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Proving a linear transform is diagonalizable

  1. May 10, 2009 #1
    1. The problem statement, all variables and given/known data

    Let U be the R-vector space consisting of all polynomials of degree at most n with
    coefficients that are real.

    The Derivative Map
    F : U [tex]\rightarrow[/tex] U
    f(x) [tex]\rightarrow[/tex] f'(x)

    Is the derivative function F diagonalizable?

    3. The attempt at a solution

    My instinct says yes but I'm not too sure why. Am I right in saying that eigenvalues are invariant after a linear transform. I'm pretty sure that before the function is applied you can have a matrix that has n+1 distinct eignenvalues but not sure if this is the right lines or if it is how to set about writing it down in a proof format.

    Any help would be great thanks !
  2. jcsd
  3. May 10, 2009 #2


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    Can you be more precise in what you mean? Eigenvalues of what? What linear transform?
  4. May 10, 2009 #3
    The linear transform is differentiation function.

    The eigenvalues I'm not so sure of , the main reason I say it is because in my notes it seems to be the only thing remotely to do with this question.

    In the following deffinitions I have , I assume he means A is the vector space you start with

    Let A be an (nxn) matrix, A is diagonalisable by similarity transform if and only if there is a basis for R^n consisting of entirely eigenvectors for A.


    If A is diagonisible then

    the sum of the distinct eigenvalues for dimension of the eigenspace = n

    Oh and I also know that eigenvalues are similarity invariant but their eigenvector aren't (altho they have a 1 to 1 correspondance after the function is applied)

    Sorry I'm pretty confused at how to go about this, so I'm sorry if it doesn't make sense
  5. May 10, 2009 #4


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    No, he meant what he said: A is an nxn matrix. Diagonalizability is not a property of vector spaces; it is a property of matrices (and of linear transformations).

    (This n, of course, does not necessarily have anything to do with the n in your original problem)

    This doesn't make sense. I know the theorem you're trying to state, but the idea you have in your head doesn't make sense.

    If a matrix is diagonalizable, then:

    It have a set of (distinct) eigenvalues.
    Each of those eigenvalues has a corresponding eigenspace.
    Each of those eigenspaces has a dimension.
    The sum of all of those dimensions is n.

    It's often useful to start with special cases. Try working through the n=0 case. Then the n=1 case. Mabe even the n=2 case. Only then worry about the general case.
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