This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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 ith 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 ....
View
Full
Document
This note was uploaded on 11/10/2011 for the course MA 1B taught by Professor Aschbacher during the Winter '08 term at Caltech.
 Winter '08
 Aschbacher
 Math

Click to edit the document details