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Linear algebra: Finding a basis for a space of polynomials

  1. Jan 21, 2016 #1
    1. The problem statement, all variables and given/known data
    Let a39c3e275c2591d05b49e54e3284b4ea.png and 1179c4f0625914c5a491a155d5528415.png are two basis of subspaces d20caec3b48a1eef164cb4ca81ba2587.png and [PLAIN]http://www.sosmath.com/CBB/latexrender/pictures/69691c7bdcc3ce6d5d8a1361f22d04ac.png. [Broken] Find one basis of http://www.sosmath.com/CBB/latexrender/pictures/38d4e8e4669e784ae19bf38762e06045.png and [PLAIN]http://www.sosmath.com/CBB/latexrender/pictures/fd36d76c568c236aaaad68e084eef495.png. [Broken]

    2. Relevant equations
    -Vector space
    -Basis
    -Polynomials

    3. The attempt at a solution

    Could someone explain the method for finding a basis for a space of polynomials.
    I know that with 2369a2488f59aa39a3fca53e0eff9f88.png we need to find RREF of an augmented matrix,
    and read a basis from matrix, but how to do it with polynomials?
     
    Last edited by a moderator: May 7, 2017
  2. jcsd
  3. Jan 21, 2016 #2

    PeroK

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    Do you understand what L and M are? Can you describe these as vector spaces?
     
  4. Jan 21, 2016 #3

    HallsofIvy

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    The first post says exactly what L and M are- it gives their bases. L is the space of polynomials spanned by [itex]\{1+ t- t^3, 1+ t+ t^2, 1- t\}[/itex] so any vector in L is of the form [itex]a(1+ t- t^3)+ b(1+ t+ t^2)+ c(1- t)= -at^3+ bt^2+ (a+ b- c)t+ (a+ b+ c)[/itex] for any number a, b, and c. M is the space of polynomials spanned by [itex]\{t^3+ t, 2- t^3, 2+ t^3\}[/itex] so any vector is M is of the form [itex]p(t^3+ t)+ q(2- t^3)+ r(2+ t^3)= (p- q+ r)t^3+ pt+ (2q+ 2r)[/itex] for any numbers p, q, and r. Any vector in L+ M is of the form [itex](-a+ p- q+ r)t^3+ bt^2+ (a+ b- c+ p)t+ (a+ b+ c+ 2q+ 2r)[/itex]. Any vector in [itex]L\cap M[/itex] can be written as either of those 2 first forms with -a= p+ q+ r, b= 0, a+ b- c= p, and a+ b+ c= 2q+ 2r. Simplify those.
     
  5. Jan 21, 2016 #4

    blue_leaf77

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    -a= p - q+ r?
     
  6. Jan 21, 2016 #5

    PeroK

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    Looks like this has turned into Halls of Ivy's homework!
     
  7. Jan 21, 2016 #6

    HallsofIvy

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    Oh, dear!
     
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