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Basis for kernel of linear transform

  1. Mar 16, 2010 #1
    Hey guys!

    I am having a major brain problem today, with this problem.

    L is a linear transform that maps L:P4[tex]\rightarrow[/tex]P4

    As such that (a1t3+a2t2+a3t+a4 = (a1-a2)t3+(a3-a4)t.

    I am trying to find the basis for the kernel and range.

    I know that the standard basis for P4 is {1,x,x2,x3}
    And the kernel is when L(u)=0, but I don't know how to find the transformation matrix, since we're not dealing with numbers in R, but in the set of polynomials. Is there another way to find the kernel/range and bases without using the T matrix?
     
  2. jcsd
  3. Mar 16, 2010 #2

    jbunniii

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    Don't worry about trying to find a matrix. It's not necessary in order to answer this problem.

    To find the kernel, just look at the defining equation for [itex]L[/itex]. What relationships must hold among [itex]a_1, a_2, a_3, a_4[/itex] in order to obtain

    [tex]L(a_1 t^3 + a_2 t^2 + a_3 t + a_4) = 0[/tex]

    To find the range, simply answer this: if you allow [itex]a_1, a_2, a_3, a_4[/itex] to take on all possible values, what are all the possible polynomials that you can produce of the form

    [tex](a_1 - a_2) t^3 + (a_3 - a_4) t[/tex]
     
  4. Mar 16, 2010 #3
    a1=a2 and a3=a4, so that a2 and a4 can be arbitrary. So plugging back into the original L(a), and factoring, you'd get that {(1+t),(t2+t3)} is the kernel of L?

    The range would be … any value of a2 or a4 for t3 and t, so the range is just {t,t3} ?
     
  5. Mar 16, 2010 #4

    jbunniii

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    That is a BASIS for the kernel of L. The kernel itself consists of all possible linear combinations of the basis elements, which is any polynomial of the form [itex]a (1+t) + b(t^2 + t^3)[/itex].

    That is a BASIS for the range of L. The range itself is once again the set of all possible linear combinations of the basis elements, i.e. anything of the form [itex]a t + b t^3[/tex].
     
  6. Mar 16, 2010 #5
    Ohhh right right. Yeah I was getting ahead of myself. I understand now! Thanks :)
     
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