Timeline for Equivalence of linear systems, solving one instead of the other

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Sep 22, 2012 at 23:58	comment	added	Geoff Oxberry		You're right that if $L$ is positive definite, the residual assumption maps nicely to a least squares problem. However, in applications (for example, in combustion, among other others), $L$ may not be positive definite (or even symmetric), and these methods are still used anyway. "Sensible" choices for $A$ are still a matter of debate (and consternation) in parts of the model reduction community. Singularity of $A^{T}LA$ depends on the application; for ODEs, the matrix $I - A^{T}LA$ is more meaningful for implicit methods, which is one reason I sidestepped the issue. One can only say so much.
Sep 22, 2012 at 19:54	comment	added	Stefano M		Nice answer, but you fail to mention that in order to guarantee that $A^TLA$ is non-singular you have to assume (a) that $L$ is positive definite or (b) that the columns of $A$ are chosen in a sensible way. If $L$ is pos. def., then $\min_y \frac12 y^TLy -y^Tb$ nicely maps to $\min_x \frac12 x^TA^TLAx - x^TA^Tb$, justifying the request of the orthogonality of the residuum to $A^T$.
Sep 22, 2012 at 18:06	comment	added	usero		Your elaboration addressed the important assumption on $A$ that make my proposal meaningful. Thanks.
Sep 22, 2012 at 12:50	vote	accept	usero
Sep 22, 2012 at 10:02	history	answered	Geoff Oxberry	CC BY-SA 3.0