $X$ and $Y$ are standard-normally distributed. Show that the random variables $\frac{X-Y}{\sqrt{2}}$ and $\frac{X+Y}{\sqrt{2}}$ are independent under these assumptions.
I know that two random variables $X$ and $Y$ are independent if and only if $$ F_{X Y}(x, y)=F_X(x) F_Y(y) \quad \forall\, x, y \in \mathbb{R}, $$ where $F_{X, Y}(x, y)$ is their joint distribution function and $F_X(x)$ and $F_Y(y)$ are their marginal distribution functions.
Besides, I also got this hint:
The matrix $\left(\begin{matrix} \frac{1}{\sqrt{2}} & \frac{-1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{matrix}\right)$ describes a rotation by $45^{\circ}$.
Any hints on how to tackle this problem would be appreciated.
Since $X$ and $Y$ are independant and standard-normally distributed variables, the couple $(X, Y)$ is a gaussian vector, wich covariance matrix $I_2$. Let $$A := \begin{pmatrix}1/\sqrt2&-1/\sqrt2\\ 1/\sqrt2 & 1/\sqrt2\end{pmatrix}.$$ The vector $(\frac{X-Y}{\sqrt2}, \frac{X+Y}{\sqrt2}) = A \cdot (X, Y)$ is also a gaussian vector with $A I_2 A^T=AA^T$ as its covariance matrix. Since $A$ is the $\pi/2$-rotation, it is an orthognal matrix and $AA^T = I_2$. Hence, $\frac{X-Y}{\sqrt2}$ and $\frac{X+Y}{\sqrt2}$ are independent.