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On idempotent invertible matrices

Let A,B \in \mathcal{M}_n(\mathbb{C}) be two idempotent matrices such that A-B is invertible and let \alpha, \beta \in \mathbb{C}. Let \mathbb{I} \in \mathcal{M}_n(\mathbb{C}) be the identity matrix. Show that:

  1. if \alpha \notin \{0,-1\}  then \mathbb{I}+\alpha AB is not necessarily invertible.
  2. if \alpha \in \{0,-1\} then \mathbb{I}+\alpha AB is invertible.
  3. A+B-AB is invertible.
  4. if \alpha \beta \neq 0 then \alpha A+\beta B is invertible.

Solution

  1. Pick

        \[A=\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} \quad , \quad B=\begin{pmatrix} -\frac{1}{\alpha} & -\frac{1}{\alpha} \\ 1+\frac{1}{\alpha} & 1+\frac{1}{\alpha}\end{pmatrix}\]

    and note that A^2=A , B^2=B and A-B is invertible but \mathbb{I}+\alpha AB is not.

  2. It’s clear for \alpha =0. For \alpha =-1, suppose that ABv=v for some v \in \mathbb{C}^n. We need to show that v=0. We have ABv=A^2Bv=Av thus Bv-v \in \ker A. But since B^2=B we also have Bv -v \in \ker B and hence

        \[Bv -v \in \ker A \cap \ker B \subseteq \ker(A-B)=\{0 \}\]

    because A-B is invertible. So Bv=v and therefore Av=ABv=v. So v \in \ker(A-B) =\{0 \}.

  3. Let A'=\mathbb{I}-A, B'=\mathbb{I}-B. Since A',B' are idempotents we conclude by (ii) that \mathbb{I} - A'B' is invertible since A'-B'=-(A-B) is invertible. The result now follows because

        \[\mathbb{I}-A'B'=\mathbb{I}- \left(\mathbb{I}-A \right) \left(\mathbb{I}-B \right)=A+B-AB\]

  4. Let \gamma=-\frac{\beta}{\alpha} \neq 0 and suppose that Av=\gamma Bv for some v \in \mathbb{C}^n. We are done if we show that v=0. Well, we have BAv=\gamma Bv=Av=A^2v and thus (A-B)Av=0 implying that Av=0 because A-B is invertible. Hence \gamma Bv=Av=0 which gives Bv=0 because \gamma \neq 0. Thus (A-B)v=0 and therefore v=0.

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