eigendecomposition of symmetric matrix

Question

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 $QQ^{\top}=I$ ?

If $Q^TQ = I$, then $Q^T = Q^{-1}$. Now $QQ^{-1} = I$, so... — , Sep 22 '11 at 00:15
it is a full rank matrix and its inverse is $Q$. I don't get your question. — , Sep 22 '11 at 00:15
You might find this useful; it explains why $AB=I$ implies $BA=I$ for general square matrices $A$ and $B$. — Nick Strehlke, Sep 22 '11 at 00:21
Thanks All. it simply answered my question. I was not thinking about it that way. — Austin, Sep 22 '11 at 01:06

score 2 · Answer 1 · answered Sep 22 '11 at 08:13

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 $

score 1 · Accepted Answer · answered Sep 22 '11 at 00:16

1

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

answered Sep 22 '11 at 00:16

J. M. ain't a mathematician

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eigendecomposition of symmetric matrix

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