cot4501fa10_hw5_sol

# cot4501fa10_hw5_sol - COT 4501 Fall 2010 Homework 5...

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Unformatted text preview: COT 4501 Fall 2010 Homework 5 Solutions 3.12: (a) • 1&2 ⇒ 3 : If A T = A and A T A = I , then A 2 = A · A = A T · A = I . • 1&3 ⇒ 2 : If A T = A and A 2 = A , then A T · A = A · A = A 2 = I . • 2&3 ⇒ 1 : If A T A = I and A 2 = I , then A T = A T · I = A T A 2 = A T A · A = I · A = A . (b) Let A be the 3 × 3 rotation matrix, A = − 1 0 − 1 1 . It satisfies all three properties. (c) Reflection matrices have this property. 3.13: If A is an orthogonal projector, then it is a symmetric matrix with A T = A and idempotent matrix with A 2 = A . Since it is also an orthogonal matrix A T A = I and therefore A 2 = I . From A 2 = A and A 2 = I , we get that A = I . 3.26: Let u = ( x y ) , v = ( x ′ y ′ ) and v = G u = ( cos θx + sin θy − sin θx + cos θy ) . The matrix G acts as rotation of θ in the clockwise direction. Given v = H u = ( − cos θx + sin θy sin θx + cos θy ) , since det( H ) = − 1 , H acts as a compound transform of a reflection about the...
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## This note was uploaded on 01/22/2012 for the course COT 4501 taught by Professor Davis during the Spring '08 term at University of Florida.

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cot4501fa10_hw5_sol - COT 4501 Fall 2010 Homework 5...

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