Results 1 to 4 of 4

Math Help - Polynomial, direct sum

  1. #1
    Newbie
    Joined
    Sep 2010
    Posts
    15

    Polynomial, direct sum

    Let P be the vector space consisting of all polynomials with degree less than or equal to 2, where the field that P is over is the set of rationals. Now let E2 be the subspace of all stuff in P that has a root of 2, let E5 be the subspace of all elements in P that have a root of 5. Show that P=E2+E5, where + is the subspace sum.

    This isn't a homework problem, but it's a problem that I've been having difficulty with, so any help would be much appreciated. Thanks
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member Haven's Avatar
    Joined
    Jul 2009
    Posts
    197
    Thanks
    8
    EDIT: sorry, i misread what you meant by subspace sum.


    So just to clarify the definitions, P = \{ f(x) \in \mathbb{Q}[x]| deg(f(x))\leq 2 \}
    E2 = \{f(x) \in P| f(2)=0 \}
    E5 = \{f(x) \in P| f(5)=0 \}
    E2 + E5 = \{ f(x)+g(x) | f(x) \in E2, g(x) \in E5 \}
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Sep 2010
    Posts
    15
    Yeah, that's right.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,693
    Thanks
    1466
    Any member of E2 is of the form a(x- 2)(x- b)= ax^2- a(2+ b)x+ 2ab for rational numbers a and b. Any member of E5 is of the form c(x- 5)(x- d)= cx^2- c(5+d)x+ 5cd for rational numbers c and d. Any polynomial in the direct sum of E2 and E5 is of the form m(ax^2- a(2+ b)x+ 2ab)+ n(cx^2- c(5+ d)x+ 5cd)= (am+ nc)x^2- (2am+ abm+ 5cn+ cdn)x+ (2ab+ 5cd) for rational numbers a, b, c, d, m, and n. Certainly each of the coefficients is a rational number so E2+ E5 is a subspace of P.

    Any member of P is of the form ux^2+ vx+ w with u, v, and w rational numbers. If you can show that, given any rational numbers u, v, and w, there exist rational numbers a, b, c, d, m, and n such that am+nc= u, 2am+ abm+ 5cn+ cdn= v, and 2ab+ 5cd= w, you are done. Since that is 3 equations in 6 numbers, it should not be two difficult to show that there exist at least one solution for every u, v, w. (You should NOT expect that solution to be unique.)
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Direct Sum
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: February 17th 2011, 07:42 PM
  2. direct sum
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: November 16th 2010, 07:39 PM
  3. Replies: 1
    Last Post: December 15th 2009, 07:26 AM
  4. Direct Sum
    Posted in the Advanced Algebra Forum
    Replies: 0
    Last Post: December 6th 2009, 08:31 AM
  5. Direct Sum of Dimensions
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: October 1st 2009, 11:49 AM

Search Tags


/mathhelpforum @mathhelpforum