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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

$$ \begin{pmatrix} -2 & 1 & 0 & \cdots & 0\\ 1 & -2 & 1 & \cdots & 0\\ 0 & 1 & -2 & \cdots & 0\\ \vdots & \vdots & \vdots & \ddots & \vdots\\ 0 & 0 & 0 & \cdots & -2 \end{pmatrix} $$

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    If $A$ is neg. def., then $-A$ is pos. def., can you get the LU factorization of $A$ from the LU factorization of $-A$? – 2017-02-10
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    Yes indeed, but wouldn't the linear system then have a different solution? I know this is a different question though. – 2017-02-10
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    Well, you can always solve $-Ax=-b$ instead of $Ax = b$. – 2017-02-10
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    Haha you're right it's simple. Do you mind posting both your comments as an answer? I'll accept it – 2017-02-10
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    a bit late, but done! – 2017-02-14

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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.

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    Fix the sign of b? – 2017-02-14
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    @Eugenio Sorry, just came back from vacation and had a rough trip home. Should be fixed now. – 2017-02-14