A matrix decomposition Suppose $N$ is a symmetric matrix, then show that it can be uniquely decomposed as $N=N^+-N^-$, where $N^+$ and $N^-$ are both nonnegative-definite, i.e. their eigenvalue are non-negative, and $N^+N^-=0$. This decomposition is not hard to get, if we note that any symmetric matrix can be similar diagonalized.  But I don't know how to show the uniqueness.
 A: For the existence, N can be diagonalised, therefore there exists $M$ and $D$ such that
$ N = M^{-1} D M$, where $D$ is diagonal.
By introducting $D^+$ and $D^-$ such that 
\begin{equation}
   \left\{
   \begin{aligned}
      & D^+_{i,i} = (D_{i,i})^+ \\
      & D^-_{i,i} = (-D_{i,i})^+
   \end{aligned}
   \right.
\end{equation}
where $x \mapsto x^+$ is the positive part function, the pair $(M^{-1}D^+M, M^{-1}D^-M)$ is one possible decomposition of $A$. This is most likely what you had in mind.
In order to show uniqueness, think of $D^+$ and $D^-$ as the "positive part of $A$ seen through the basis change $M$" : if you can find another decomposition of $A$ exists, it should be in contradiction with the fact the related matrices are semi-definite positive (SDP).
Let us note $(N^+, N^-)$ the decomposition we exhibited above and consider another possible decomposition. It has to be of the following form : $(N^+ + E, N^- + E)$, where $E$ is SDP. Consequently
\begin{equation}
   (N^+ + E)(N^- + E) = N^+E + EN^- + E^2 = 0
\end{equation}
\begin{equation}
   E^2 = - N^+E - EN^-
\end{equation}
The left-hand side is SDP, while the right-hand side is SDN, therefore the equality can only hold if $E=0$.
