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Linear Algebra (Change of Basis)

  1. Jul 20, 2009 #1
    1. The problem statement, all variables and given/known data

    Let E={1, x, x2,x3} be the standard ordered basis for the space P3. Show that G={1+x,1-x,1-x2,1-x3} is also a basis for P3, and write the change of basis matrix S from G to E.

    2. Relevant equations


    3. The attempt at a solution

    Here's what I got:

    [tex]
    S_E^G=\left(
    \begin{array}{cccc}
    1 & 1 & 1 & 1 \\
    1 & -1 & 0 & 0 \\
    0 & 0 & -1 & 0 \\
    0 & 0 & 0 & -1
    \end{array}
    \right)
    [/tex]

    Now, to prove that this is also a basis, I just need to show that it has an inverse, right?

    Here's the problem. If the above is correct, then when you multiply it by G you should get E, right? After all, it is the "change of basis matrix S from G to E". However, this isn't the case:

    [tex]
    \left(
    \begin{array}{cccc}
    1 & 1 & 1 & 1 \\
    1 & -1 & 0 & 0 \\
    0 & 0 & -1 & 0 \\
    0 & 0 & 0 & -1
    \end{array}
    \right).\left(
    \begin{array}{c}
    1+x \\
    1-x \\
    1-x^2 \\
    1-x^3
    \end{array}
    \right)=\left(
    \begin{array}{c}
    4-x^2-x^3 \\
    2 x \\
    -1+x^2 \\
    -1+x^3
    \end{array}
    \right)
    [/tex]

    Am I doing something wrong, or am I just confused about what a change of basis matrix is supposed to do?
     
  2. jcsd
  3. Jul 20, 2009 #2

    Office_Shredder

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Your vectors in your representation in R4 are going to have just numbers, not polynomials in their entries. So (1+x, 1-x, 1-x2, 1-x3) isn't a vector, instead the vectors are supposed to be representing polynomials, for example 1-x is represented as (1,-1,0,0).

    Your change of basis matrix is correct. Don't use it to show G is a basis though... you know P3 has dimension 4, so show G is linearly independent and hence since it has 4 elements it must span the space also
     
  4. Jul 20, 2009 #3
    I see. so a valid test would be:

    [tex]
    \left(
    \begin{array}{cccc}
    1 & 1 & 1 & 1 \\
    1 & -1 & 0 & 0 \\
    0 & 0 & -1 & 0 \\
    0 & 0 & 0 & -1
    \end{array}
    \right).\left(
    \begin{array}{c}
    0 \\
    0 \\
    1 \\
    0
    \end{array}
    \right)=\left(
    \begin{array}{c}
    1 \\
    0 \\
    -1 \\
    0
    \end{array}
    \right)
    [/tex]

    And the fact that I get (1,0,-1,0) says that I have 1-x^2, which is what I would expect.

    Cool, I think I got it. Thanks!
     
  5. Jul 21, 2009 #4
    Although that is quite valid, in this problem I would find it simpler to show that G spans P3 and thus is linearly independent.
     
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