Let $Z\sim N(0,\Sigma)$ with $$ \Sigma=\begin{pmatrix}\sigma_1^2 & p\sigma_1\sigma_2\\p\sigma_1\sigma_2 & \sigma_2^2\end{pmatrix} $$ whereat $\sigma_i^2=\text{var}(Z_i), i=1,2$ and $p=\text{cov}(Z_1,Z_2)/(\sigma_1\sigma_2)$. Show that by rotation $Z$ can be transformed into a bivariate normal distribution $Y\sim N(0,D)$ with a diagonal covariance matrix $D=\text{diag}(\lambda_1,\lambda_2)$ (resp. that $Z$ emanates from such a distribution by the inverse rotation).
Hello everybody, if I got that right, I have to find a $2\times 2$-matrix $R$ which is a rotation matrix and fullfills $$ R\Sigma=\begin{pmatrix}\lambda_1 & 0\\0 & \lambda_2\end{pmatrix}. $$
I do not know how to do so. Can you please help me?
EDIT (based on the help I got in the comments below)
So the task is to find a rotation matrix $R$ with $RZ=Y$. So the covariance matrix $D$ of $Y$ will be of the form $D=R\Sigma R^T$.
So I started with $R=\begin{pmatrix}\cos\alpha & -\sin\alpha\\\sin\alpha & \cos\alpha\end{pmatrix}$ and determined the entries of $R\Sigma R^T$: $$ \begin{pmatrix}\sigma_1^2 \cos^2\alpha-2\sin\alpha \cos\alpha p\sigma_1\sigma_2+\sigma_2^2 \sin^2\alpha & \cos\alpha\sin\alpha(\sigma_1^2-\sigma_2^2)+p\sigma_1 \sigma_2 (\cos^2\alpha-\sin^2\alpha)\\ \sin\alpha\cos\alpha (\sigma_1^2-\sigma_2^2)+p\sigma_1\sigma_2(\cos^2\alpha-\sigma^2\alpha) & \sigma_1^2 \sin^2\alpha+2p\sigma_1\sigma_2\sin\alpha\cos\alpha+\sigma_2^2\cos^2\alpha\end{pmatrix} $$
Now I set the entries that are not on the diagonal to $0$ and solved it for $\alpha$: For both equations I got that $$ \alpha_k=-\frac{1}{2}\arctan\left(\frac{2p\sigma_1\sigma_2}{\sigma_1^2-\sigma_2^2}\right)+k\cdot\frac{\pi}{2}, k\in\mathbb{Z} $$
are the possible solutions. Because it is enough to find one appropriate rotation matrix I chose $$ \alpha_0=-\frac{1}{2}\arctan\left(\frac{2p\sigma_1\sigma_2}{\sigma_1^2-\sigma_2^2}\right). $$ With this result I calculated the entries on the diagonal, just putting $\alpha_0$ in the above entries: $$ \lambda_1:=\sigma_1^2\cos^2\alpha_0-2\sin\alpha_0 \cos\alpha_0p\sigma_1\sigma_2+\sigma_2^2\sin^2\alpha_0 $$ $$\lambda_2:=\sigma_1^2\sin^2\alpha_0+2\sin\alpha_0\cos\alpha_0 p\sigma_1\sigma_2+\sigma_2^2 \cos^2 \alpha_0 $$ So with the found $\alpha_0$ it is $$ \begin{pmatrix}\cos\alpha_0 & -\sin\alpha_0 \\ \sin\alpha_0 & \cos\alpha_0\end{pmatrix}\Sigma \begin{pmatrix}\cos\alpha_0 & \sin\alpha_0 \\ -\sin\alpha_0 & \cos\alpha_0\end{pmatrix}=\begin{pmatrix}\lambda_1 & 0 \\ 0 & \lambda_2\end{pmatrix}=D. $$
I think that's it?
With greetings
Miro