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Unformatted text preview: 1 Math 1b Practical Isometries March 04, 2011 Recall that an orthogonal matrix is a matrix square matrix U so that U > U = I (which holds if and only if UU > = I ). In other words, U is orthogonal if and only if its columns form an orthonormal basis of RR n , which is the case if and only if its rows form an orthonormal basis of RR n . [You might think that such a matrix should be called an orthonormal matrix, not just orthogonal. But it is by tradition called orthogonal. This is confusing, but nothing can be done about it.] Theorem 0. A linear mapping from RR n to itself is an isometry if and only if its matrix Q is an orthogonal matrix. Proof: Suppose first that Q is an orthogonal matrix. Then, for any vector x , || Q x || 2 = ( Q x ) > Q x = x > ( Q > Q ) x = x > I x = || x || 2 . So Q (or the corresponding linear mapping) preserves distance. Now assume that || Q x || = || x || for every vector x . Let u 1 , u 2 , . . . , u n be the standard basis (column) vectors. Then Q u i has the same length as u i , namely 1. But Q u i is the i-th column of Q . Thus every column of Q has length 1. Also, for i 6 = j , 2 = || u i + u j || 2 = || Q ( u i + u j ) || 2 = || Q ( u i ) + Q ( u j ) || 2 , and so 2 = h Q u i + Q u j , Q u i + Q u j i = h Q u i , Q u i i + h Q u i , Q u j i + h Q u j , Q u i i + h Q u j , Q u j i = 1 + 2 h Q u i , Q u j i + 1 ....
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