How is the complexity of PCA O(min(p^3,n^3))?

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旧时难觅i
旧时难觅i 2021-02-04 04:48

I\'ve been reading a paper on Sparse PCA, which is: http://stats.stanford.edu/~imj/WEBLIST/AsYetUnpub/sparse.pdf

And it states that, if you have n data poin

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  •  醉梦人生
    2021-02-04 05:17

    Assuming your dataset is $X \in \R^{nxp}$ where n: number of samples, d: dimensions of a sample, you are interested in the eigenanalysis of $X^TX$ which is the main computational cost of PCA. Now matrices $X^TX \in \R^{pxp}$ and $XX^T \in \R^{nxn}$ have the same min(n, p) non negative eigenvalues and eigenvectors. Assuming p less than n you can solve the eigenanalysis in $O(p^3)$. If p greater than n (for example in computer vision in many cases the dimensionality of sample -number of pixels- is greater than the number of samples available) you can perform eigenanalysis in $O(n^3)$ time. In any case you can get the eigenvectors of one matrix from the eigenvalues and eigenvectors of the other matrix and do that in $O(min(p, n)^3)$ time.

    $$X^TX = V \Lambda V^T$$

    $$XX^T = U \Lambda U^T$$

    $$U = XV\Lambda^{-1/2}$$

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