Is square root of a matrix necessarily Cholesky decomposition? Or other decomposition satisfies?

Question

If $a$ is a scalar, $a = \sqrt{a}\sqrt{a}$, then if A is a matrix, $A=\sqrt{A}\sqrt{A}$, now I think the square root operator is applied to the matrix instead of a scalar. Is square root of a matrix necessarily Cholesky decomposition? Or other decomposition satisfies, like QR decomposition? Because $A=QR$ where Q and R are two terms like $\sqrt{A}\sqrt{A}$. Or does the square roots have to satisfy the condition that one is the other's conjugate transpose as in Cholesky decomposition?

Answer 1

If $B$ can at all be called a square root of $A$, then $B^2=A$, whether we are taking about matrices or anything else. For matrices, there can be infinitely many distinct square roots (even for the zero matrix), but they all satisfy the same equation and no other conditions are necessary.