Hello fellow users of this forum:
Show that for any orthogonal matrix Q, either det(Q)=1 or -1.
Thanks
$\endgroup$ 11 Answer
$\begingroup$Not sure what's wrong with using the transpose, but here it goes.
Since $Q$ is orthogonal, $QQ^T = I = Q^TQ$ by definition. Using the fact that $\det(AB) = \det(A) \det(B)$, we have $\det(I) = 1 = \det(QQ^T) = \det(Q) \det(Q^T) = \det(Q) \det(Q) = [\det(Q)]^2$.
Since we have $[\det(Q)]^2 = 1$, then $\det(Q) = \pm \sqrt{1} = \pm 1$.
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