How is the matrix square root defined? I am wondering how the square root of a positive definite matrix is formally defined.
The square root of a positive definite matrix $A$ is $A^{1/2}$ if $A^{1/2}A^{1/2}= A$. However, I have also seen definitions that has $A^{1/2}$ as the square root of a matrix A if ${A^{1/2}}^T A^{1/2} = A$, which is confusing to me. 
Furthermore, if I define the square root of a positive definite matrix $A$ to be $A^{1/2}$, and the square root of $A^{-1}$ is $A^{-1/2}$, is it necessarily true that $A^{1/2}A^{-1/2} = I$? Or would it be ${A^{1/2}}^TA^{-1/2} = I$? thanks.
 A: The Cholesky factorization $A=R^{T}R$ where $R$ is upper triangular with positive entries on the diagonal can often be used as an effective "square root" of a symmetric and positive definite matrix $A$.  It's relatively fast to compute.  
If $A$ is symmetric and positive definite, it will also have a symmetric and positive definite matrix square root $S$, with $A=S^{2}$.  This can be computed from the eigenvalue decomposition of $A$ as explained in Jacky Chong's answer, or by some special purpose algorithms.  The symmetric matrix square root is quite a bit slower to compute than the Cholesky factorization.  
It is unfortunate that some authors choose to refer to the Cholesky factorization as a "square root", but this is fairly common.  It would be better if these authors used the term Cholesky factorization instead.  
A: Since $A$ is positive definite, then we have that
\begin{align}
A = Q^TDQ
\end{align}
where $D$ is a diagonal matrix with non-negative entries. Then define
\begin{align}
A^{1/2} = Q^TD^{1/2}Q
\end{align}
i.e. we look at the square root of each entries in the diagonal matrix. 
Then observe
\begin{align}
A^{1/2}A^{1/2} = Q^TD^{1/2}QQ^TD^{1/2}Q = Q^TD^{1/2}D^{1/2}Q = Q^TD Q= A. 
\end{align}
