Linear Algebra - Jordan forms

[daniel_i_l]

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 \bar{J} 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 \bar{J}. If that where true than it would be easy to answer the question
(Since then there would be a matrix M to that [T] = M^{-1}JM ,
[T^{*}] = M^{-1} \bar{J} M and then
[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^{*}]
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.

[morphism]

Could you clarify what you mean by a unitarian space?

[daniel_i_l]

A unitarian space is a vector space over the complex field with a defined inner-product.

[morphism]

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*.)