| Vincent Rabaud
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10-07-2004 04:09 AM ET (US)
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cool paper. A bit long, but half of it is just basics of linear algebra so that is fine. So, this seems cool to solve Ax=b when A is symmetric positive definite, but what is done when it is not the case ? Is there a way of modifying a bit the method like it was done in this fantastic talk on tuesday on LM ? (like you "add" enough to A to have it symmetric positive definite; you find a solution; then you do something with the original A). Or maybe something can only be done if the matrix is purely symmetric, purely positive... Briefly: how generalizable is it ? (I've heard of biconjugate gradient method but I am not sure of its application)
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