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Fun with idempotent matrices

Let $$A$$ and $$B$$ be $$n\times n$$ commuting, idempotent matrices such that $$A - B$$ is invertible. Prove that $$A + B$$ is the $$n\times n$$ identity matrix.

(Mathematics Magazine Problem Q987 (2009).)

Note by Scott Kominers
3 years, 10 months ago

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Start \begin{align}\left(\mathbf{A}+\mathbf{B}\right)\left(\mathbf{A}-\mathbf{B}\right) &= \mathbf{A}^2-\mathbf{B}^2-\mathbf{A}\mathbf{B}+\mathbf{B}\mathbf{A}\\ &=\mathbf{A}^2-\mathbf{B}^2+\left[\mathbf{B},\mathbf{A}\right]\end{align}

That $$\mathbf{A}$$ commutes with $$\mathbf{B}$$ is to say that $$\left[\mathbf{B},\mathbf{A}\right]=\mathbf{B}\mathbf{A} - \mathbf{A}\mathbf{B}=0$$.

Idempotency is to say $$\mathbf{A}^2=\mathbf{A}$$ and likewise for $$\mathbf{B}$$, so

$\left(\mathbf{A}+\mathbf{B}\right)\left(\mathbf{A}-\mathbf{B}\right) =\mathbf{A}-\mathbf{B}$

Acting with the right inverse of $$\mathbf{A}-\mathbf{B}$$, which is stated to exist, we have

\begin{align}\left(\mathbf{A}+\mathbf{B}\right)\left(\mathbf{A}-\mathbf{B}\right)\left(\mathbf{A}-\mathbf{B}\right)^{-1} &= \left(\mathbf{A}-\mathbf{B}\right)\left(\mathbf{A}-\mathbf{B}\right)^{-1} \\ \left(\mathbf{A}+\mathbf{B}\right) &= \mathbb{1}\end{align}

Staff - 3 years, 10 months ago