1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Modular arithmetic on vector spaces

  1. Feb 23, 2016 #1
    1. The problem statement, all variables and given/known data
    Let [itex]U[/itex] is the set of all polynomials [itex]u[/itex] on field [itex]\mathbb F[/itex] such that [itex]u(3)=u(-2)=0[/itex]. Check if [itex]U[/itex] is the subspace of the set of all polynomials [itex]P(x)[/itex] on [itex]\mathbb F[/itex] and if it is, determine the set [itex]W[/itex] such that [itex]P(x)=U\oplus W[/itex].

    2. Relevant equations
    -Polynomial vector spaces
    -Subspaces
    -Modular arithmetic

    3. The attempt at a solution
    [itex]U=\{u(x):u(x)\mod (x-3)=0 \land u(x)\mod (x+2)= 0\}[/itex]

    [itex]U[/itex] is the subspace of [itex]P(x)[/itex] iff

    [itex]1)[/itex] [itex]\forall u_1,u_2\in U\Rightarrow u_1+u_2\in U[/itex]

    [itex]2)[/itex] [itex]\forall u\in U,\forall \alpha\in \mathbb F\Rightarrow \alpha u\in U[/itex]

    How to check if [itex]U[/itex] is the subspace of [itex]P(x)[/itex]?

    Assuming [itex]U[/itex] is the subspace of [itex]P(x)\Rightarrow[/itex]

    [itex]P(x)=U\oplus W=U+W \mod n[/itex]

    where [itex]n[/itex] should be the total number of polynomials in [itex]U[/itex] and [itex]W[/itex].

    This means that [itex]W[/itex] is the set of all polynomials [itex]u(x)[/itex] defined as
    [itex]W=\{u(x): u(3)\neq u(-2)\neq 0 \lor u(3)\neq u(-2)=0\lor u(3)=u(-2)\neq 0\}[/itex]

    Is this correct?
     
  2. jcsd
  3. Feb 23, 2016 #2

    Mark44

    Staff: Mentor

    How does modular arithmetic come into this problem? In the problem statement, you say only that u(3) = 0 and u(-2) = 0.
    Minor point, but ##\Rightarrow## is not appropriate in the above. It should be used when one statement implies another statement.##\forall u \in U## is not a statement (i.e., a sentence whose truth value can be determined).
     
  4. Feb 23, 2016 #3

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Any polynomial, u(x), of degree n, such that u(3)= u(-2)= 0 is of the form (x- 3)(x+ 2)v(x) where v is a polynomial of degree n- 2. To show this is a subspace of the space of all polynomials, you only need to show that this set is closed under addition and scalar multiplication- If u(x)= (x- 3)(x+ 2)v(x) and w(x)= (x- 3)(x+ 2)y(x), what can you say about u+ w? What can you say about au where a is a number?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Modular arithmetic on vector spaces
  1. Modular arithmetic (Replies: 3)

  2. Modular arithmetic (Replies: 3)

  3. Modular arithmetic (Replies: 6)

  4. Modular Arithmetic (Replies: 23)

  5. Modular arithmetic (Replies: 3)

Loading...