I'm trying to show that \begin{equation} P^H ( I_M + PBP^H) ^{-1} P = \big( (P^H P)^{-1} + B \big)^{-1}, \end{equation} where $P$ is an $M$-by-$N$ matrix, $I_M$ is the $M$-by-$M$ identity matrix, $B$ is an $N$-by-$N$ matrix, and $(P^H P)$ is invertible.

I've used various versions of matrix inversion lemmas, but I'm stuck.

How can the above equality be shown?

  • $\begingroup$ Have you tried some low-dimensional cases with explicit values for $M$ and $N$? $\endgroup$ – Deane Jan 3 '14 at 18:16
  • $\begingroup$ I verified that equation by Matlab, and it seems to be true for arbitrary integers M and N. Alsi, it can be verified mathematically with Ardakov's answer below. $\endgroup$ – user45020 Jan 3 '14 at 20:53

Let $X = P^H(I + PBP^H)^{-1}P$ and $Y = (P^HP)^{-1} + B$. Since $X$ and $Y$ are both square matrices of the same size ($N$-by-$N$), it's enough to show that $XY = I$ say.

Now $Y = (I + BP^H P)(P^HP)^{-1}$ and $P(BP^HP) = (PBP^H)P$. So

$XY = P^H ( I + PBP^H)^{-1}P (I + BP^HP)(P^HP)^{-1} = P^H(I + PBP^H)^{-1}(I + PBP^H)P (P^HP)^{-1} = P^HP (P^HP)^{-1} =I.$

  • $\begingroup$ Thanks! Even though a matrix inversion lemma is not used, this verification may be enough :) $\endgroup$ – user45020 Jan 3 '14 at 20:52
  • 2
    $\begingroup$ I don't see why this "may be enough" as it answers the question you asked. $\endgroup$ – Terry Loring Jan 4 '14 at 16:40
  • $\begingroup$ not 'may be enough', but 'definitely enough'. $\endgroup$ – user45020 Jan 5 '14 at 18:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.