How to decompose covariance matrix for a bivariate normal distribution

Question

Let $$\boldsymbol{x} = \begin{bmatrix} x\\ y\\ \end{bmatrix}, \boldsymbol{\mu} = \begin{bmatrix} \mu_x\\ \mu_y\\ \end{bmatrix}, \boldsymbol{\Sigma}=\begin{bmatrix} \sigma_x & \rho\sigma_x\sigma_y\\\ \rho\sigma_x\sigma_y & \sigma_y\\ \end{bmatrix}$$ where $\rho$ is the Pearson correlation between $x$ and $y$. Suppose I have the following bivariate normal distribution

$$\boldsymbol{x} \sim N(\boldsymbol{\mu},\boldsymbol{\Sigma})$$

Let

$$R= \begin{bmatrix} \cos(\theta) & \sin(\theta)\\ \sin(\theta) & -\cos(\theta)\\ \end{bmatrix}$$

I want to decompose the covariance matrix $\boldsymbol{\Sigma}$ such that the resulting decomposition uses rotational matrices. According to this article

https://www.visiondummy.com/2014/04/geometric-interpretation-covariance-matrix/

I can decompose the covariance matrix via eigendecomposition

$$\Sigma = VLV^{-1}$$ and more specifically, I can write it as

$$\Sigma = RSSR^{-1}$$

where $S=\sqrt{L}$ and $R=V$. Unfortunately, I'm unfamiliar with eigendecomposition so I'm having trouble finding how to do this. The example in the article is for an uncorrelated case.

Can someone show me how to eigendecompose the covariance matrix into the $RSSR^{-1}$ form?

I tried for instance to figure out what $L$ is by doing

$$R^{-1}\Sigma R = L$$

But if I plug that into Wolfram Alpha (to check if I'm going in the right direction), I get something horribly ugly, see here

Answer 1

The covariance matrix is PSD and symmetric, so a standard eigen decomposition results in $$\Sigma=VDV^T$$ where $V=V^T$. $V$ is orthonormal matrix, and for it to be rotation matrix, you also need $|V|=1$ where you can arrange it by suitable choice of signs for the eigenvectors.

Note: $\Sigma$ matrix above should contain $\sigma_x^2,\sigma_y^2$ and your example rotation matrix has $|R|=-1$, you can use the one in wikipedia.