= Positive-pivot Givens QR factorization of an orthogonal matrix
{c}
{title2=$R=\operatorname{diag}(1,\ldots,1,\det A)$}
Use determinant-one <Givens rotations> and choose every nonzero eliminated pair to leave a positive pivot. If the input square <matrix> $A$ is orthogonal, its resulting <upper triangular> factor $R$ is orthogonal too, and hence diagonal with entries of absolute value one. Positivity fixes the first $n-1$ entries to one, while <determinant> preservation fixes the last to $\det A$. A negative last entry is unavoidable when the input reverses orientation.
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