When $A$ is positive semi-definite, the positive semi-definite square root $B$, which is $BB=A$ is unique.
I have heard that this holds true for a symmetric matrix. What about under weaker conditions? In particular, is it true when $A,\ B$ may not be symmetric but $x^TAx>0,\ x^TBx>0$? or may not be symmetric and may not be $x^TAx>0,\ x^TBx>0$ but all eigenvalues are positive?
It's hard for me to think of a way to prove. For example, I guess this is true for the normal matrix of real numbers such as $$A=\begin{pmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 1& 0 & 1\end{pmatrix},\ B=\frac{1}{3}\begin{pmatrix}\sqrt{2}+\sqrt{3} & \sqrt{2} & \sqrt{2}-\sqrt{3}\\ \sqrt{2}-\sqrt{3} & \sqrt{2}+\sqrt{3}& \sqrt{2} \\\sqrt{2} &\sqrt{2}-\sqrt{3} & \sqrt{2}+\sqrt{3}\end{pmatrix}$$
