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Let $A_n$ be a symmetric square positive definite matrix. (A variance and covariance matrix.)

So, can I say $A^{-\frac{1}{2}}A(A^{-\frac{1}{2}})^T = I_n$?

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Yes, since for every symmetric non-negative definite matrix $A$, there exists a square-root factor $A^{1/2}$ such that $ A^{1/2}\left(A^{1/2}\right)^T = A$. If $A$ is positive definite, $A$ is invertible and $A^{1/2}$ is invertible as well. Therefore, you have $$A^{-1/2} A \left(A^{-1/2}\right)^T =A^{-1/2} A^{1/2}\left(A^{1/2}\right)^T \left(A^{-1/2}\right)^T = I_n I_n = I_n. $$

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