Does exists a matrix $B$ such that $A^TA=A^TB+B^TA$? with $B^TB$ being a diagonal matrix and $A$ an incidence matrix $A$ is a incidence matrix for some undirected graph.
$A^TA$ is a positive definite matrix, so I know that we can factorize it as $A^TA = C + C^T$
There exists always a matrix $C$ such that $C = A^TB$?
Satisfying the next requirement, $B^TB = cI$ is a diagonal matrix, being $I$ the identity matrix and $c$ a scalar, and $B$ also is related to the incidence matrix $A$ in the sense that for each edge, we select one node.
$A$ might not be square (more edges than nodes).
For instance,
$
A = \begin{bmatrix}1 & 0 & -1\\-1 & 1 & 0\\0 &-1 & 1\end{bmatrix} \quad
B = \begin{bmatrix}0 & 0 & -1 \\ -1 & 0 & 0 \\ 0 & -1 & 0\end{bmatrix},
$
satisfies $A^TA=A^TB+B^TA$ and $B^TB = I$
Note that here I have taken for $B$ all the $-1$ from $A$.
Can I find such $B$ for instance for complete undirected graphs? what about for more general undirected grahps?
 A: If you think of this in terms of columns vectors, write $A = \begin{bmatrix} v_1 & \dots v_n \end{bmatrix}$ and $B = [ b_1 /2, 
\dots, b_n/2 ]$ (the one-half factor is just to normalize stuff), so that
$$
A^{\top}A = [(v_i \, | \, v_j)], \qquad A^{\top} B + B^{\top} A = [(v_i \, | \, b_j/2)] + [ (b_j/2 \, | \, v_i) ] = [(v_i \, | \, b_j)].
$$
where $( \cdot \, | \, \cdot )$ denotes inner product. So you want to find $n$ column vectors $b_1,\dots, b_n$ ($n$ is the number of edges in your graph, the vectors lie in $\mathbb R^k$ where $k$ is the number of vertices) such that 
$$
(v_i \, | \, v_j) = (v_i \, | \, b_j).
$$
In particular, if $k \le n$ and $\{ v_1, \dots, v_k \}$ forms a basis of $\mathbb R^k$, your vectors $b_j$ are completely determined by the first $k$ vectors, i.e. any set of edges which gives rise to $k$ linearly independent vectors in $\mathbb R^k$ (perhaps there is a graph-theoretic interpretation to such a constraint) determines if there exists such a matrix $B$, and what it is if it exists. 
This puts a lot of constraints on the possible graphs which would acheive this, so I don't expect you to be able to find such a matrix $B$ for every graph if it has more edges than vertices in general. Otherwise these equations tell you how to compute the vectors $b_j$ ; it reduces to solve a linear system given your matrix $A$. 
Hope that helps,
