Suppose the following matrix $A$ is given $\begin{pmatrix} 0.500 & -0.333 & -0.167\\ -0.500 & 0.667 & -0.167\\ -0.500 & -0.333 & 0.833\end{pmatrix}$ with its transpose $A^T$. The product $A^TA=G$ yields $\begin{pmatrix} 0.750 & -0.334 & -0.417\\ -0.334 & 0.667 & -0.333\\ -0.417 & -0.333 & 0.750\end{pmatrix},$
where $G$ is a Laplacian matrix. Note that matrices $A$ and $G$ are of rank 2, with the zero eigenvalue corresponding to eigenvector $1_n=[1~ 1 ~1]^T$.
I wonder what would be the way to obtain $A$ if only $G$ is given. I tried eigendecomposition $G=UEU^T$, and then set A'=UE^{1/2}, but obtained different result. I guess this has to do with rank deficiency. Could someone explain this? Clearly, the above example is for illustration; you could consider general Laplacian matrix decomposition of the above form.
Since, for instance, Cholesky decomposition could be used to find $G=LL^T$, the decomposition on $G$ could yield many solution. I'm interested in the solution that could be expressed as $A=(I-1_nw),$ where $I$ is a $3\times 3$ identity matrix, $1_n=[1~ 1~ 1]$, and $w$ being some vector satisfying $w^T1_n=1$. If it simplifies matters, you could assume that the entries of $w$ are non-negative.