Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Linear operators

  1. Feb 16, 2008 #1
    Let T be a linear operator on a finite dimensional vector space V, over the field F.
    Suppose TU = I, where U is another linear operator on V, and I is the Identity operator.
    It can ofcourse be shown that T is invertible and the invese of T is nothing but U itself.

    What I want to know is an example explicitly to show that the above is false if V is not finite dimensional.

    Thank You.
  2. jcsd
  3. Feb 16, 2008 #2


    User Avatar
    Science Advisor
    Homework Helper

    It suffices to find an operator T that is surjective but not injective.

    A standard example of this phenomenon is the backwards shift on [itex]\ell^2[/itex], the space of square summable sequences: T(x_1, x_2, ...) = (x_2, x_3, ...). I'll let you verify this.
  4. Feb 16, 2008 #3
    This space is new to me !
    Work me through this example.
    Can't we show something on the vector space of R (reals) over rationals Q, or maybe even on vector space of all polynomials over any field.

    Thank you.
  5. Feb 16, 2008 #4


    User Avatar
    Science Advisor
    Homework Helper

    We actually don't really need the space to be l^2, but this is the space this example usually comes up in.

    Let's reduce it to a more familiar space: F[x], the space of polynomials over F. Define T:F[x]->F[x] by T(a0 + a1x + ... + anxn) = a1 + a2x + ... + anxn-1. I'll let you verify that this is linear and surjective, but not injective.
  6. Feb 16, 2008 #5


    User Avatar
    Science Advisor

    To add to what morphism just said, let U(a0+ a1x+ ...+ anxn)= 1+ a0x+ a1x2+ ...+ anxn+1 and it should be clear that TU= I but UT is not equal to I!
  7. Feb 16, 2008 #6


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    You've (hopefully) already seen this example, in another context. The construction the others have described precisely corresponds to finding a (set) function from the set of natural numbers to itself that is injective but not surjective.

    This is actually a quite general method useful in many contexts -- all you have to do is find a way to relate natural numbers to the kind of structure you're studying.
  8. Feb 17, 2008 #7
    finally got it !
    yes, used similar mappings in ring theory !
  9. Feb 17, 2008 #8
    Well I've come across another question.
    It's from Hoffman and Kunze's text on linear algebra.
    Let T be a linear operator on F^n, let A be the matrix of T in the standard ordered basis for F^n, and let W be the subspace of F^n spanned by the column vectors of A. What does W have to do with T?

    W is simply the space spanned by the range of T, right ? And in case the columns of A are linearly independent, they would form a basis for F^n and so W would be F^n itself.
    In this case the transformation T would be necessarily injective, since it is preserving linear independence.
    In case the columns of A are linearly dependent, W might or might not span F^n. But the columns of A would in this case not form a basis.

    Is there something I've missed here ?
  10. Feb 17, 2008 #9


    User Avatar
    Science Advisor

    If the columns are not independent, then T does not have an inverse and its range is not all of R^n. Since W does span the range of T, it cannot span F^n. The "might" part of "might or might not" is incorrect.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook