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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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What if your symmetric matrix is not invertible? – Chris Godsil Jan 10 '13 at 19:40
Replace definite with semi-definite and the answer is yes (or if the original matrix is non-singular then all is definite) – adam W Jan 10 '13 at 19:42
Why does it matter $ 0 = 1 + (- 1) $. – Arin Chaudhuri Jan 10 '13 at 19:51
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. – whuber Jan 10 '13 at 20:21
up vote 6 down vote accepted

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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Great answer, thanks! – Erik M Jan 11 '13 at 1:22

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.

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+1 This is absolutely the simplest answer, $X+\lambda I$ is positive definite and $-\lambda I$ is negative definite for some $\lambda$ – adam W Jan 13 '13 at 23:41

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