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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, 7 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

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

Josh Silverman Staff - 3 years, 7 months ago

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