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Homework Help: Let f1(t)=e^t, f2(t)=te^t, f3(t)=t^2e^t, and let V=Span(f1,f2,f3) in

  1. Apr 11, 2012 #1
    Let f1(t)=e^t, f2(t)=te^t, f3(t)=t^2e^t, and let V=Span(f1,f2,f3) in the infinite continuous functions. Let T:V-->V be give by T(f)=f''-2f'+f. Decide whether T is diagonalizable.

    We learned a theorem that this will be diagonalizable if and only if the geometric multiplicity of each eigenvalue equals its algebraic multiplicity.

    What I am having trouble with is translating this into a way to find geometric and algebraic multiplicity. I'm not entirely sure what to do when I'm not given a matrix, since that's how we did it in class.
  2. jcsd
  3. Apr 11, 2012 #2
    Re: Diagonalization

    Try to find the matrix associated to this linear map.

    Choose a basis of V, and try to construct the matrix.
  4. Apr 11, 2012 #3
    Re: Diagonalization

    So would I choose f1, f2, f3 as a basis for V and then construct a matrix
    [f1 f2 f3]? Something like that?
  5. Apr 11, 2012 #4
    Re: Diagonalization

    You need to construct a matrix with respect to the basis [itex](f_1,f_2,f_3)[/itex]. Do you know how to do that?
  6. Apr 12, 2012 #5
    Re: Diagonalization

    I know I should know how to do that, but I get tripped up in functions, especially involving e. After looking at it, I can't figure it out.
  7. Apr 12, 2012 #6


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    Re: Diagonalization

    You are given the basis vectors for a space, V, and a linear transformation from V to itself. To find the matrix representation of the linear transformation, apply the transformation to each basis vector, in turn, and write the result as a linear combination of the basis vectors. The coefficients give a column of the matrix.

    Here, the given basis is [itex]\{e^t, te^t, t^2e^t\}[/itex] and the linear transformation is T(f)= f''- 2f'+ f. Applying that to, say, [itex]t^2e^t[/itex], gives [itex]2e^t= 2(e^t)+ 0(te^t)+ 0(t^2e^t)[/itex] so that the third column of the matrix is
    [tex]\begin{bmatrix}2 \\ 0 \\ 0\end{bmatrix}[/tex]
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