solution2

# solution2 - M341 Homework Solution 2 Section 1.3 6(a If x y...

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M341 Homework Solution 2 Section 1.3 - 6(a) : If x + y = x and x is orthogonal to y , then we have x 2 = x + y 2 = ( x + y ) · ( x + y ) = x 2 + 2( x · y ) + y 2 = x 2 + y 2 . It follows y 2 = 0 . But y 2 = 0 implies y = 0 , therefore if x + y = x then y = 0 or x is not orthogonal to y . Section 1.3 - 7 : If x + y 2 = x 2 + y 2 , then we have x 2 + y 2 = x + y 2 = ( x + y ) · ( x + y ) = x 2 + 2( x · y ) + y 2 . It follows x · y = 0 . Therefore if x · y ̸ = 0 , then x + y 2 ̸ = x 2 + y 2 . Section 1.3 - 8(b) : Contrapositive: "If y ̸ = proj x y , then x is not parallel to y ." Converse: "If y = proj x y , then x is parallel to y ." Inverse:"If x is not parallel to y , then y ̸ = proj x y ." Section 1.3 - 10(a) : (i) The converse of the given statement is "If y = z , then x · y = x · z " (ii) The converse is true.
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## This note was uploaded on 12/05/2011 for the course M 341 taught by Professor Hietmann during the Spring '08 term at University of Texas.

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