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I want to know, since the covariance matrix is symmetric, positive, and semi-definite, then if I calculate its eigenvectors what would be the properties of the space constructed by those eigenvectors (corresponds to non-close-zero eigenvalues), is it orthogonal or anything else special? Suppose this eigenvector matrix is called U, then what would be the properties with U*transpose(U)?

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One can always set things up such that the matrix of eigenvectors of a symmetric positive semidefinite matrix is an orthogonal matrix, zero eigenvalues or not. –  Guess who it is. Apr 16 '11 at 19:08
How to set things up? and what would be the U*transpose(U) –  baboonWorksFine Apr 16 '11 at 19:18
Most eigenroutines would generate an orthogonal matrix of eigenvectors. Mathematica and MATLAB (due to how LAPACK routines are set up) do. Remember that multiplying an orthogonal matrix with its transpose gives an identity matrix. –  Guess who it is. Apr 16 '11 at 19:20
@J.M.: Do you know what the eigenvectors of a symmetric positive semidefinite matrix mean if the matrix is not necessarily a covariance matrix? –  Mitch Apr 17 '11 at 4:17
@Mitch: I tend to think of those things geometrically, much like Qiaochu's answer here... or you had something else in mind? –  Guess who it is. Apr 17 '11 at 4:42

2 Answers 2

A symmetric matrix has orthogonal eigenvectors (irrespective of being positive definite - or zero eigenvalues). Hence, if we normalize the eigenvectors, U * transpose(U) = I

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This assumes, of course, that orthogonal bases have been selected for any degenerate eigenspaces. For instance, $(1,0)$ and $(\sqrt{1/2}, \sqrt{1/2})$ form a normalized eigenbasis for the 2 by 2 identity matrix but $U U^T \ne I$. –  whuber Apr 17 '11 at 17:56
@whuber: You're right. –  leonbloy Apr 17 '11 at 21:26

The eigenvectors correspond to the principal components and the eigenvalues correspond to the variance explained by the principal components.

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