Show that any orthogonal matrix has determinant 1 or -1

Question

Hello fellow users of this forum:

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

Thanks

Hello fellow user, please search before asking. – Mar 03 '15 at 02:11 — , Mar 03 '15 at 02:11

score 23 · Accepted Answer · answered Mar 03 '15 at 02:14

23

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 $\det(Q) = \pm \sqrt{1} = \pm 1$.

answered Mar 03 '15 at 02:14

Xoque55

4,384

2

Very nice and neat proof. Thank you Xoque55 – AVP Mar 03 '15 at 04:05
To see why $x^2=1$ implies $x=\pm 1$, see this answer. – Filippo Jun 26 '23 at 08:37

Show that any orthogonal matrix has determinant 1 or -1

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