linear algebra — are eigenvectors orthogonal? 
Are eigenvectors always orthogonal? The short answer is no. Eigenvectors of an arbitrary (but not degenerate) square real matrix 
are sure to be independent (if there are no repeated eigenvalues), however, they are not necessarily orthogonal.
Are eigenvectors of a symmetric matrix always orthogonal? The short answer is no. Gilbert Strang gives the following definition: A real matrix has perpendicular eigenvectors if and only if 
. This means that for a symmetric real matrix (i.e., 
) an orthogonal eigen basis is guaranteed to exist, however, "it is also possible to choose a non-orthogonal basis" (see the user feedback below). Thus eigenvectors of a symmetric real matrix are not necessarily perpendicular. However, any two eigenvectors (with distinct eigenvalues) are orthogonal. Another nice property of symmetric matrices is that their eigenvalues are real.
Reference
Gilbert Strang, "Introduction to Linear Algebra, 4th Edition", Wellesley-Cambridge Press, 2009.
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October 24th, 2011 at 4:34 am
"It follows that eigenvectors of a symmetric real matrix A (i.e., A=A^T) are perpendicular"
This is incorrect, the guarantee is that an orthogonal set of eigenvectors exist for symmetric matrices, but it is also possible to choose a non-orthogonal basis for subspaces corresponding to eigenvalues with multiplicity > 1. For example, the identity matrix in R^2 is symmetric, and all vectors in R^2 are its eigenvectors, we could choose any non-orthogonal pair to form the basis for the eigenvectors - if a given algorithm returned such an eigenvectors basis, this is perfectly correct, but we users may incorrectly assume basis is orthogonal.
October 31st, 2011 at 8:41 am
Thank you for the correction! I have updated the post.