Square root of matrix product My linear algebra abilities are somewhat limited, so this may be a very basic question. Suppose we have two positive definite matrices A and B. Is $(AB)^{1/2}=A^{1/2}B^{1/2}$? Furthermore, is $(AB^{-1})^{1/2}=A^{1/2}B^{-1/2}$?
 A: I suppose that what you want to know is this: if you multiply a square root of $A$ by a square root of $B$, do you get a square root of $AB$? The answer is negative in general. Suupose that $A=B=\left[\begin{smallmatrix}1&0\\0&1\end{smallmatrix}\right]$. Now, take$$\alpha=\begin{bmatrix}2 & 1 \\ -3 & -2\end{bmatrix}\text{ and }\beta=\begin{bmatrix}-1 & 0 \\ 0 & 1\end{bmatrix}.$$Then $\alpha^2=\operatorname{Id}=A$, $\beta^2=\operatorname{Id}=B$, but$$(\alpha.\beta)^2=\begin{bmatrix}-2 & 1 \\ 3 & -2\end{bmatrix}\neq\operatorname{Id}=A.B.$$
A: The OP  mixes all mathematical notions; in a first step, the square root is obtained with the Choleski decomposition (!); in a second step, the square root of $A$ is a matrix $B$ s.t. $B^2=A$. The accepted answer uses the second definition that obviously cannot satisfy the studied relation:
for example, $n=1$, $A=B=1,A^{1/2}=-1,B^{1/2}=1,(AB)^{1/2}=1$.
Yet @Hw Chu  gave, in his two comments, the essence of the argument!
$\textbf{Definition}$. Let $C$ be  a  real matrix whose the eigenvalues are $> 0$ and that is diagonalizable. If $C=PDP^{-1}$ where $D$ is diagonal $>0$, then, by definition, $C^{1/2}=PD^{1/2}P^{-1}$; it can be proved that $C^{1/2}$ is a polynomial in $C$ and it is the only square root of $C$ whose eigenvalues are $>0$.
Note that if $A,B$ are symm. $>0$, then $AB$ has $>0$ eigenvalues and is diagonalizable.
$\textbf{Proposition 1}.$ Let $A,B$ be symm. $>0$ matrices which commute. Then $A^{1/2}B^{1/2}=(AB)^{1/2}$.
$\textbf{Proof}$. Since $A^{1/2},B^
{1/2}$ are polynomials in $A,B$, we deduce that $A^{1/2},B^{1/2}$ commute and $(A^{1/2}B^{1/2})^2=AB$.
Of course, the converse is false; yet, one has the following
$\textbf{Proposition 2}.$ Let $P,Q$ be random matrices; then $P,Q$ are invertible with probability $1$; the matrices $A=P^TP,B=Q^TQ$ are symm. $>0$. Then $A^{1/2}B^{1/2}\not= (AB)^{1/2}$ with probability $1$.
