Results 1 to 4 of 4

Math Help - Transformation raised to power

  1. #1
    Junior Member
    Joined
    Jun 2011
    Posts
    45

    Transformation raised to power

    This problem comes from our Final Review. It states:

    Suppose V is a vector space over a field F, and T: V-->V is a linear map. Suppose also that, for a vector v in V, that T^(k+1) * v = 0 for some positive integer k. If (T^k) is not equal to zero, show:
    i.) {v, Tv, (T^2)v .. (T^k)v} is linearly independent;
    ii.) The span of {v, Tv, (T^2)v .. (T^k)v} is a T-invariant subspace of V.

    Our professor hasn't been using a textbook and google doesn't understand maths .
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Mar 2011
    From
    Tejas
    Posts
    3,392
    Thanks
    759

    Re: Transformation raised to power

    i)let v be a vector in V such that T^{k+1}(v) = 0 but T^k(v) \neq 0. suppose that c_1v + c_2T(v) + \dots + c_{k+1}T^k(v) = 0.

    taking T^k of both sides, every term but c_1T^k(v) vanishes. since T^k(v) \neq 0, we must have c_1 = 0.

    thus we have c_2T(v) + \dots + c_{k-1}T^k(v) = 0, and now we take T^{k-1} of both sides to show that c_2 = 0.

    continuing in this way, we eventually find that c_1 = c_2 = \dots = c_{k+1} = 0, which shows that \{v,T(v),T^2(v),\dots,T^k(v)\} is linearly independent.

    ii) if w \in \text{span}(\{v,T(v),T^2(v),\dots,T^k(v)\}) , then

    w = c_1v + c_2T(v) + \dots + c_{k+1}T^k(v) so that

    T(w) = T(c_1v + c_2T(v) + \dots + c_{k+1}T^k(v))

    = c_1T(v) + c_2T^2(v) + \dots + c_kT^k(v) (since T^{k+1}(v) = 0),

    which is certainly in \text{span}(\{v,T(v),T^2(v),\dots,T^k(v)\}), hence this span is a T-invariant subspace (if we call it W, T(W) is contained in W).
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor FernandoRevilla's Avatar
    Joined
    Nov 2010
    From
    Madrid, Spain
    Posts
    2,162
    Thanks
    45

    Re: Transformation raised to power

    I'd like to comment the following: if W=\textrm{span}\;[v,T(v),\ldots,T^k(v)] , \widehat{T}:W\to W is the restriction of T to W , and we consider the basis of W given by B=\{T^k(v),\ldots , T(v),v\} then,

    [\;\widehat{T}\;]_B=\begin{bmatrix} 0 & 1 & 0 & \ldots & 0 & 0\\ 0 & 0 & 1 & \ldots & 0 & 0 \\ \vdots&&&&&\vdots \\ 0 & 0 & 0 & \ldots & 0 & 1\\0 & 0 & 0 & \ldots & 0 & 0\end{bmatrix}

    important result in the theory of canonical forms of Jordan.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Mar 2011
    From
    Tejas
    Posts
    3,392
    Thanks
    759

    Re: Transformation raised to power

    yes, since T is nilpotent on W, of degree k+1. while it is not always true that we can find a diagonal matrix D such that a change of basis diagonalizes A, for a given matrix A

    that is PAP^{-1} = D, we CAN find a matrix D and a matrix N, where D is diagonal, and N is nilpotent, such that PAP^{-1} = D+N.

    (well, almost. to be sure we can do so, all the eigenvalues of A, have to lie in the field F, so it is usually assumed one is working within an algebraically closed field,

    such as the complex numbers. that is, the eigenvalues of a REAL matrix, may be complex, so to get the Jordan form, we need to "enlarge the field").

    what happens is, when we look at a factor (x - \lambda)^k of det(xI - A), it may be that we have less than k linearly independent eigenvectors.

    but (λI - A) will be nilpotent, of degree ≤ k, so the block of PAP^{-1} corresponding to the eigenspace E_{\lambda}

    will be "almost diagonal", it will be a diagonal matrix, with some 1's on the super-diagonal.

    a vector v that works as in your problem (where (\lambda I - A)^m(v) \neq 0 but (\lambda I - A)^{m+1}(v) = 0)

    is called a "generalized eigenvector corresponding to λ" (regular eigenvectors correspond to m = 0).
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 9
    Last Post: October 2nd 2011, 12:08 PM
  2. Replies: 2
    Last Post: May 1st 2011, 08:03 AM
  3. Eigenvalues and Eigenvectors raised to a power.
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: March 20th 2010, 12:55 PM
  4. Log raised to a power
    Posted in the Algebra Forum
    Replies: 5
    Last Post: January 5th 2010, 08:09 AM
  5. Trying to integrate x raised to e power
    Posted in the Calculus Forum
    Replies: 3
    Last Post: April 26th 2009, 07:13 PM

Search Tags


/mathhelpforum @mathhelpforum