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for any symmetric real matrix $S$, the following eigendecomposition exists:

$ S = Q \Lambda Q^{\top} $

where $Q$ is a unitary matrix, consisting of the eigenvectors of $S$ wikipedia . By definition of unitary, we have $Q^{\top}Q=QQ^{\top}=I$. Given an orthonormal set of eigenvectors, $Q^{\top}Q=I$, is trivial. How can one show $QQ^{\top}=I$ ?

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    Thanks All. it simply answered my question. I was not thinking about it that way.2011-09-22

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If $\mathbf Q^\top\mathbf Q=\mathbf I$, then $\mathbf Q\mathbf Q^\top\mathbf Q=\mathbf Q$. Then consider postmultiplying both sides by $\mathbf Q^{-1}$...

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Hmm, I think, the more interesting part in the original question is in the direction: "why is it with a symmetric matrix S, that the diagonalization $ \small S = Q \Lambda Q^{-1} $ provides a unitary matrix Q , such that $ \small Q^T = Q^{-1} $ ?" Which, of course, can be answered by considering the equality of S with its transpose: $\small S = S^T=(Q^T)^{-1} \Lambda Q^T = Q \Lambda Q^{-1} \to Q=(Q^T)^{-1} $ and $ \small Q^{-1}=Q^T $