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Inner product with (1,1) tensors: Diff. Geometry/ Lin algebra

  1. Apr 22, 2010 #1
    1. The problem statement, all variables and given/known data
    Given g[tex]\equiv g_{ij} = [/tex]
    [-1 0;
    0 1]

    Show that A= [tex]A^{i}_{j}[/tex] =
    [1 2
    -2 1]

    is symmetric wrt innter product g, has complex eigenvalues, but eigenvectros have zero length wrt the complex inner product.


    3. The attempt at a solution

    Im sure this is just a simple linear algebra problem but im having trouble figuring out how to compute the dot procut with this 1,1 tensor.

    My guess would be to break the matrix [tex]A^{i}_{j}[/tex] intro rows and compute { (row1)g(row2)^t} then to show this is symmetric calculate { (row2)g(row1)^t}. But that seems wrong.

    I can calculate the eigenvalues (they come out to [tex]\pm i \sqrt{3}[/tex] )

    Also i am lost on showing the eigenvecotrs are 0 wrt the this inner product. I would have no idea how to approach this even if i knew how to calculate inner product.

    Thanks in advance
     
  2. jcsd
  3. Apr 22, 2010 #2
    The metric g lets you lower the indices of A_i^j to get a tensor A_{ij}. Then the claim is that A_{ij} = A_{ji}. For the other part, I'm not quite sure what inner product they mean. I'm guessing g as a complex inner product?
     
  4. Apr 22, 2010 #3
    Ok but as written, how do you compute that complex inner product?
     
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