I would like to know if when we read that $A$ is a square matrix and $A^{-1}$ appears at some points without being explicitly defined, is there a "mutual agreement" to know that $A^{-1}$ is the inverse matrix of $A$ that is the one satisfying $AA^{-1}$=$A^{-1}A=I$ ? And is this applies also to the notation $A^{-1/2}$ to say that $A^{-1/2}$ is the inverse matrix of $A^{1/2}$ where $A^{1/2}$ satisfies $A^{1/2}A^{1/2}=A$ and $A^{-1/2}A^{-1/2}=A^{-1}$ or there is condition in order to use this kind of notation ?
I mean, this notation makes directly sense in $\mathbb{R}$ for exemple but for matrices I find it tricky.
Thank you
Yeah, $A^{-1}$ is usually the inverse matrix (when $A$ is an invertible square matrix, otherwise it isn't defined).
For $A^{-1/2}$, I would guess it means $(A^{1/2})^{-1} = (A^{-1})^{1/2}$, but... the square root of the matrix is not always defined (for matrices with coefficients in $\mathbb R$). For example, even a simple example like
$$\left(\begin{array}{cc} 1&0\\ 0&1\end{array}\right)$$
has infinitely many "square roots" in the sense you mean, for example
$$\left(\begin{array}{cc} 1&0\\ 0&1\end{array}\right)=\left(\begin{array}{cc} 1&0\\ 0&1\end{array}\right)^2=\left(\begin{array}{cc} -1&0\\ 0&1\end{array}\right)^2=\left(\begin{array}{cc} 1&0\\ 0&-1\end{array}\right)^2=\left(\begin{array}{cc} 0.8&0.6\\ 0.6&-0.8\end{array}\right)^2$$
Usually, the good definition is given by the following theorem:
Theorem Let $A$ be a positive semidefinite $n\times n$ matrix (with coefficients in $\mathbb R$). There exists a unique positive semidefinite $n\times n$ matrix $B$ such that $B^2=A$.
So if $A$ is positive semidefinite, you can define $A^{1/2}$ as the unique $B$ as before, that is still positive semidefinite.