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Hello fellow users of this forum:

Show that for any orthogonal matrix Q, either det(Q)=1 or -1.

Thanks

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    $\begingroup$ Hello fellow user, please search before asking. $\endgroup$
    – user147263
    Mar 3, 2015 at 2:11

1 Answer 1

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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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    $\begingroup$ Very nice and neat proof. Thank you Xoque55 $\endgroup$
    – AVP
    Mar 3, 2015 at 4:05

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