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Suppose I have the LU factorization for a given matrix $A$ ($A$ is not symmetric positive definite), then is there a faster way to solve for $x$ in the following equation, as compared to doing LU all over again?

$$ (I - A)x = b $$

where $I$ is the identity matrix.

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    $\begingroup$ $I-A$ need not be invertible to start with $\endgroup$ – user251257 Dec 29 '16 at 16:52
  • $\begingroup$ @user251257 I am sorry, can you please elaborate. $\endgroup$ – Sandeep Dec 29 '16 at 20:01
  • $\begingroup$ If $1$ is an eigenvalue of $A$, then $I-A$ is not invertible and the system $(I-A) x = b$ has no solution. That doesn't say that it isn't possible to solve $(I-A) x= b$ using a $LU$ factorization of $A$, it is just likely not trivial... $\endgroup$ – user251257 Dec 30 '16 at 3:44

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