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Computing Matrix, finding kernel and image

  1. Nov 2, 2010 #1
    Let T: R[x]2[tex]\rightarrow[/tex] R[x]3 be defined by T(P(x))=xP(x). Compute the matrix of x with respect to bases {1,x,x2} and {1,x,x2,x3}. Find the kernel and image of T.

    I know how to do this when given bases without exponents, however I do not know exactly what this is saying and therefore am having a hard time starting it.
     
  2. jcsd
  3. Nov 2, 2010 #2

    micromass

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    Let's compute the first column.

    You take the first basis element, that is 1. Now you'll have to express T(1)=x in terms of the basis {1,x,x²,x³}. That would be (0,1,0,0). So the first column would consist out of

    [tex] \left(\begin{array}{ccc}
    0 & ? & ? \\
    1 & ? & ? \\
    0 & ? & ? \\
    0 & ? & ?
    \end{array} \right) [/tex]

    Now for the second and third column, you'll have to express T(x) and T(x²) in terms of the basis {1,x,x²,x³}.
     
  4. Nov 2, 2010 #3
    Okay so matrix T(x)=x2 and with respect to the basis {1,x,x²,x³} the second column would be (0,0,1,0), T(x2)=x3 and with respect to the basis {1,x,x²,x³} would be (0,0,0,1).
    I would then say that
    T=(0 0 0)
    (1 0 0)
    (0 1 0)
    (0 0 1)
    with respect to {1,x,x²,x³} right? Not with respect to {1,x,x²} and {1,x,x²,x³}
    so then to find the im(T) I just use
    (0 0 0)
    (1 0 0)
    (0 1 0)
    (0 0 1)
    and solve it with respect to the basis {1,x,x²,x³} right?
    Then how would I go about solving for the ker(T)?
     
  5. Nov 2, 2010 #4

    micromass

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    For the Ker(T). Take your matrix A. Then wonder when Ax=0. This you should be able to solve easily (it's a system with 3 unknowns and 4 equations)
     
  6. Nov 2, 2010 #5

    micromass

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    So, if (x,y,z) is in the kernel, then you must have

    [tex]
    \left( \begin{array}{ccc}
    0 & 0 & 0\\
    1 & 0 & 0\\
    0 & 1 & 0\\
    0 & 0 & 1
    \end{array} \right) \left( \begin{array}{c}
    x\\
    y\\
    z
    \end{array} \right) = 0
    [/tex]

    It's not hard to see that this can only be the case iff x=y=z=0
     
  7. Nov 2, 2010 #6
    right, that just seemed way too easy I thought I was doing it wrong.
    Then the im(T) is the span of e2, e3, and e4
     
  8. Nov 2, 2010 #7

    micromass

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    Yes, your image is correct to.

    And remember, math doesnt have to be difficult :biggrin:
     
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