KindiakMath

A Surprising Determinant

Problem 1. Let \mathbf{A} be a real square matrix. Suppose \mathbf{A} \mathbf{A}^{\mathrm T} = \mathbf{I} and \det(\mathbf{A}) < 0. Compute \det(\mathbf{A} + \mathbf{I}).

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Solution. Since \mathbf{A}\mathbf{A}^{\mathrm T} = \mathbf{I}, we have

\begin{aligned} \mathbf{A} + \mathbf{I} &= \mathbf{A} + \mathbf{A}\mathbf{A}^{\mathrm T} \\ &= \mathbf{A}(\mathbf{I} + \mathbf{A}^{\mathrm T}) \\ &= \mathbf{A}(\mathbf{A} + \mathbf{I}^{\mathrm T}) \\ &= \mathbf{A}(\mathbf{A}^{\mathrm T} + \mathbf{I}^{\mathrm T}) \\ &= \mathbf{A}(\mathbf{A} + \mathbf{I})^{\mathrm T}. \end{aligned}

Taking determinants on both sides,

\begin{aligned} \det(\mathbf{A} + \mathbf{I}) &= \det (\mathbf{A} (\mathbf{A} + \mathbf{I})^{\mathrm T}) \\ &= \det (\mathbf{A}) \det((\mathbf{A} + \mathbf{I})^{\mathrm T}) \\ &= \det (\mathbf{A}) \det(\mathbf{A} + \mathbf{I}). \end{aligned}

Therefore, \det(\mathbf{A} + \mathbf{I}) ( 1 - \det(\mathbf{A}) ) = 0. Since \det(\mathbf{A}) < 0, we have

1 - \det(\mathbf{A}) > 1 - 0 = 1 > 0 \quad \Rightarrow \quad \det(\mathbf{A} + \mathbf{I}) = 0.

—Joel Kindiak, 9 Feb 25, 1322H

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