1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Kernal and Range of a Linear Transformation

  1. Apr 21, 2011 #1
    Let L:p2 >>> p3 be the linear transformation defined by L(p(t)) = t^2 p'(t).
    (a) Find a basis for and the dimension of ker(L).
    (b) Find a basis for and the dimension of range(L).

    The hint that I get is to begin by finding an explicit formula for L by determining
    L(at^2 + bt + c).
    Does this hint mean let p(t) = at^2 + bt + c?
    Then, I find that t^2 p'(t) = 2at^3 + bt^2.
    Then, I conclude that the basis for ker(L) = {1}.
    Is it right?
    Also, how to find range(L)?

    Thanks
     
  2. jcsd
  3. Apr 21, 2011 #2

    hunt_mat

    User Avatar
    Homework Helper

    Factorise...

    [tex]
    L(at^{2}+bt+c)=2at^{3}+bt^{2}=t^{2}(2at+b)
    [/tex]

    This will be zero when t=0 or t=-b/2a, so...
     
  4. Apr 21, 2011 #3

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    That doesn't have much to do with the problem. hkus10 correctly has ker(L)={1} and having written L(p(t))=2at^3+bt^2 the answer to the dimension and a basis of range(L) should be pretty obvious. Why isn't it hkus10? What's a basis for p3?
     
  5. Apr 21, 2011 #4
    is the basis for ker(L) {t, 1} and the basis for range(L) {t^3, t^2}?
     
  6. Apr 21, 2011 #5

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Two steps forward, one step backward. Yes, that's a basis for range(L). But now your basis for ker(L) is wrong. I liked your ker(L)={1} a lot better. Why did you put t in? Is t in the kernel?
     
  7. Apr 21, 2011 #6

    Char. Limit

    User Avatar
    Gold Member

    I believe you have a workable basis for range(L). However, I think your basis for ker(L) has too many entries.
     
  8. Apr 22, 2011 #7
    Since L(at^2 + bt + c) = 2at^3 + bt^2
    No matter what value of t and 1, 2a^3 + bt^2 should always give me 0 vector. So, I have a question why Ker(L) does not have t as a basis?
    Another question is dim Ker(L) + dim range(L) = dim (p3) by thm. since the dim range(L) = 2 and dim (p3) = 4, why dim ker(L) not equal to 2?
     
  9. Apr 22, 2011 #8

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    The rank nullity theorem tells you dim ker(L)+dim range(L)=dim(p2). Not dim(p3). dim(p2)=3. t is not in ker(L) because L(t) is not zero. L(t)=t^2. t^2 is not zero.
     
  10. Apr 22, 2011 #9
    Thanks
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Similar Discussions: Kernal and Range of a Linear Transformation
Loading...