Let $A$ be $N\times N$ symmetric positive definite, and let $U$ be $N\times (N-1)$ with full column rank, i.e. $U^\top U$ invertible. $U^\top A^{-1} U$ is invertible according to this answer, but is there any way to relate $(U^\top A^{-1} U)^{-1}$ to $A$ ? I tried the Sherman-Morrison formula, but I did not manage there.

  • $\begingroup$ Just because $U^TU$ and $A$ are invertible doesn't mean $U^TAU$ is. Are you assuming that it is the case? $\endgroup$ – Arnaud D. Jan 6 '17 at 18:16
  • $\begingroup$ yes we have to assume this. I'll update the post, tx! $\endgroup$ – yannick Jan 6 '17 at 18:26
  • $\begingroup$ One of the few rules I know to open up inverses with tall dimensions inside is $(X Y)^+ = (X^+ X Y)^+ (X Y Y^+)^+$, where $(\cdot)^+$ denotes the pseudo-inverse. I'm not very optimistic it will help you much but you could try (apply it twice). $\endgroup$ – Florian Jan 6 '17 at 20:00

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