Computing Matrix, finding kernel and image

  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. micromass

    micromass 18,689
    Staff Emeritus
    Science Advisor
    Education Advisor

    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. 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. micromass

    micromass 18,689
    Staff Emeritus
    Science Advisor
    Education Advisor

    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. micromass

    micromass 18,689
    Staff Emeritus
    Science Advisor
    Education Advisor

    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. 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. micromass

    micromass 18,689
    Staff Emeritus
    Science Advisor
    Education Advisor

    Yes, your image is correct to.

    And remember, math doesnt have to be difficult :biggrin:
     
Know someone interested in this topic? Share a link to this question via email, Google+, Twitter, or Facebook

Have something to add?