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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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  • $\begingroup$ If $A$ is neg. def., then $-A$ is pos. def., can you get the LU factorization of $A$ from the LU factorization of $-A$? $\endgroup$ – Nigel Overmars Feb 10 '17 at 14:21
  • $\begingroup$ Yes indeed, but wouldn't the linear system then have a different solution? I know this is a different question though. $\endgroup$ – Eugenio Feb 10 '17 at 14:23
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    $\begingroup$ Well, you can always solve $-Ax=-b$ instead of $Ax = b$. $\endgroup$ – Nigel Overmars Feb 10 '17 at 14:27
  • $\begingroup$ Haha you're right it's simple. Do you mind posting both your comments as an answer? I'll accept it $\endgroup$ – Eugenio Feb 10 '17 at 14:49
  • $\begingroup$ a bit late, but done! $\endgroup$ – Nigel Overmars Feb 14 '17 at 16:49
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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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  • $\begingroup$ Fix the sign of b? $\endgroup$ – Eugenio Feb 14 '17 at 16:51
  • $\begingroup$ @Eugenio Sorry, just came back from vacation and had a rough trip home. Should be fixed now. $\endgroup$ – Nigel Overmars Feb 14 '17 at 16:52

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