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Math Help - Testing for diagonalizability and then finding a basis beta for V

  1. #1
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    Testing for diagonalizability and then finding a basis beta for V

    For the linear operator T on a vector space V, test for diagonalizibility and if T is diagonalizable then find a basis beta for V such that [T]_beta is a diagonal matrix.

    V=P_3(R) and T is defined by T(f(x))=f'(x)+f"(x).

    so far i have:

    {1,x,x^2,x^3}: T(1)=0, T(x)=1, T(x^2)=2x+2, T(x^3)=3x^2+6x

    and then I found [T]_beta to be the matrix
    0120
    0026
    0003
    0000

    I found that the characteristic polynomial is (0-t)^4 so the eigenvalues are t=0 (Im using t instead of lamda because I always mess up the latex) so t=0 has multiplicity of 4 and since everything factors we know it splits and then it is NOT diagonalizable because n-rank(A-tI)=1 and 1 does not equal 4 or is there something else that i'm missing. I know those are the two tests we need to look for, but I found a different answer that looks weird to what I found in terms of diagonalizibilty.
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  2. #2
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    Quote Originally Posted by alice8675309 View Post
    For the linear operator T on a vector space V, test for diagonalizibility and if T is diagonalizable then find a basis beta for V such that [T]_beta is a diagonal matrix.

    V=P_3(R) and T is defined by T(f(x))=f'(x)+f"(x).

    so far i have:

    {1,x,x^2,x^3}: T(1)=0, T(x)=1, T(x^2)=2x+2, T(x^3)=3x^2+6x

    and then I found [T]_beta to be the matrix
    0120
    0026
    0003
    0000

    I found that the characteristic polynomial is (0-t)^4 so the eigenvalues are t=0 (Im using t instead of lamda because I always mess up the latex) so t=0 has multiplicity of 4 and since everything factors we know it splits and then it is NOT diagonalizable because n-rank(A-tI)=1 and 1 does not equal 4 or is there something else that i'm missing. I know those are the two tests we need to look for, but I found a different answer that looks weird to what I found in terms of diagonalizibilty.

    Looks fine to me...you could also mention that as T is nilpotent then it is diagonalizable iff it is the zero matrix (why? Look at the minimal polynomial)

    Tonio
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  3. #3
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by alice8675309 View Post
    ... or is there something else that i'm missing
    You are completely right, f is not diagonalizable and its canonical form is:

    J=\begin{bmatrix}{0}&{1}&{0} &{0}\\{0}&{0}&{1}&{0}\\{0}&{0}&{0}&{1}\\{0}&{0}&{0  }&{0}\end{bmatrix}

    Regards.

    Fernando Revilla
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