{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

M115A-AssignmentOneExtraSolutions

M115A-AssignmentOneExtraSolutions - MATH 115A Assignment...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
MATH 115A - Assignment One - Solutions of Select Non-Graded Problems Paul Skoufranis October 3, 2011 § 1.2 Question 13) Let V denote the set of ordered pairs of real numbers. If ( a 1 , a 2 ) and ( b 1 , b 2 ) are elements of V and c R , define ( a 1 , a 2 ) + ( b 1 , b 2 ) = ( a 1 + b 1 , a 2 b 2 ) and c · ( a 1 , a 2 ) = ( ca 1 , a 2 ) Is V a vector space over R with these operations? Justify your answer. Solution : The set V with the above operations is not a vector space. We will provide two ways to see this: 1. Property (VS 8) in the definition of a vector space fails for V with these operations. To see this, we notice that if ~v = (1 , 2), then (1 + 1) · ~v = 2 · ~v = (2 , 2) yet 1 · ~v + 1 · ~v = 1 · (1 , 2) + 1 · (1 , 2) = (1 , 2) + (1 , 2) = (2 , 4) so (1 + 1) · ~v 6 = 1 · ~v + 1 · ~v . Hence V is not a vector space. 2. Suppose V is a vector space. Then V must have a unique zero vector. Since ( a, b ) + (0 , 0) = ( a + 0 , b (0)) = ( a, 0) , ~v + (0 , 0) = ~v for all ~v V . Hence (0 , 0) must be the zero vector of V . However, we notice that (1 , 0) + ( - 1 , 0) = (0 , 0) = (1 , 0) + ( - 1 , 1) . Therefore ( - 1 , 0) and ( - 1 , 1) are distinct additive inverse of (1 , 0). Since every element of a vector space has a unique additive inverse, we have a contradiction. Therefore V is not a vector space. Thus we have demonstrated that V is not a vector space in two different ways. § 1.2 Question 17) Let V = { ( a 1 , a 2 ) | a 1 , a 2 F } , where F is a field. Define addition of elements of V coordinate wise, and for c F and ( a 1 , a 2 ) V , define c · ( a 1 , a 2 ) = ( ca 1 , 0) Is V a vector space over F with these operations? Justify your answer. Solution : The set V with the above operations is not a vector space. The easiest way to see this is to note (using the know properties of fields) that 1 · (1 , 1) = (1 , 0). Since 1 6 = 0 in any field, (1 , 1) 6 = (1 , 0) so 1 · (1 , 1) 6 = (1 , 1). Hence property (VS 5) in the definition of a vector space fails for V with these operations.
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
§ 1.2 Question 21) Let V and W be vector spaces over a field F with respect to the operations + V , + W , · V , and · W . Let Z = { ( ~v, ~w ) | ~v V, ~w W } Prove that Z is a vector space with the operations ( ~v 1 , ~w 1 ) + ( ~v 2 , ~w 2 ) = ( ~v 1 + V ~v 2 , ~w 1 + W ~v 2 ) and c · ( ~v, ~w ) = ( c · V ~v, c · W ~w ) for all ~v,~v 1 ,~v 2 V , ~w, ~w 1 , ~w 2 W , and c F . Proof : To show that Z is a vector space over F with the above operations, we will show that Z satis- fies the definition of a vector space. It is clear that the operations + and · are well-defined (that is, if ~v,~v 1 ,~v 2 V , ~w, ~w 1 , ~w 2
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}