Matrix factorization to get the right solution 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.
 A: One looks for $A=I-1_3w^T$ such that $G=A^TA$, with $1_3=[1\ 1\ 1]^T$ and $w=[w_1\ w_2\ w_3]^T$ such that $w^T1_3=1$. Using $1_3^T1_3=3$, this yields
$$
G=(I-w1_3^T)(I-1_3w^T)=I-w1_3^T-1_3w^T+3ww^T,
$$ 
that is, for every $i\ne j$,
$$
G_{ii}=1-2w_i+3w_i^2,\quad G_{ij}=-w_i-w_j+3w_iw_j.
$$
Introducing $v=3w-1_3$, one gets $v^T1_3=0$ and, for every $i\ne j$,
$$
v_i^2=3G_{ii}-2,\quad v_iv_j=3G_{ij}+1.\tag{$\ast$}
$$
One sees that, if a solution $A=I-1_3w^T$ exists, then $G_{ii}\geqslant\frac23$ for every $i$, and, for every $i\ne j$,
$$
(G_{ij}+\tfrac13)^2=(G_{ii}-\tfrac23)(G_{jj}-\tfrac23).
$$
In the present case, these conditions are met and the system of equations $(\ast)$ reduces to $v_1^2=v_3^2=\frac14$, $v_2=0$ and $v_1v_3=-\frac14$, hence $v_1=\pm\frac12$, $v_2=0$, $v_3=\mp\frac12$. 
Since $w_i=\frac13(1+v_i)$ for every $i$, the solutions are $w=[\frac12\ \frac13\ \frac16]^T$ and $w=[\frac16\ \frac13\ \frac12]^T$.
In dimension $n$, the same technique applies, which yields the conditions
$$
nG_{ii}\geqslant n-1,\qquad
(nG_{ij}+1)^2=(nG_{ii}-n+1)(nG_{jj}-n+1),
$$
and the solutions
$$
w_i=\tfrac1n\left(1\pm\sqrt{nG_{ii}-n+1}\right).
$$
