Cholesky-type factorization of positive definite matrices

Question

Let $\boldsymbol{A}$ be an $n$-by-$n$ positive definite matrix. I'm interested in finding $\boldsymbol{X}$ such that $\boldsymbol{X}^T\boldsymbol{X}=\boldsymbol{A}$, subject to $\boldsymbol{X}$ being $n$-by-$n$ also. The Cholesky decomposition is one way of achieving this. Eigen-decomposition is another. However, I'm interested to know whether

(1) the number of solutions is finite

(2) and if so, is there a way to find all of them?

Thirdly, if we allow $\boldsymbol{X}$ to be $m$-by-$n$ where $m > n$, how do the answers to (1) and (2) differ? I presume in this case, the number of solutions is likely not finite, and if so, what are possible ways to find some of the solutions for given $m$ and $n$?

Many thanks for your help.

Answer 1

Question 1 : assume R being a rotation-matrix, thus being orthogonal, and $R^T = R^{-1}$ and thus $R^T \cdot R = I$ (where $I$ is the identity matrix). Then let $X = R \cdot Y $. Then $ X^T \cdot X = (Y^T \cdot R^T ) \cdot (R \cdot Y) = Y^T \cdot (R^T \cdot R) \cdot Y = Y ^T \cdot Y$ and any rotation matrix $R$ leaves this constant. So the number of solutions is infinite (even continuous/uncountable)

Answer 2

It is not unique. Note that for any $m\times m$ ($m\geq n$) Diagonal matrix $D$ with non-zero diagonal entries, $(X^TD)D^{-1}X=A$.