LADR Assorted Solutions

LADR Assorted Solutions - Math 441/541 Assignment 1:...

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Unformatted text preview: Math 441/541 Assignment 1: Solutions Chapter 1 3. Here we use the associativity of scalar multiplication and the fact that- 1 v =- v .- (- v ) =- 1 (- v ) =- 1 (- 1 v ) = (- 1 - 1) v = 1 v = v. 4. Let a F and v V and suppose av = 0. If a = 0 then were done. So assume a 6 = 0. Then we need to show that v = 0. Since a 6 = 0, we know from the fact that F is a field that a has a multiplicative inverse 1 a F , such that a 1 a = 1. Hence we have 0 = av = 1 a 0 = 1 a ( av ) = 1 a a v = 1 v = v. Thus, v = 0 as required. 5. a) { ( x 1 ,x 2 ,x 3 ) F 3 | x 1 + 2 x 2 + 3 x 3 = 0 } . 0 + 2(0) + 3(0) = 0, hence the zero vector 0 is in the set. If x = ( x 1 ,x 2 ,x 3 ) and y = ( y 1 ,y 2 ,y 3 ) are in the set, then x + y = ( x 1 + y 1 ,x 2 + y 2 ,x 3 + y 3 ), so check ( x 1 + y 1 ) + 2( x 2 + y 2 ) + 3( x 3 + y 3 ) = [ x 1 + 2 x 2 + 3 x 3 ] + [ y 1 + 2 y 2 + 3 y 3 ] = 0 + 0 = 0. Finally, if a F , then ax = ( ax 1 ,ax 2 ,ax 3 ) so we check ( ax 1 ) + 2( ax 2 ) + 3( ax 3 ) = a [ x 1 + 2 x 2 + 3 x 3 ] = a 0 = 0 . So this set is a subspace. b) { ( x 1 ,x 2 ,x 3 ) F 3 | x 1 + 2 x 2 + 3 x 3 = 4 } . This set is not a subspace because 0+2(0) +3(0) = 0 6 = 4, hence the zero vector is not an element of it. (in fact, the other two conditions fail as well!) c) { ( x 1 ,x 2 ,x 3 ) F 3 | x 1 x 2 x 3 = 0 } . The zero vector is in this set, and actually it is closed under scalar multiplication as well, but it is not closed under addition. Counterexample: (1 , , 0) and (0 , 1 , 1) are both in the set, but the sum (1 , 1 , 1) is not. d) { ( x 1 ,x 2 ,x 3 ) F 3 | x 1 = 5 x 3 } . 0 = 5(0), hence the zero vector is in this set. Given vectors x and y in the set, we know that x 1 = 5 x 3 and y 1 = 5 y 3 , hence x 1 + y 1 = 5 x 3 +5 y 3 = 5( x 3 + y 3 ), so it is closed under addition. Similarly, if x 1 = 5 x 3 then ax 1 = a ( x 1 ) = a (5 x 3 ) = 5( ax 3 ), so it is also closed under scalar multi- plication. Thus it is a subspace. 6. The lattice Z 2 = { ( x,y ) R 2 | x,y Z } inside of R 2 is closed under addition and inverses (because the sum of any two integers is another integer and the additive inverse of an integer is an integer), but it is not closed under scalar multiplication. As an example (1 , 0) Z , but 1 2 (1 , 0) = ( 1 2 , ) 6 Z 2 . 1 8. Let U 1 ,U 2 ,...,U n V be subspaces. Then U 1 U 2 U n is a subspace of V too. Proof: Since each U i is a subspace we know that 0 U i for each i . Hence 0 U 1 U n too. If x,y U 1 U n , then specifically x,y U i for each i . Hence x + y U i for each i , which implies x + y U 1 U n ....
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LADR Assorted Solutions - Math 441/541 Assignment 1:...

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