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Chapter 1

# Chapter 1 - Chapter 1 Vector Spaces 1.1 Let us multiply by...

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Unformatted text preview: Chapter 1: Vector Spaces 1.1 Let us multiply by the complex conjugate, a- bi : 1 a + bi a- bi a- bi = a- bi a 2 + b 2 = a a 2 + b 2 + b a 2 + b 2 i = c + di Hence, we want c = a/ ( a 2 + b 2 ) and d = b/ ( a 2 + b 2 ). 1.2 Let us cube this number:- 1 + √ 3 i 2 ! 3 =- 1 + √ 3 i 2 ! 2- 1 + √ 3 i 2 ! =- 2- 2 √ 3 i 4 ! - 1 + √ 3 i 2 ! = 2- 2 √ 3 i + 2 √ 3 i + 6 8 = 1 Since the cube of this number is 1, it is a cube root of 1. 1.3 We wish to do this problem without the fact that (- 1)(- 1) = 1, since that might not be true for the field F that this vector space V is over. To show that two thngs are equal, we can equivalently show that the sum of one and the inverse of the other is 0. Hence, we want to show- (- v )- v = 0. Using the distributive property, we have- 1(- v + v ) =- 1(0) = 0, showing that- (- v )- v = 0, and hence- (- v ) = v . 1.4 Suppose that av = 0. If both a = 0 and v = 0, then we are done. Suppose that v 6 = 0; then since av = 0, we can use the definition of a vector space to obtain 0 = av = av- av = ( a- a ) v = 0 v = ⇒ av = 0 = 0 v Hence, we must have a = 0. Finally, suppose that a 6 = 0; once again, since av = 0, we have (using the definition of a vector space): 0 = av = av- av = a ( v- v ) = a 0 = ⇒ av = 0 = a Hence, we must have v = 0. 1.5 a. This is indeed a subspace of F 3 . To see this, let x = ( x 1 ,x 2 ,x 3 ) and y = ( y 1 ,y 2 ,y 3 ) be elements of this set, and a ∈ F . Then since a ( x 1 + 2 x 2 + 3 x 3 ) = a 0 = ⇒ ax 1 + 2 ax 2 + 3 ax 3 = 0, we have that ax is an element of this set for all scalars a . Furthermore, since x 1 +2 x 2 +3 x 3 = 0 + y 1 +2 y 2 +3 y 3 = 0 ( x 1 + y 1 ) +2 ( x 2 + y 2 ) +3 ( x 3 + y 3 ) = 0 Ths sum x + y is also in this subset. Hence, it is a subspace. b. No, this is not a subspace. Let x = ( x 1 ,x 2 ,x 3 ) and y = ( y 1 ,y 2 ,y 3 ) be elements of this set; then x 1 +2 x 2 +3 x 3 = 4 + y 1 +2 y 2 +3 y 3 = 4 ( x 1 + y 1 ) +2 ( x 2 + y 2 ) +3 ( x 3 + y 3 ) = 8 6 = 4 So the set isn’t closed under addition, and hence is not a subspace....
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Chapter 1 - Chapter 1 Vector Spaces 1.1 Let us multiply by...

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