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Let matrices $P_0$ and $P_1$ be symmetric and positive definite.

Is it possible to find a symmetric matrix $S$ such that $SP_0S=P_1$?

If $P_0=I$, this is always possible: $S=\sqrt{P_1}$.

Rodrigo de Azevedo
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Pete
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1 Answers1

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This is quite surprising! In general, I would suggest you to refrain from making conjectures based on the fact that the statement holds for the identity matrix. You might use some of the very strong hidden properties of the identity matrix (e.g., it commutes with everything) when you check the result on it.

However, in this particular case, you are spot on!

Solution: $S:= P_0^{-1/2}(P_0^{1/2}P_1P_0^{1/2})^{1/2}P_0^{-1/2}$.

First of all, products of the form $ABA$ with $A,B$ symmetric, are also symmetric. So $S$ is symmetric.

Moreover, $SP_0S = P_0^{-1/2}(P_0^{1/2}P_1P_0^{1/2})^{1/2}P_0^{-1/2} P_0 P_0^{-1/2}(P_0^{1/2}P_1P_0^{1/2})^{1/2}P_0^{-1/2} = P_0^{-1/2}(P_0^{1/2}P_1P_0^{1/2})^{1/2}(P_0^{1/2}P_1P_0^{1/2})^{1/2}P_0^{-1/2} = P_0^{-1/2}P_0^{1/2}P_1P_0^{1/2}P_0^{-1/2} = P_1$

A. Pongrácz
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