Finding $X$ When $Y=XX'$ Consider matrices $Y\in\mathbb{R}^{n\times n}$
 and $X\in\mathbb{R}^{n\times m}$
 where $m\geq n$. $X$ is unknown but $Y=XX'$, which implies that $Y$ is positive definite (I see no reason why this couldn't alternately be expressed as a positive semi-definite problem with $Y=X'X$, a different $Y\in\mathbb{R}^{m\times m}$ would still be known). 
What the easiest method to find $X$? I was thinking of minimizing the Frobenius norm, but wasn't sure if there was some relatively straightforward thing that I'm missing. 

Consider $X = I$, then $X' = I$ and $Y = I$. On the other hand, if $X = -I$, then $X' = -I$, but $Y = I$ still. Are there any other assumptions you can make about $X$?
In fact, for any $X$, $(-X)(-X)' = XX'$, so there are always at least two solutions.
Additionally, if $$X_1 = \left(\begin{array}{cc} a_1 & b_1 \\ 0 & 0\end{array}\right),\quad X_2 = \left(\begin{array}{cc} a_2 & b_2 \\ 0 & 0\end{array}\right)$$
with $a_1^2 + b_1^2 = a_2^2 + b_2^2$ then $Y = X_1 X_1' = X_2 X_2'$.
I think maybe something needs to be said about the singular values of $X$ to get better uniqueness - maybe that they're all non-degenerate?
