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Dimension, Image, Kernels etc - Linear Algebra stuff

  1. Aug 23, 2008 #1
    Does linear algebra go in this thread??
    Anyway,
    1. The problem statement, all variables and given/known data
    Let n be an integer at least 1, and x1,....,xn be distinct points in R. For any integer m>=1, let P denote the vector space of polynomials of degree at most m. Define a linear transformation T:P->R^n by
    [f(x1)]
    [f(x2)]
    T(f)= [ . ] <--- (That's meant to be one big bracket)
    [ . ]
    [ . ]
    [f(xn)]

    What is the dimension of the kernel of T? What is the dimension of the range of T? (your answer will depend on m and n)

    2. Relevant equations
    dim V = dim Im + dim ker

    3. The attempt at a solution
    Well I've considered the case when n>m. In this case, dim V= m+1 and dim ker = 0 as it's not possible for any polynomial of degree m+1 to have more than m roots. So I find that dim Im = m+1.
    Now for m>=n, I don't seem to be getting anywhere. Is the dimension of the kernel m-1 cos there can be m-1 roots?

    Any help would be really appreciated :)
     
  2. jcsd
  3. Aug 23, 2008 #2
    Argh it messed up my bracket.
    Well T(f) is meant to be
    [f(x1)]
    [f(x2)]
    [f(x3)] <- it's meant to be a single long bracket extending downwards
    [......]
    [......]
    [......]
    [f(xn)]
     
  4. Aug 24, 2008 #3

    HallsofIvy

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    So T(f) maps f into (f(x1), f(x2), f(x3)...)?

    Okay, f is in the kernel of T if and only if f(x1)= 0, f(x2)= 0, ...

    (y1, y2, y3, ...) is in the image of T if and only if there exist a polynomial f such that f(x1)= y1) f(x2)= y2, ...
     
  5. Aug 24, 2008 #4
    Yea that's right. So when m>n, the polynomial will have m roots. In this case is the dimension of the kernel 1 and the dimension of the image m? I'm kinda confuse about how I can tell what the dimension is.
    (The dimension just means the number of vectors in the basis)
     
  6. Aug 24, 2008 #5

    Defennder

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    I'd like to clarify something here. Does the transformation T operate only on (x1,x2...xn) where all the xi's are distinct? Suppose for eg. m=2, then f(x) has two roots and the vectors in ker(T) are of the form (x1,x2,x1,x1...) <-any permutation of x1 and x2, which grows to an exponentially large number.
     
  7. Aug 24, 2008 #6

    HallsofIvy

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    The transformation T operates on polynomials. I'm going to assume that x1, x2, etc. are distinct.

    For example, suppose [itex]x_1= 0[/itex], [itex]x_2= 1[/itex] and m= 2 so P is the vector space of polynomials of order at most 2.

    Then T(p(x))= (p(0), p(1)). The kernel is the subspace of all such polynomials such that p(0)= 0, p(1)= 0. Since p(x)= a+ bx+ cx2, p(0)= a= 0 and p(1)= b+ c= 0. Since c= -b, we are free to choose b and then a= 0, c= -b so the kernel has dimension 1. Since P itself has dimension 3, it follows that the image of T has dimension 2.

    More generally, P has dimension m+ 1. Setting P(x1)= 0, ..., P(xn)= 0, we have n equations for those m+ 1 variables. That should leave m+1- n independent variables so it looks to me like the kernel of T has dimension m-n+1 and the image has dimension m+1- (m-n+1)= n.

    Of course, that's not a proof and may be completely wrong.
     
  8. Aug 24, 2008 #7
    That seems correct :). But what if m+1<n? The kernel can't have a negative dimension, can it?
     
  9. Aug 24, 2008 #8

    HallsofIvy

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    If that is the case, then we have more points than we have coefficients: but we will still have the trivial polynomial, p(x)= 0 for all x. The smallest the kernel can be is the 0 vector only and that has dimension 0. As long as [itex]m+1\le n[/itex], the kernel will have dimension 0 and the image will have dimension m+1.
     
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