Given the spectral decomposition $A = PDP^T$ where $D$ is the diagonal matrix of eigenvalues, I can define the following:
$$ A^{\frac{1}{2}} = Q D^{\frac{1}{2}}Q^T $$
Is the following also true: if $A = P D P^{-1}$ where $D$ is again diagonal of eigenvalues then:
$$ A^{\frac{1}{2}} = Q D^{\frac{1}{2}}Q^{-1} $$
I tested it numerically and it is true - just not sure if these two equations are absolutely equivalent.
Yes, any diagonalizable matrix has a square root (provided you deal with complex numbers). If $A=PDP^{-1}$ and $E^2=D$, then $$ (PEP^{-1})^2=PEP^{-1}PEP^{-1}=PEIEP^{-1}=PE^2P^{-1}=PDP^{-1}=A. $$