If $A, G, H$ ∈ $M_n$ be positive definite , $GAG = HAH$ then $G = H$

Question

Let $A, G, H$ ∈ $M_n$ be positive definite and suppose that $GAG = HAH$.

Why does $G = H$?

It is well known that the square root of a positive definite matrix is unique (see Square root of Positive Definite Matrix), but this only solves the case $A=\rm{id}$. Does the general case also follow from this?

Take $A = [1, 0; 0, 1]$, $G = [1, 0; 0, 1]$, and $H = [-1, 0; 0, -1]$. $GAG = HAH$ holds, but $G = -H$! — m0_as, Oct 15 '15 at 04:41

score 2 · Accepted Answer · answered Oct 16 '15 at 01:54

The general case also follows from the uniqueness of square roots as follows:

If we multiply your assumption by $A^{1/2}$ from left and right, we get $$ (A^{1/2} G A^{1/2} )(A^{1/2} G A^{1/2} ) = (A^{1/2} H A^{1/2} ) (A^{1/2} H A^{1/2} ). $$

Since both $A^{1/2} G A^{1/2} $ and $A^{1/2} H A^{1/2} $ are positive definite (why?), uniqueness of square roots shows that they are identical. By multiplying with $A^{-1/2} $ from the left and right, we get $G=H$ as desired.

If $A, G, H$ ∈ $M_n$ be positive definite , $GAG = HAH$ then $G = H$

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