I know in general that if a matrix $A$ is positive definite, then there exists a (unique?) square root matrix $B$, which is also positive definite, such that $BB=A$.
Therefore, suppose $A$ is positive definite. It is invertible, and its inverse is also positive definite. Therefore I know there exists $C$ (possibly unique?) so that $CC=A^{-1}$. For simplicity, I will call $C=A^{-1/2}$, so that in this notation $A^{-1/2}A^{-1/2}=A^{-1}$.
My question is, is it also true that $A^{-1/2}AA^{-1/2}=I$? Or is this not necessarily true?
I have posted an attempted solution below — please let me know what you think.
Thanks!
Edit It occurs to me that at the essence of this question is whether or not $B=C^{-1}$; is this necessarily true? That is, is square root of the inverse of a matrix equal to the inverse of the square root of the matrix? (This also partly depends on whether or not $B$ and $C$ are themselves invertible; would this be true?)
Edit 2 There's now a related topic which asks what kind of matrix decomposition we're using here (and the differences between these different approaches to matrix decomposition. That question can be found here.
Best Answer
It is generally true that if $A$ is an $n\times n$ invertible and if $A^{-1}$ has a "square root" $C$, also $n\times n$, such that:
$$ A^{-1} = C^2 $$
then $C^{-1} A^{-1} C^{-1} = I$ holds.
The first fact we need is that since $A$ is invertible, $A^{-1}$ is invertible, and this implies $C$ is invertible. For if not, then there would exist a nonzero vector $x$ in the nullspace of $C$, and $Cx=0$ would imply $A^{-1}x=0$, contradicting the invertibility of $A^{-1}$. Thus $A = (C^2)^{-1} = (C^{-1})^2$.
The second fact we need is that a one-sided inverse of a matrix is a two-sided inverse, so that:
$$ A C^2 = I \; \implies \; C A C = I $$
That is, using associativity of matrix multiplication, the left hand side tells us $(AC)C = I$, so that $C$ is a (right) inverse of $AC$. Thus it must also be a (left) inverse of $AC$, which is what the right hand equation states.
Finally by the same reasoning:
$$ C A C = I \; \implies \; C^{-1} A^{-1} C^{-1} = I $$
In this discussion/proof we have not invoked the symmetry of $A$ nor the uniqueness of a symmetric positive definite square root $C$ for $A^{-1}$, which also would be symmetric positive definite. The reasoning above is correct even if $A$ is not symmetric, and even if $C$ is not positive definite, and relies only on $A^{-1} = C^2$.