2D rotation of standardized PCA, how to compute new eigen values, eigen vectors, and loadings ?

I want to apply a 2D rotation of a $\theta$ angle to my two first principal components of a PCA. What I understood from this post is that I have to apply a rotation matrix R : $$R_\theta = \left( \begin{array}{ c c } \cos{\theta} & -\sin{\theta} \\ \sin{\theta} & \cos{\theta} \end{array} \right)$$

If the coordinates of my observations in the factorial plan are the matrix $Y$ of dimension $n*2$, then the rotated coordinates are $Y_R = YR$. With this new matrix, I can plot my observation in the rotated factorial plan.

In the same way, rotation of loadings $L = V S$, with $V$ the eigen vectors of the correlation matrix and $S$ the square-root matrix of the eigen values, are obtained such as $L_R = L R = V S R$. With this rotated loading, I can plot the new circle of correlation between original variables and rotated components.

I also understood that the new eigen values $S_R^2$ (variance explained by each component) are obtained doing $S^2_R = R^T S^2 R$.

What I suppose is that the rotated eigen vectors are $V_R = VR$. But if I follow loading definition ($L = V S$) then $L_R = V_R S_R = V R \sqrt{S²_R} = V R \sqrt{R^T S^2 R}$.

However, I find that $V R \sqrt{R^T S^2 R} \neq VSR$.

So why rotated loadings are defined as $L_R = L R = V S R$ and not as $L_R = V_R S_R$ ? Which definition of $L_R$ should I use and why ?

Thanks a lot =)

• You are not correct thinking that rotation of loadings L is equivalent with the same rotation of eigenvectors V (and multiplied by the modified eigenvalues, as you did). Actually, as shown on a flow chart, rotation of loadings implies the same rotation of the left eigenvectors U known also as standardized pr. component scores. – ttnphns Nov 22 '17 at 12:22
• @ttnphns I checked your flow chart and I understand that I tried to compare different things. I also realised that the way I compute the rotated raw principal components scores (Yr in my post) is wrong. If I well understood, I should get them from the rotated left eigenvectors U by giving them back their scale deposited (and accordingly rotated), but I'm not sure how to proceed. Does it mean that I should multiply U by the Sr (= to root square of R^t . S² . R) ? – Cucus Nov 22 '17 at 15:07
• As my answer and the accompanying flow chart say, in PCA and FA we typically rotate loadings (A=VS) (to achieve some wanted property of the loadings) by a rotation matrix Q. Corresponding rotation of the data points is given by UQ and then giving the data back their variances in the new axes, which equal the column sums of squares of the rotated loadings, the AQ matrix. – ttnphns Nov 22 '17 at 15:44
• ...more precisely to say, sqrt(n-1)UQ, because st. pr. component scores have sums of squares n-1, not 1, since we typically process in PCA the covariance or the correlation matrix that is already divided by n-1, and not the scatter matrix. – ttnphns Nov 22 '17 at 15:59