In Principal component analysis, why the symmetric covariance matrix's eigenvector points in the most “significant” direction of the data set?
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What exactly don't you understand? Do you know what an eigenvector is? It's not just any eigenvector that gives the "most significant direction." It's an eigenvector associated with the largest eigenvalue. Possible duplicate: http://math.stackexchange.com/questions/23596/why-is-the-eigenvector-of-a-covariance-matrix-equal-to-a-principal-component?rq=1 – symplectomorphic May 14 '16 at 01:55
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@symplectomorphic : in your link they don't say that it is because the $k$ first eigenvectors are the solution of $\min_P |X - X P P^T |^2_F$ subject to $ P : n \times k \quad$ (with $ X : m \times n$ being the matrix of your $m$ points) – reuns May 14 '16 at 02:06