LADR Assorted Solutions

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

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

View Full Document Right Arrow Icon
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 we’re 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 , 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 , 0 ) 6∈ Z 2 . 1
Image of page 1

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

View Full Document Right Arrow Icon
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 . Finally, if a F and x U 1 ∩ · · · ∩ U n , then specifically x U i for each i . Thus, ax U i for each i , which implies that ax U 1 ∩ · · · ∩ U n . 12. Addition of subspaces does have an additive identity, namely the zero subspace { 0 } . Since U + { 0 } = { u + 0 | u U } = { u | u U } = U . Given any subspaces U and W we know that U + W contains both U and W as subspaces. Hence, U + W = { 0 } implies that U = { 0 } and W = { 0 } . Hence, the only subspace with an additive inverse is the zero subspace { 0 } and its inverse is itself.
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 ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern