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Linear Algebra - Linear Operators

  1. Sep 20, 2015 #1
    1. The problem statement, all variables and given/known data

    True or false?
    If T: ℙ8(ℝ) → ℙ8(ℝ) is defined by T(p) = p', so exists a basis of ℙ8(ℝ) such that the matrix of T in relation to this basis is inversible.

    2. Relevant equations

    3. The attempt at a solution

    So i think that my equations is of the form:
    A.x = x'
    hence A is non-singular and therefore is inversible. But my exercise anwser say the opposite.
    Could somebody elucide me?

    P.s: (sorry for bad english, I'm from Brazil and still learning it.)
    Last edited: Sep 20, 2015
  2. jcsd
  3. Sep 20, 2015 #2


    Staff: Mentor

    You need to give a reason that A is nonsingular, if in fact this is a true statement.

    BTW, we say "invertible" if an inverse exists.
    What does this transformation do to the standard basis for Ps? Also, what is Ps? Was this supposed to be P5, the space of polynomials of degree < 5?
  4. Sep 20, 2015 #3
    Hello, my friend! Thanks for your feedback.

    I thought that A is non-singular because if it was, A.x = x' could not be true, because it would have infinite solutions. But thinking right now i don't see how this is true at all.
    No, it's not P5, it's P8. I thought it was a generic polynomial space.
    P8 is the space of real polynomials of degree < 8.
    I'll edit first post.
  5. Sep 20, 2015 #4
    I see why this is false!!!!

    My matrix A is a echeloned matrix with trace 0. So it's det is 0 and it's not invertible.
    Sorry for not posting how i found this. I will try to edit and post my anwser!

    Thanks PF.
  6. Sep 21, 2015 #5


    User Avatar
    Science Advisor

    You seem to have "singular" and "non-singular" reversed! This linear operator is singular precisely because it is not one- to- one.
    If the difference between polynomials p and q is a constant, then Ap= p'= q'= Aq so A inverse of p' is not unique.

  7. Sep 21, 2015 #6
    Oooooh, so logic and true! Thank you.
    So, a matrix is singular if it's determinant is zero hence it have no inverse, is it right?
  8. Sep 21, 2015 #7


    Staff: Mentor

    Regarding the problem in your first post, you need to show that if B is any basis for P8, the matrix for the transformation is not invertible. What you've shown is the matrix in terms of the standard basis is noninvertible.
  9. Sep 23, 2015 #8
    Thanks everyone who helped! I think i figured how to do it and will try soon and post my result!
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