# Result for $A^{-\frac{1}{2}}A(A^{-\frac{1}{2}})^T$

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

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

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 $$A^{-1/2} A \left(A^{-1/2}\right)^T =A^{-1/2} A^{1/2}\left(A^{1/2}\right)^T \left(A^{-1/2}\right)^T = I_n I_n = I_n.$$