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I know that If $U$ is invertible, then the following are equivalent.

(i) $E$ and $U+E$ are idempotent;

(ii) $E(-U^{-1})E=E$ and $(I-E)U^{-1}(I-E)=I-E;$

(iii) $-EU^{-1}$ and $(I-E)U^{-1}$ are idempotent;

(iv) $-EU^{-1}$ and $U^{-1}-EU^{-1}$ are idempotent.

I have an invertible matrix $U=F-E$ and an idempotent matrix $E=\begin{pmatrix} 0 & 0 \\ 0 & I_{r} \end{pmatrix}$ for some $0<r<n$ and $F$ is and idempotent matrix. Using one of those equivalent propositions above I am suppose to show that $U^{-1}= \begin{pmatrix} I_{s} & C_{s\times r} \\ B_{r\times s} & -I_{r} \end{pmatrix}$ for some matrices $B$ and $C$.

The first thing I would like to do is to put $U^{-1}$ alone in one side of one those equations.

Any hint?

pshmath0
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1 Answers1

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Consider the following approach, let $U^{-1}$ be of the form \begin{equation} U^{-1} = \begin{pmatrix} A & C \\ B & D \end{pmatrix}. \end{equation} Bullet $(ii)$ implies directly that $D=-I_r$ and that $A=I_s$. To continue, bullet $(iii)$ and $(iv)$ are exactly the same, looking only at the first part of $(iii)$, from $-EU^{-1}$ being idempotent we obtain that there are no constraints on $B$. Similarly, from $(I-E)U^{-1}$ being idempotent, we do not obtain any further constraints on $C$. The trick is to use some parametric form of $U^{-1}$.

WalterJ
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