Question #61931

Show that an orthogonal map of the plane is either a reflection, or a rotation.
1

Expert's answer

2016-09-13T09:34:03-0400

Answer on Question #61931 – Math – Linear Algebra

Question

Show that an orthogonal map of the plane is either a reflection, or a rotation.

Solution

If T ⁣:R2R2T \colon R^2 \to R^2 is an orthogonal linear operator, then the standard matrix for TT can be expressed in the form


Rθ=[cosθsinθsinθcosθ]R_{\theta} = \begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix}


(RθR_{\theta} rotates vectors by θ\theta radians, counterclockwise)

or


Hθ2=[cosθsinθsinθcosθ].H_{\frac{\theta}{2}} = \begin{bmatrix} \cos \theta & \sin \theta \\ \sin \theta & -\cos \theta \end{bmatrix}.


The determinant can be used to distinguish between the two cases:


detRθ=cos2θ+sin2θ=1\det R_{\theta} = \cos^2 \theta + \sin^2 \theta = 1detHθ2=(cos2θ+sin2θ)=1.\det H_{\frac{\theta}{2}} = -(\cos^2 \theta + \sin^2 \theta) = -1.


Thus, a 2×22 \times 2 orthogonal matrix represents a rotation about the origin if det=1\det = 1 and a reflection about a line through the origin if det=1\det = -1. This sentence may be a definition of an orthogonal map of plane. If the definition of 2×22 \times 2 matrix is ATA=IA^T A = I, where II is an identity matrix, then the determinant of an orthogonal matrix is equal to 1 or 1-1, because using properties of determinants and equality ATA=IA^T A = I, obtain det(A)=det(AT)\det(A) = \det(A^T) and


1=det(I)=det(ATA)=det(AT)det(A)=(det(A))2, hence det(A)=1 or det(A)=1.1 = \det(I) = \det(A^T A) = \det(A^T) \det(A) = (\det(A))^2, \text{ hence } \det(A) = 1 \text{ or } \det(A) = -1.


Lemma. Every orthogonal transformation of the Euclidean plane is a reflection or a rotation.

We consider the effect of an orthogonal transformation TT on the two vectors α=(1,0)\alpha = (1,0) and

β=(0,1)\beta = (0,1). The orthogonality of TT forces T(α)T(\alpha) and T(β)T(\beta) also to be of unit length and orthogonal to each other. Let θ\theta and θ^\hat{\theta} be the counter-clockwise angles formed by T(α)T(\alpha) and T(β)T(\beta) with the xx-axis. Then θ^=θ±π2\hat{\theta} = \theta \pm \frac{\pi}{2}. In the case θ^=θ+π2\hat{\theta} = \theta + \frac{\pi}{2}, TT is a rotation through angle θ\theta. In the case θ^=θπ2\hat{\theta} = \theta - \frac{\pi}{2}, TT is a reflection through the dashed line forming an angle of θ2\frac{\theta}{2} with the xx-axis. In the pictures we have



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