PCA, covariance, eigenvector matrix and rotation
Having 2 uncorrelated normal variables $x_1 \sim N(0, 1)$ and $x_2 \sim N(0, 1)$, one can correlate them through a linear transformation:
$$A = \begin{bmatrix}a &amp; b \\ c &amp; d\end{bmatrix}$$
$$X' = AX$$
The covariance matrix of the transformed correlated variables $X'$ is given by:
$$\Sigma' = A A^T$$
For a rotation by an angle $\theta$, and scaling for adding individual standard deviations, the transform is  (sign is same as that given):
$$A =  \begin{bmatrix}cos(\theta) &amp; sin(\theta) \\ -sin(\theta) &amp; cos(\theta)\end{bmatrix} \begin{bmatrix}\sigma_1 &amp; 0 \\ 0 &amp; \sigma_2\end{bmatrix}$$
This produces the covariance matrix $\Sigma'$ as:
$$\Sigma' = \begin{bmatrix}\sigma_1cos(\theta) &amp; \sigma_2sin(\theta) \\ -\sigma_1sin(\theta) &amp; \sigma_2cos(\theta)\end{bmatrix} \begin{bmatrix}\sigma_1cos(\theta) &amp; -\sigma_1sin(\theta) \\ \sigma_2sin(\theta) &amp; \sigma_2cos(\theta)\end{bmatrix}$$
The cross-variance component is given by:
$$\Sigma'_{12} = \sigma_1\sigma_2(cos^2(\theta)-sin^2(\theta))+(\sigma_2^2-\sigma_1^2)cos(\theta)sin(\theta)$$
This is the relation of cross-variance on angle $\theta$. Setting $\theta = 45^\circ$ we get:
$$\Sigma'_{12} = \frac{\sigma_2^2-\sigma_1^2}{2}$$
Setting $\sigma_1 = 0.333$ and $\sigma_2 = 1$ then $\Sigma'_{12} = 0.445$
Note $n = \frac{1}{tan(\theta)} \ne \Sigma'_{12}$ does not represent the covariance component as shown by analysis above.
PCA decomposes the covariance matrix into un-correlated components. In other words it undoes the rotation (the scaling remains), since the rotation correlated the components in the first place.
Explicitly PCA decomposes the covariance matrix into:
$$\Sigma' = U S U^T$$
where $U$ is an orthogonal matrix of eigen-vectors and $S$ is a diagonal matrix which represents individual variances.
Going back to $\Sigma' = A A^T$ and representing the rotation by $R$ we have:
$$\Sigma' = R \begin{bmatrix}\sigma_1^2 &amp; 0 \\ 0 &amp; \sigma_2^2\end{bmatrix} R^T$$
Thus the orthogonal matrix of eigen-vectors $U$ corresponds to the (orthogonal) rotation matrix $R$.
Thus the eigen-vectors should (precisely in theory, approximately in practise) correspond to the rotation matrix (up to some sign factor and permutation which remains arbitrary).
For further info see:

Bivariate normal distribution
Understanding the Covariance Matrix
Interesting Properties of the Covariance Matrix