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.
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$\begingroup$ What if your symmetric matrix is not invertible? $\endgroup$ – Chris Godsil Jan 10 '13 at 19:40
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2$\begingroup$ Replace definite with semi-definite and the answer is yes (or if the original matrix is non-singular then all is definite) $\endgroup$ – adam W Jan 10 '13 at 19:42
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2$\begingroup$ Why does it matter $ 0 = 1 + (- 1) $. $\endgroup$ – Arin Chaudhuri Jan 10 '13 at 19:51
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1$\begingroup$ Intuition: Sylvester's Law of Inertia implies there always exists a basis in which a Real symmetric nondegenerate matrix is diagonal with $p$ ones and $q$ $-1$'s on the diagonal; $(p,q)$ is its signature and is a property of the matrix as an endomorphism of a vector space (that is, it is the same for all such bases). The matrix with the $-1$'s replaced by zeros is obviously positive semidefinite and the matrix with the $1$'s replaced by zeros is obviously negative semidefinite; their sum is the original matrix. $\endgroup$ – whuber Jan 10 '13 at 20:21
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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 $D = D_p + D_n$.
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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.
answered Jan 10 '13 at 19:46
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$\begingroup$ +1 This is absolutely the simplest answer, $X+\lambda I$ is positive definite and $-\lambda I$ is negative definite for some $\lambda$ $\endgroup$ – adam W Jan 13 '13 at 23:41
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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.
