Suppose $A$ is hermitian matrix such that $A = A^H$.
Exponentiating matrix $A$ is defined as $e^A = \sum_{n=0}^\infty \frac{1}{n!}A^n$
Where $A^n = \underbrace{A \cdots A}_{n-\text{times}}$.
Since $A^2 = AA$ we might reasonably expect $A^{1/2} = Q$ where $QQ = A$ but instead it's defined as the Q that solves $A = QQ^H$.
I was thinking that perhaps $A^H = A \iff Q = Q^H$ so that it doesn't matter but I don't think this is true.
Can I get some intuition (or even a numerical example) for why $A^{1/2} = Q$ where $QQ^H = A$?
For matrices over the real numbers, Hermitian matrices are precisely the symmetric matrices. That is, $Q^H= Q$ so $QQ^H= QQ$.
Complex numbers do not have real square roots but $xx^T$, where $x^T$ is the complex conjugate of x, is a real number so we can take the square root if it. Similarly, if Q is a matrix over the complex numbers we need $QQ^H$ to get real numbers.