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Let $M$ be a positive definite symmetric $n\times n$ real matrix. Suppose the real matrix $A$ satisfies $MA M^{-1}=A^t$. Show that there exist real matrix satisfying $P^tMP=I_n$ such that $P^{-1}AP$ is diagonal.

Here is some of my approach: First, decompose $M=SDS^t$ by orthogonal diagonalizable,since M be positive definite, all eigenvalue are positive, we can multiple B, diagonal entries be square root of the inverse of the $D$, and easily get that $I_n=BDB=BS^tMSB$ and we now definte P=SB, and we want to show that $P^{-1}AP$ is diagonal, how can we use the fact $MA=A^tM$ to get that result? I have tried many substitution, but they all make thing more complicate, any one can give me some hint?

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Let $P=M^{-1/2}Q$. If we want $P^tMP=I$, we must have $Q^tQ=I$, i.e., $Q$ must be orthogonal. The requirement that $P^{-1}AP$, or identically, $Q^tM^{1/2}AM^{-1/2}Q$, is diagonal thus amounts to the requirement that $M^{1/2}AM^{-1/2}$ is orthogonally diagonalisable, but this is true because $M^{1/2}AM^{-1/2}$ is symmetric: $$ M^{1/2}AM^{-1/2}=M^{-1/2}(MAM^{-1})M^{1/2}=M^{-1/2}A^tM^{1/2}=(M^{1/2}AM^{-1/2})^t. $$

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