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Homework Help: Linear Algebra: does it form a basis?

  1. Dec 10, 2011 #1
    1. The problem statement, all variables and given/known data
    See attachment.

    3. The attempt at a solution

    I already did parts i and ii (correctly, I hope). On part iii I found 2 linearly independent elements to be: t+1, t^2 - 1.

    However, I don't understand how to show that these form a basis of W. Because W is a subspace of P2, and P2 has dimension 3, we know W has at most dimension 3. But I only have 2 elements, not 3, so I can't use this to show that I have a basis. Perhaps the constraint that p(-1) = 0 reduces the dimension, but I don't know how to prove this.


    Attached Files:

  2. jcsd
  3. Dec 11, 2011 #2


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    A polynomial in W is can be represented by an ordered pair of numbers (a0, a1): p(t)=a0+a1t+(a1-a0)t^2. So the set W is isomorphic with the set of ordered pair of numbers ...

  4. Dec 11, 2011 #3
    the constraint that -1 is a root gives

    so then


    since a is the coefficient on t^2 and b is the coefficient on t we can let
    [itex]a=a_2[/itex] and [itex]b=a_1[/itex]
    to find
    [itex]a_2(t^2-1)+a_1(b+1)=a_2 t^2+a_1 t +(a_1-a_2)[/itex]
    from the constraint we found [itex]a_0=a_1-a_2[/itex]
    and so

    you follow?

    I should also add, every constraint you add to a system reduces the dimension of it by 1, adding another root would make it a 1 dimensional problem
  5. Dec 11, 2011 #4


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    Not really...:smile: I would do it just on the opposite way.

    Those polynomials which have root -1, are of form pa,b(t)=at^2+bt+(b-a), that is pa,b(t)=a(t^2-1)+b(t+1), which means all such polynomials are linear combination of the basic "vectors" p1=t^2-1 and p2=t+1. Two basic vectors define a two-dimensional subspace.

    That is right.
    With one more constraint, there would be only one free parameter, so the space would become one-dimensional.

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