How can I most simply show, without referring to advanced theorems, that the Schur complement is better-conditioned than the original SPD matrix itself?
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Rodrigo de Azevedo
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HBHSU
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2Not sure which theorems are qualified as advanced, but you may see https://math.stackexchange.com/q/2749464 for a proof. – user1551 Feb 05 '19 at 05:24
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I would like to avoid using the Courant-Fischer principle. – HBHSU Feb 05 '19 at 05:25
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1It's always there in one way or [another](https://math.stackexchange.com/questions/1994633/spektre-schur-complement). – Algebraic Pavel Feb 06 '19 at 17:50