# positive definte symmetric real matrix.

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?

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: $$M^{1/2}AM^{-1/2}=M^{-1/2}(MAM^{-1})M^{1/2}=M^{-1/2}A^tM^{1/2}=(M^{1/2}AM^{-1/2})^t.$$