Decomposing a symmetric matrix as a sum of nilpotent matrices Assume that a real-valued symmetric matrix $M$ with trace zero can be written as
$$
M = A + A^T,
$$
with $A^2=0$. 

Given that $M$ is known, how (if possible) can $A$ be found? 

The diagonal elements of $A$ are just half those of $M$, that is
$$
A_{ii}=M_{ii}/2.
$$
But for the non diagonal ones, considering only the decomposition above, the number of unknowns is the double of the number of equations:
$$
M_{ij}=A_{ij}+A_{ji},
$$
with $i\neq j$, implying that is not possible to get an unique solution. 
Using the nilpotency property of $A$, nonlinear equations such as $M^2=AA^T+A^TA$ or $A^TMA=0$ can be generated, but I can't see how using these nonlinear relations can lead to an unique solution.
 A: I just read the user1551's solution (that is perfect as usual). We can also say 
since $S,K^2$ commute, we may assume (up to an orthogonal change of basis) that $S=diag((\lambda_i)),K^2=diag((-\lambda_i^2))$.
Since $K,S^2$ commute, we may assume (up to a permutation of the new basis) that $S^2=diag(\lambda_1^2 I_{i_1},\cdots,\lambda_k^2 I_{i_k})$, where the $(\lambda_i)$ are distinct  and $K=diag(K_1,\cdots,K_k)$, where $K_i$ is skew with the correct dimension. After, it's not difficult to conclude (as user1551) that $S=Pdiag(S_1,\cdots,S_k,0_{n-2k})P^T$ where $P=[q_1,q'_1\cdots,q_k,q'_k,\cdots]$ is orthogonal and $S_i=\begin{pmatrix}\lambda_i&0\\0&-\lambda_i\end{pmatrix}$ (where the $(\lambda_i)$ are not necessarily distinct).
Then, a particular solution is $K=Pdiag(L_1,\cdots,L_k,0_{n-2k})P^T$ where $L_i=\begin{pmatrix}0&\lambda_i\\-\lambda_i&0\end{pmatrix}$.
In other words, if $S=\sum_{i=1}^k\lambda_i(q_i{q_i}^T-q'_i{q'_i}^T)$, then a particular solution is $K=\sum_{i=1}^k \lambda_i (q_i{q'_i}^T-q'_i{q_i}^T)$.
