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Linear Algebra

  1. Aug 13, 2009 #1
    1. The problem statement, all variables and given/known data
    let [tex]T:V \to V[/tex] be a linear transformation which satisfies [tex]T^2 = \frac{1}{2} (T + T^*) [/tex] and is normal. Prove that [tex]T=T^*[/tex].


    2. Relevant equations



    3. The attempt at a solution
    I think we should start like this:
    Let [tex]\mathbf{A}=[T]_B[/tex] be the matrix representation of T in the orthonormal base [tex]B[/tex]. We look at A as a matrix in the Complex plane. If we prove that for every Eigenvalue, [tex]\lambda[/tex], it happens that [tex]\lambda \in \mathbb{R}[/tex], then [tex]T = T^*[/tex] by a theorem and we finished.
    Now, be [tex]\mathbf{x}[/tex] an Eigenvector that satisfies [tex]\mathbf{Ax} = \lambda \mathbf{x}[/tex], then [tex]<\mathbf{Ax},\mathbf{x}> = \lambda ||\mathbf{x}||^2[/tex] and [tex] \lambda = \frac{<\mathbf{Ax},\mathbf{x}>}{||\mathbf{x}||^2}[/tex]. If only I could show that [tex]<\mathbf{Ax},\mathbf{x}>[/tex] is a real number, I solved it.
     
  2. jcsd
  3. Aug 13, 2009 #2
    Prove that the transformation is hermitian, once you do it's pretty east to prove that eigenvalues of a hermitian transformation are real.
     
  4. Aug 13, 2009 #3
    It doesn't help me, because I plan to show that the transformation is Hermitian by the following theorem:
    "If T is a normal transformation whose Characteristic polynomial can be completely factored into linear factors over [tex]\mathbb{R}[/tex], then T is Hermitian".
    And then it follows, of course, that T = T*
     
  5. Aug 13, 2009 #4
    Can someone help me please? I work on it two days. Nothing works. Please. I spent hours.
     
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