Solution05_121 - Solutions to PS 5 (Math 121) J. Scholtz...

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Solutions to PS 5 (Math 121) J. Scholtz November 15, 2006 Question 1: 2.5/7 in R 2 let L be a line y = mx , m 6 = 0. Find an expression for the following T ( x,y )’s. First let us figure out a change of basis matrix from the standard basis to the basis a = { ( m, 1) , ( - 1 ,m ) } , because that is the basis with a vector along the line and a vector perpendicular to the line. We need to express the standard basis vectors in terms of the new basis vectors: (1 , 0) = 1 m 2 +1 ( m ( m, 1) - ( - 1 ,m )) and similarly (0 , 1) = 1 m 2 +1 (( m, 1) + m ( - 1 ,m )) Then the change of matrix P will look like: P = 1 m 2 + 1 ± m 1 - 1 m ² Then the inverse to P is: P - 1 = 1 m 2 + 1 ± m - 1 1 m ² part a) A reflection along L The reflection along the new basis just takes a form: [ T ] β 0 = ± 1 0 0 - 1 ² So in the standard basis: [ T ] β = P - 1 AP = 1 m 2 + 1 ± m 2 + m m + 1 m - m 2 1 - m ² part b) A projection onto L Similarly in the β 0 basis: [ T ] β 0 = ± 1 0 0 0 ² Hence in the β -standard basis: [ T ] β = P - 1 AP = 1 m 2 + 1 ± m 2 m m 1 ² Question 2: 2.6/13 Let us first define an annihilator S 0 of a subset of a vector space S V . S 0 = { f V * : f ( x ) = 0 x S } 1. Prove that S 0 is a subspace of V * . 1
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(a) Clearly 0 S 0 . (b) If a ( x ) ,b ( x ) S 0 , then this means that a ( x ) = b ( x ) = 0 for all x S . Therefore so will ( a + b )( x ). (c) If a ( x ) inS 0 so will ca ( x ) by the same argument. 2. If W V and x / W , then show there exists f W 0 such that f ( x ) 6 = 0 . Pick a basis for W , β W = { v 1 ,...,v k } and extend it onto V , β V = { v 1 ,...,v k ,v k +1 ,...,v n } . Then form a dual basis to this basis: ( β V ) * = { f 1 ,...,f k ,f k +1 ,...,f n } . Clearly W 0 ∩{ f 1 ,...,f k } since f i ( v i ) = 1 6 = 0. On the other hand we know that f k +1 ,...,f n W 0 , since v k +1 ,...,v n / W . But since v / W , then this means that v = w + a i v i , where i > k and at least one of the a i ’s is non zero, let it be a l . But then this means that f l ( v ) = a l 6 = 0, therefore we have found one.
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This note was uploaded on 07/03/2011 for the course MATH 110 taught by Professor Gurevitch during the Spring '08 term at University of California, Berkeley.

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Solution05_121 - Solutions to PS 5 (Math 121) J. Scholtz...

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