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College Algebra Exam Review 221

College Algebra Exam Review 221 - U is an isometry The...

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4.4. LINEAR ISOMETRIES 231 v i and v j is the same as the inner product of the i th and j th rows of A . Hence, the v i ’s are an orthonormal basis if and only if the rows of A are an orthonormal basis. By Exercise 4.4.3 , this is true if and only if A is orthogonal. n Theorem 4.4.7. Let be a linear map on R n . The following are equiva- lent: (a) is an isometry. (b) preserves inner products. (c) For some orthonormal basis f f 1 ; : : : ; f n g of R n , the set f . f 1 /; : : : ; . f n / g is also orthonormal. (d) For every orthonormal basis f f 1 ; : : : ; f n g of R n , the set f . f 1 /; : : : ; . f n / g is also orthonormal. (e) The matrix of with respect to some orthonormal basis is or- thogonal. (f) The matrix of with respect to every orthonormal basis is or- thogonal. Proof. Condition (a) implies (b) by Lemma 4.4.2 . The implications (b) ) (d) ) (c) are trivial, and (c) implies (e) by Lemma 4.4.6 . Now assume the matrix A of with respect to the orthonormal basis F is orthogonal. Let U. x / be the coordinate vector of x with respect to F ; then by Remark
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Unformatted text preview: , U is an isometry. The linear map ± is U ± 1 ı M A ı U , where M A means multiplication by A . By Lemma 4.4.5 , M A is an isometry, so ± is an isometry. Thus (e) H) (a). Similarly, we have (a) H) (b) H) (d) H) (f) H) (a). n Proposition 4.4.8. The determinant of an orthogonal matrix is ˙ 1 . Proof. Let A be an orthogonal matrix. Since det .A/ D det .A t / , we have 1 D det .E/ D det .A t A/ D det .A t / det .A/ D det .A/ 2 : n Remark 4.4.9. If ± is a linear transformation of R n and A and B are the matrices of ± with respect to two different bases of R n , then det .A/ D det .B/ , because A and B are related by a similarity, A D VBV ± 1 , where V is a change of basis matrix. Therefore, we can, without ambiguity, de-fine the determinant of ± to be the determinant of the matrix of ± with respect to any basis....
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