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For a given centered configuration of points $X\in\mathbb{R}^{n\times 3}$, the covariance matrix is denoted by $S=\frac{1}{n}X^TX$. Recall that the 2D PCA solution is obtained by $Y=X\cdot U$, where $U\in\mathbb{R}^{3\times 2}$ is a matrix of orthonormal eigenvectors of $S$ associated with its 2 dominant eigenvalues $\lambda_1, \lambda_2$.

Now, suppose that $Y_1\in\mathbb{R}^{n\times 2}$ is obtained by normalizing each axis of $Y$. The normalization factor would be $1/\sqrt{\lambda_i}$, $i\in\{1,2\}$ (please correct me if I'm wrong here).

Also, suppose that $Z$ is obtained by normalizing each axis of $X$ to unit length. A 2D PCA solution on $Z$ is denoted by $Y_2$. Are $Y_1$ and $Y_2$ the same? I obtain that the results are different, but it hold that the axes of both $Y_1$ and $Y_2$ are orthonormal. If so, what is the distinguishing characteristic of $Y_2$ relative to $Y_1$?

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Normalizing the axes of $X$ by the singular values of $Y$ is arbitrary and hence meaningless. ($X$ might just be $Y$ with the ordering of the three reversed.) No sensible relation can be expected, neither is the use of the second recipe sound.

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  • $\begingroup$ Please note the edit I made (configuration $Z$ is $X$ with all axes having unit lenght (1)). What property would have a 2D PCA solution from $Z$ relative to PCA solution from $X$. (they will not be the same, and not just differ by the aspect ratio, but I'm interested in the actual properties and possibly a relation to PCA on intact $X$) $\endgroup$ – usero Jul 27 '12 at 19:56

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