what does it by raising a matrix to the power of $1/2$? I came across the following which I did not understand at all.
Let $A$ be a positive semi-definite.
If $A(I-B)$ is positive definite, then the eigenvalues of $$A^{1/2}(I-B)A^{-1/2} = I -A^{1/2}BA^{-1/2}\tag 1$$ are all positive, which implies:
 $$\text {the eigenvalues of} \ B \ \text{are real and less than 1} \tag 2$$
(i) What are $A^{1/2}$ and $A^{-1/2}$ in (1)?  
(ii) How does the positive definiteness of $A(I-B)$ imply the eigenvalues of (1) are all positive?
(iii) Also, why does (1) imply (2)?
I tried to figure these out by myself, I could not find a good reference to look up. Any help would be greatly appreciated!
 A: Answer to (1):
If $A$ is positive definite, then $A^{1/2}$ denotes the unique positive definite square root of $A$. That is, $A^{1/2}$ is the unique positive definite matrix $M$ satisfying $M^2 = A$.
Because $A^{1/2}$ is positive definite, it is invertible.  $A^{-1/2}$ denotes the inverse of $A^{1/2}$.

Answer to (2):
If $A(I - B)$ is positive definite, then it has positive eigenvalues.  This implies that for any invertible matrix $S$, the congruent matrix $S^*[A(I-B)]S$ will also be positive definite.
Setting $S = A^{-1/2}$ (noting that $S = S^*$), we conclude that $A^{1/2}(I-B)A^{-1/2}$ must be positive definite, from which we may conclude that it has positive eigenvalues.

Answer to (3):
We can write $A^{1/2}(I-B)A^{-1/2} = S^{-1}(I-B)S$, where $S$ is as above.  That is, $A^{1/2}(I-B)A^{-1/2}$ is similar to $I-B$.
Note, however, that $\lambda$ is an eigenvalue of $B$ if and only if $1 - \lambda$ is an eigenvalue of $I-B$.
If both $I-B$ has positive eigenvalues, then for every eigenvalue $\lambda$ of $B$, $1-\lambda$ is (real and) positive.  Thus, the eigenvalues of $B$ must be real less than $1$.
