problemset1solutions_1 - MAS 213 Linear Algebra II...

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MAS 213: Linear Algebra II. Solutions for problem list for Week #1. Tutorial on 7th September. Tutorial problems: Problem 1: (Problem 1.2.13, 1.2.17-19 in [FIS].) Let V denote the set of ordered pairs of real numbers. In the following 4 parts, we are going to attempt to put an R -vector space structure on V by making different definitions of vector addition and scalar multiplication. For each of the following examples, either check that all the axioms hold, or find an axiom which fails to hold. I’ll go through the first part in detail, but will be much quicker with the other parts. Solution to problem 1, part 1. Here we tried to define the operations by ( 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 ) . I’ll go through each of the axioms as far as possible. The clearest fail is in Axiom VIII. Axiom III fails too, but the explanation there is a bit harder. I. Commutativity is obvious: ( a 1 , a 2 ) + ( b 1 , b 2 ) = ( a 1 + b 1 , a 2 b 2 ) = ( b 1 + a 1 , b 2 a 2 ) = ( b 1 , b 2 ) + ( a 1 , a 2 ). II. Associativity is also obvious: (( a 1 , a 2 ) + ( b 1 , b 2 )) + ( c 1 , c 2 ) = ( a 1 + b 1 , a 2 b 2 ) + ( c 1 , c 2 ) = (( a 1 + b 1 ) + c 1 , ( a 2 b 2 ) c 2 ) = ( a 1 + ( b 1 + c 1 ) , a 2 ( b 2 c 2 )) = ( a 1 , a 2 ) + ( b 1 + c 1 , b 2 c 2 )) = ( a 1 , a 2 ) + (( b 1 , b 2 ) + ( c 1 , c 2 )) 1
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III. Existence of a zero? Well, we need an element ( p, q ) such that ( p, q ) + ( x, y ) = ( x, y ) for all pairs ( x, y ). x + p = x and yq = y for all x, y R . Choosing the zero to be the element (0 , 1) is the only possible choice, and when you make that choice, Axiom III holds. IV. Existence of an additive inverse? This is where it goes wrong for the first time. Our vectors are allowed to have a zero in the second position e.g. ( x, 0), but there is no vector you can add to that (using the definition of addition in this part) to get the zero vector, which has to be (0 , 1). V. This axiom is OK. Scalar multiplication by 1 does not change any vector. VI. Associativity of scalar multiplication is obviously OK. Observe: c 1 ( c 2 ( a 1 , a 2 )) = c 1 ( c 2 a 1 , a 2 ) = ( c 1 ( c 2 a 1 ) , a 2 ) = (( c 1 c 2 ) a 1 , a 2 ) = ( c 1 c 2 )( a 1 , a 2 ) . VII. Distribution in the vector position. This axiom holds. Observe: x (( a 1 , a 2 ) + ( b 1 , b 2 )) = x ( a 1 + b 1 , a 2 b 2 ) = ( x ( a 1 + b 1 ) , a 2 b 2 ) = ( xa 1 + xb 1 , a 2 b 2 ) = ( xa 1 , a 2 ) + ( xb 1 , b 2 ) = x ( a 1 , a 2 ) + x ( b 1 , b 2 ) . VIII. Distribution in the scalar position. This axiom fails . To show that an axiom fails we need to find a particular example where the equation is false. A simple one is the case where the two scalars are x = y = 1 and the vector v = (1 , 2). Then the left hand side of the axiom becomes (1 + 1)(1 , 2) = 2(1 , 2) = (2 , 2). And the right hand side becomes 1(1 , 2) + 1(1 , 2) = (1 , 2) + (1 , 2) = (2 , 2 . 2), which is different. 2
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Solution to problem 1, part 2. In this part we defined vector addition in the usual way, and scalar multi- plication by c ( a 1 , a 2 ) = ( a 1 , 0). Of course Axioms I-IV are going to be OK, because these four axioms just refer to addition, and here we have just defined addition in the normal way for R 2 . Axiom V fails , because multiplication by 1 changes the vector if the second position is non-zero.
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