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Linear Algebra - Jordan forms

  1. Aug 22, 2008 #1

    daniel_i_l

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    1. The problem statement, all variables and given/known data
    V is a unitarian space of finite dimensions and T:V->V is a linear transformation.
    Every eigenvector of T is an eigenvector of T* (where (Tv,u) = (v,T*u) for all u and v in V).
    Prove that T(T*) = (T*)T.


    2. Relevant equations



    3. The attempt at a solution
    First of all, since the space in unitarian both T and T* can be expressed as a jordan matrix. It's easy to show that if J is the jordan matrix of T then [tex]\bar{J}[/tex] is the jordan matrix of T*. My idea is that since every eigenvector of T is an eigenvector of T* then there exists some base of V where T is expressed as J and T* is expressed as [tex]\bar{J}[/tex]. If that where true than it would be easy to answer the question
    (Since then there would be a matrix M to that [tex] [T] = M^{-1}JM [/tex] ,
    [tex][T^{*}] = M^{-1} \bar{J} M [/tex] and then
    [tex] [T^{*}][T] = M^{-1} \bar{J} M M^{-1} J M = M^{-1} \bar{J} J M =
    J M^{-1} \bar{J} M = [T][T^{*}] [/tex]
    but I can't prove that such a base exists. In all the examples I've tried there's a base like that.
    Is this the right direction? If so, how do I prove that a base exists?
    Is there a better way to approach the problem?
    Thanks.
     
  2. jcsd
  3. Aug 22, 2008 #2

    morphism

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    Could you clarify what you mean by a unitarian space?
     
  4. Aug 23, 2008 #3

    daniel_i_l

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    A unitarian space is a vector space over the complex field with a defined inner-product.
     
  5. Aug 27, 2008 #4

    morphism

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    In that case it would be better to use Schur decomposition rather than Jordan decomposition. That is, get a unitary matrix U such that U*TU is upper triangular. Note that T commutes with T* iff U*TU commutes with (U*TU)*=U*T*U. Moreover, note that x is an eigenvector of T iff U*x is an eigenvector of U*TU. Thus we may assume without loss of generality that T itself is upper triangular. In this case T has (1,0,..,0) as an eigenvector. Consequently, so does T*. Proceed inductively to conclude that T must be diagonal. (We've essentially 'chopped off' the upper left corners of both T and T*.)
     
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