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Generalized Eigenspace and JOrdan Form

  1. May 25, 2010 #1
    1. The problem statement, all variables and given/known data
    For each linear operator T, find a basis for each generalized eigenspace of T consisting of a union of disjoint cycles of generalized eigenvectors. The find a Jordan canonical form J of T.

    a) T is the linear operator on P2(R) defined by T(f(x)) = 2f(x) - f '(x)


    2. Relevant equations



    3. The attempt at a solution

    OK, so I know the matrix rep of this transformation on the standard basis {1, x, x2}:
    T(1) = 2
    T(x) = -1 + 2x
    T(x2) = -2x + 2x2

    [tex][T]_{\beta}[/tex] =
    2 -1 0
    0 2 -2
    0 0 2 and the eigenvalue of this matrix is [tex]\lambda[/tex] = 2 with a multiplicity of 3.

    I know the JOrdan form will be :

    2 1 0
    0 2 1
    0 0 2 , just not sure how to get the basis or how to get the J from the [T]
     
  2. jcsd
  3. May 25, 2010 #2

    HallsofIvy

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    Staff Emeritus
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    Any eigenvector corresponding to eigenvalue 2 must satisfy
    [tex]\begin{bmatrix}2 & -1 & 0 \\0 & 2 & -2 \\ 0 & 0 & 2\end{bmatrix}\begin{bmatrix}x \\ y \\ z \end{bmatrix}[/tex]
    [tex]= \begin{bmatrix}2a- b & 2b- 2c & 2c\end{bmatrix}=\begin{bmatrix} 2x \\ 2y \\ 2z\end{bmatrix}[/tex]

    So we have 2a- b= 2a, 2b- 2c= 2b, and 2c= 2c. The first says that b= 0 and the second that c= 0. All eigenvectors are multiples of [tex]\begin{bmatrix}1 \\ 0 \\ 0\end{bmatrix}[/tex].

    The characteristic equation for this operator is [itex](\lambda- 2)^3= 0[/itex] and, since a matrix always satifies its own characteristic equation, we must have [itex](T- 2I)^3v= 0[/itex] for all vectors v. Of course if Tv= 2v, that is if v is an eigenvector, the (T- 2I)v= 0 so [itex](T- 2I)^3v= 0[/itex]. But the eigenvectors for this T only span a one-dimensional subspace.

    We need to find v such that (T-I)v is not 0 but is in that eigen-space: then we would have [itex](T- I)^2((T-I)v)= (T-I)^2v= 0[/itex]- a "generalized eigenvector".

    That is, we look for x, y, z such that
    [tex]\begin{bmatrix}0 & -1 & 0 \\ 0 & 0 & -2 \\ 0 & 0 & 0\end{bmatrix}\begin{bmatrix} x \\ y \\ z \end{bmatrix}[/tex][tex]= \begin{bmatrix} -y \\ -2z \\ 0 \end{bmatrix}= \begin{bmatrix}1 \\ 0 \\ 0\end{bmatrix}[/tex].

    It is easy to see that
    [tex]\begin{bmatrix}0 \\ -1 \\ 0\end{bmatrix}[/tex]
    works.

    But now we need a vector such that neither (T- 2I)v= 0 nor [itex](T- 2I)^2v= 0[/itex] but such that [itex](T- 2I)^3v= 0[/itex]. Now we need [itex](T- 2I)^2v[/itex] to be an eigenvector which will be the case if
    [tex]\begin{bmatrix}0 & -1 & 0 \\ 0 & 0 & -2 \\ 0 & 0 & 0\end{bmatrix}\begin{bmatrix}x \\ y \\ z\end{bmatrix}[/tex][tex]= \begin{bmatrix}-y \\ -2z \\ 0\end{bmatrix}= \begin{bmatrix}0 \\ -1 \\ 0\end{bmatrix}[/tex]

    That gives
    [tex]\begin{bmatrix}0 \\ 0 \\ \frac{1}{2}\end{bmatrix}[/tex]

    Those three vectors form the basis. If P is the matrix having those vectors as columns, then [itex]P^{-1}TP[/itex] will be in Jordan Normal Form.
     
    Last edited: May 25, 2010
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