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Unformatted text preview: CS205 Homework #2 Solutions Problem 1 [Heath 3.29, page 152] Let v be a nonzero nvector. The hyperplane normal to v is the (n1)dimensional subspace of all vectors z such that v T z = . A reflector is a linear transformation R such that Rx = x if x is a scalar multiple of v , and Rx = x if v T x = . Thus, the hyperplane acts as a mirror: for any vector, its component within the hyperplane is invariant, whereas its component orthogonal to the hyperplane is reversed. 1. Show that R = 2P I , where P is the orthogonal projector onto the hyperplane normal to v . Draw a picture to illustrate this result 2. Show that R is symmetric and orthogonal 3. Show that the Householder transformation H = I 2 vv T v T v , is a reflector 4. Show that for any two vectors s and t such that s 6 = t and k s k 2 = k t k 2 , there is a reflector R such that Rs = t Solution 1. We can obtain the reflection Rx of a vector x with respect to a hyperplane through the origin by adding to x twice the vector from x to Px , where Px is the projection of x onto the same hyperplane (see figure 1). Thus Rx = x + 2 ( Px x ) = ( 2Px x ) = ( 2P I ) x Since this has to hold for all x we have R = 2P I . An alternative way to derive the same result is to observe that the projection Px lies halfway between x and its reflection Rx . Therefore 1 2 ( x + Rx ) = Px ⇒ Rx = ( 2Px x ) = ( 2P I ) x which leads to the same result. 2. A reflection with respect to a hyperplane through the origin does not change the magnitude of the reflected vector (see figure 1). Therefore we have k Rx k = k x k ⇒ x T R T Rx = x T x ⇒ x T ( R T R I ) x = 1 Figure 1: Reflector for any vector x . If we could show that x T ( R T R I ) x = implies R T R I = we would have proven the orthogonality of R . Furthermore, since reflecting a vector twice just gives the original vector we have R 2 = I . Therefore we would have R T R = I ⇒ R T R 2 = R ⇒ R T = R which shows that R is symmetric. In order to show that R T R I = it suffices to show that for a symmetric matrix C , x T Cx = for all x implies C = (since R T R I is symmetric). To show that, we note that e T i Ce j = C ij where e k is the kth column of the identity matrix. We have e T i Ce i = C ii = for any i and furthermore...
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This homework help was uploaded on 01/29/2008 for the course CS 205A taught by Professor Fedkiw during the Fall '07 term at Stanford.
 Fall '07
 Fedkiw

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