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College Algebra Exam Review 223

# College Algebra Exam Review 223 - .² sin.² cos.² 3 5 so...

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4.4. LINEAR ISOMETRIES 233 Proof. Let A 2 SO . 3 ; R / , let be the linear isometry x 7! A x , and let v be an eigenvector with eigenvalue ˙ 1 . If the eigenvalue is C 1 , there is nothing to do. So suppose the eigenvalue is 1 . The plane P D P v orthogonal to v is invariant under A , because if x 2 P , then h v ; A x i D h A v ; A x i D h v ; x i D 0 . The restriction of to P is also orthogonal, and since 1 D det . / D . 1/. det . j P / , j P must be a reflection. But a reflection has an eigenvalue of C 1 , so in any case A has an eigenvalue of C 1 . n Proposition 4.4.13. An element A 2 O . 3 ; R / has determinant 1 if and only if A implements a rotation. Proof. Suppose A 2 SO . 3 ; R / . Let denote the corresponding linear isometry x 7! A x . By the lemma, A has an eigenvector v with eigen- value 1 . The plane P orthogonal to v is invariant under , and det . j P / D det . / D 1 , so j P is a rotation of P . Hence, is a rotation about the line spanned by v . On the other hand, if is a rotation, then the matrix of with respect to an appropriate orthonormal basis has the form 2 4 1 0 0 0 cos . /
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Unformatted text preview: .²/ sin .²/ cos .²/ 3 5 ; so ± has determinant 1. n Proposition 4.4.14. An element of O . 3 ; R / n SO . 3 ; R / implements either an orthogonal reﬂection, or a reﬂection-rotation, that is, the product of a reﬂection j ˛ and a rotation about the line spanned by ˛ . Proof. Suppose A 2 O . 3 ; R / n SO . 3 ; R / . Let ± denote the corresponding linear isometry x 7! A x . Let v be an eigenvector of A with eigenvalue ˙ 1 . If the eigenvalue is 1 , then the restriction of ± to the plane P orthogo-nal to v has determinant ± 1 , so is a reﬂection. Then ± itself is a reﬂection. If the eigenvalue is ± 1 , then the restriction of ± to P has determinant 1 , so is a rotation. In this case ± is the product of the reﬂection j v and a rotation about the line spanned by v . n...
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