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Given a symmetric matrix A. Is there any orthogonal matrix P that makes $PAP^{-1}$ diagonal?

I've found at wikipedia this:

The finite-dimensional spectral theorem says that any symmetric matrix whose entries are real can be diagonalized by an orthogonal matrix. More explicitly: For every symmetric real matrix A there exists a real orthogonal matrix Q such that $D = Q^{T}AQ$ is a diagonal matrix. Every symmetric matrix is thus, up to choice of an orthonormal basis, a diagonal matrix.

I think it fits with my problem, but I am not sure since I need this $PAP^{-1}$ (1) and what I get is this: $Q^{T}AQ$

I know that since P is an orthogonal matrix, this is true: $ P^{-1} = P^{T}$

So I get this: $Q^{-1}AQ$ (2) And I am not sure if expressions (1) and (2) are identical.

Thank you for your time! (emphasizing on the fact that

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I understand now where your confusion stems from. The thing is we don't really care about whether we say that there is a matrix $P$ such that $PAP^{-1}$ is diagonal, or that there is a matrix $P$ such that $P^{-1}AP$ is diagonal, because both definitions are equivalent, replacing $P$ by $P^{-1}$. The same thing happens with the concept of orthogonal diagonalization: there is a matrix $Q$ such that $QAQ^T$ is diagonal if and only if there is a matrix $Q$ such that $Q^TAQ$ is diagonal. Both definitions are equivalent replacing $Q$ by $Q^T$, so don't worry about where you write the transpose, or the inverse. Just be consistent once you decide upon one way to write that a matrix is diagonalizable or orthogonally diagonalizable.

Gerry's answer shows exactly how one would be consistent once written down the definitions.

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That was exactly my confusion! :) – Chris Jun 11 '12 at 0:13

Given $Q^tAQ=D$, let $P=Q^t$; then $P$ is orthogonal, and $PAP^{-1}=D$.

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Thank you Gerry! – Chris Jun 11 '12 at 0:12

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