Can a symmetric matrix always be represented as the sum of a positive-definite and negative-definite matrix? I was wondering if it is possible to decompose any symmetric matrix into a positive definite and a negative definite component. I can't seem to think of a counterexample if the statement is false.
 A: Yes, see one of my questions with the details. I will type up some more:
Given $A$ such that $A = A^\top$, $A$ with both positive and negative eigenvalues, the LDU factorization will have $U=L^\top$ (follows directly from symmetry) and $D$ diagonal with both positive and negative values. So
$$A=L(D_p + D_n)L^\top$$
where $D$ is separated into the positive portion $D_p$ and the negative portion $D_n$. They have all positive or all negative values and zeros. Thus when  the matrix is decomposed as
\begin{align}
 A &= LD_pL^\top + LD_nL^\top \\
   &= P + N \\
\end{align}
it is separated with $P$ symmetric positive semidefinite, and $N$ symmetric negative semidefinite.
As was pointed out in the comments $0=-1 + 1$. Thus to obtain definiteness for both, do something to $D_p + D_n$ to make it happen while retaining the value of $D = D_p + D_n$.
A: This answer is years late but I like this proof. 
The set of positive definite matrices is open in the set of symmetric matrices and must thus contain $\frac{n(n+1)}{2}$ matrices which are linearly independent. Take that as a basis for the vector space of symmetric matrices. Then every element can be written as a linear combination of positive definite matrices, and we can simply group the basis elements by the positive coefficients and negative coefficients, giving us that a symmetric matrix M can be written as the sum of a positive definite matrix and the negation of a positive definite matrix.
A: If $X$ is symmetric then $X = (X + \lambda I) - \lambda I$. Since the eigenvalues of $X + \lambda I $ are $ \lambda_i + \lambda$ where $\lambda_i$'s are the eigenvalues of X we can find a positive $\lambda$ such that $(X + \lambda I)$ is positive definite.
