While studying LU decomposition from this book I came across the statement that pivoting in LU decomposition is not necessary in some cases, as for example when the matrix is symmetric positive definite.
Then what about symmetric negative definite matrices? I couldn't find a counter-example. My specific concern is for the symmetric definite negative matrix that arises from the discretization of the stiff equation, i.e. a matrix like
$$ \begin{pmatrix} -2 & 1 & 0 & \cdots & 0\\ 1 & -2 & 1 & \cdots & 0\\ 0 & 1 & -2 & \cdots & 0\\ \vdots & \vdots & \vdots & \ddots & \vdots\\ 0 & 0 & 0 & \cdots & -2 \end{pmatrix} $$
We can construct the LU factorization of a negative definite matrix $A$ by considering its positive definite counterpart $-A$. So $-A = LU$ (since $-A$ is positive definite) and hence $A = L(-U)$.
Alternatively, you can also solve $-Ax = -b$ directly, which is equivalent to the original question.