1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook