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Knowing that $C$ is a member of $M_{m\times n}(F)$, $D$ is a member of $M_{n\times m}(F)$, and assuming that $I_m + CD$ is invertible, how would you show that $$D(I_m + CD)^{−1} = (I_n + DC)^{−1}D$$

Since I'm given information about non-square matrices, but I am dealing with square matrices I'm not sure what to do. My first thoughts were that I could change $I_m + CD$ to be $C(C^{-1} + D)$, however, since $C$ is part of a $m\times n$ matrix, I wasn't sure if I would be able to apply this. I would greatly appreciate help with solving this and understanding matrices better!

Paulo Estêvão
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ms121399
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  • $C^{-1}$ is not assumed to exist. It even cannot exist if $m\ne n$. – Anne Bauval Feb 08 '24 at 18:35
  • Your question is a multiduplicate: https://approach0.xyz/search/?q=AND%20site%3Amath.stackexchange.com%2C%20OR%20content%3A%24D(I%2BCD)%5E%7B-1%7D%3D(I%2BDC)%5E%7B-1%7DD%24&p=1 – Anne Bauval Feb 08 '24 at 18:38
  • Multiply on the right by $(I_m+CD)$. You are allowed to do this since it is invertible. – Andrei Feb 08 '24 at 18:40
  • Are we sure that the invertibility of $I+CD$ implies that of $I+DC$? @Andrei and ms121399 (the posts listed above via Approach$0$ seem to all assume both are invertible). – Anne Bauval Feb 08 '24 at 18:46
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    @AnneBauval -- they have the same determinants, ref https://math.stackexchange.com/questions/17831/how-can-we-prove-sylvesters-determinant-identity – user8675309 Feb 08 '24 at 19:08
  • What would be the meaning of $(I_n+DC)^{-1}$ otherwise? – Andrei Feb 08 '24 at 19:09
  • @user8675309 Great! (I think I knew but forgot, loathsome old age...) – Anne Bauval Feb 08 '24 at 19:12
  • @Andrei The meaning is: first (difficult) prove that it exists, and then check (easy) the equality. – Anne Bauval Feb 08 '24 at 19:13

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