Is the Square Root of an Inverse Matrix Equal to the Inverse of the Square Root Matrix? 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.
 A: 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 $A^{-1} = C^2$.
A: We know
\begin{equation}
A^{-1}=CC
\end{equation}
Inverting both sides (Is this allowed? Depends on whether $C$ is invertible, which I do not know...), we would have
\begin{equation}
A=C^{-1}C^{-1}
\end{equation}
Multiply both sides on the left by $C$, and then on the right by $C$, so we get
\begin{equation}
CAC=I
\end{equation}
or, in the "$1/2$" notation,
\begin{equation}
A^{-1/2}AA^{-1/2}=I
\end{equation}
A: Answering the question in the title of your question:
Yes, suppose $A$ is real symmetric positive definite and its (unique!) square root is $B := A^{\frac{1}{2}}$ (also real symmetric positive definite), i.e. $A = B B = B^2$.
The inverse of $A$, $A^{-1}$ is real symmetric positive definite and thus has a unique real symmetric positive definite square root $D := (A^{-1})^{\frac{1}{2}}$.
We have $D = B^{-1}$, i.e. $(A^{-1})^{\frac{1}{2}} = (A^{\frac{1}{2}})^{-1}$ again by the unicity of the square root: $A = B^2$ implies $A^{-1} = (B^{-1})^2$.
Since $A^{-1} = D^2$ and square roots are unique $B^{-1} = D$, as required.
