I have a bivariate normal distribution:
$$\begin{pmatrix} X\\Y\end{pmatrix} \sim \mathcal{N}(\mu, \Sigma)$$ where: $$\mu = \begin{pmatrix}\mu_X \\ \mu_Y\end{pmatrix}$$ $$\Sigma = \begin{bmatrix} \sigma_X^2 & \rho\sigma_X\sigma_Y \\ \rho\sigma_X\sigma_Y & \sigma_Y^2\end{bmatrix}$$
Now I'm supposed to show that even though $X$ and $Y$ are dependent the sum $X+Y$ and the difference $X-Y$ are independent to each other no matter what $\rho$ is. However, I can't seem to show this.
My working so far:
Let the transformation matrix $A=\begin{bmatrix} 1 & 1 \\ 1 & -1\end{bmatrix}$, then in general:
$$A\begin{pmatrix} X\\Y\end{pmatrix} = \begin{pmatrix} X+Y\\X-Y\end{pmatrix} \sim \mathcal{N}(A\mu, A\Sigma A^T) $$
But the covariance terms are not zero and thus there is no independence: $$A\Sigma A^T = \begin{bmatrix} \sigma_X^2+2\rho \sigma_X\sigma_Y+\sigma_Y^2 & \sigma_X^2 - \sigma_Y^2\\ \sigma_X^2 - \sigma_Y^2 & \sigma_X^2-2\rho \sigma_X\sigma_Y+\sigma_Y^2 \end{bmatrix}$$
Where am I making a mistake?