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Eigenpairs and Hermitian matrices

  1. Oct 22, 2009 #1
    1. The problem statement, all variables and given/known data

    Choose [tex]\lambda_{1}, \lambda_{2}, \lambda_{3} [/tex] along with a set of vectors [tex]{v_{1},v_{2},v_{3}}[/tex] and construct an Hermitian matrix H with the eigenpairs [tex] (\lambda_{1},v_{1}),(\lambda_{2},v_{2}),(\lambda_{3},v_{3})[/tex]

    2. Relevant equations



    3. The attempt at a solution
    [tex] \lambda_{1} = 2[/tex]
    [tex]\lambda_{2} = -2[/tex]

    [tex]v_{1} = \widehat{i}+2\widehat{j}[/tex]
    [tex]v_{2} = \widehat{i}-2\widehat{j}[/tex]

    [tex]u_{1} = \frac{1}{\sqrt{5}}\left(\widehat{i} + 2\widehat{j} \right)[/tex]
    [tex]u_{2} = \frac{1}{\sqrt{5}}\left(\widehat{i} - 2\widehat{j}\right)[/tex]

    [tex]v_{3} = u_{1} \times u_{2} = -\frac{4}{5}\widehat{k}[/tex]

    [tex]u_{3} = -\widehat{k}[/tex]

    [tex]Basis: \ U = \left\{ u_{1},u_{2},u_{3} \right}\}[/tex]

    [tex]H = \left| \begin{array}{ccc} 2 & 0 & 0 \\ 0 & -2 & 0 \\ 0 & 0 & 1 \end{array} \right| [/tex]

    How does this look?
     
  2. jcsd
  3. Oct 23, 2009 #2

    lanedance

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    can you choose any values here?

    why not just
    [tex] \lambda_{1} = 1 [/tex]
    [tex] \lambda_{2} = 2 [/tex]
    [tex] \lambda_{3} = 3 [/tex]

    and
    [tex] u_{1} = (1,0,0)[/tex]
    [tex] u_{2} = (0,1,0)[/tex]
    [tex] u_{3} = (0,0,1)[/tex]

    then
    [tex]H = \left| \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 2& 0 \\ 0 & 0 & 3 \end{array} \right| [/tex]
     
  4. Oct 23, 2009 #3
    If that's all this is, am I misunderstanding or is this a really trivial question?
     
  5. Oct 23, 2009 #4

    lanedance

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    maybe it asking if you are given arbitrary
    [tex] u_1, u_2, u_3 [/tex]
    and corresponding eigenvalues
    [tex] \lambda_1, \lambda_2, \lambda_3 [/tex]
    can you find a hermitian matrix H, with the given eigenpairs?

    one way you could do is assume you have found H and it can be diagonalised (which you know it can, from the eigenvectors), then the diagonal form of H is
    [tex] D = P^{-1}.H.P[/tex]

    where
    [tex]D = \left| \begin{array}{ccc} \lambda_1 & 0 & 0 \\ 0 & \lambda_2& 0 \\ 0 & 0 & \lambda_3 \end{array} \right| [/tex]

    now the matrix P is simply given by the eigenvectors
    where
    [tex]P = \left| \begin{array}{ccc} u_1, u_2, u_3 \end{array} \right| [/tex]

    then H is
    [tex] H = PDP^{-1}[/tex]

    that gives the eigenpairs, however I doubt it would automatically be hermitian... for that you need, where the star means conjugate transpose
    [tex] H = H^* [/tex]


    so running that through
    [tex] PDP^{-1} = (PDP^{-1})^* = P^{-1}^*D^*P^* = P^{-1}^*DP^* [/tex]

    hmmm... not totally sure where to go now
     
    Last edited: Oct 23, 2009
  6. Oct 23, 2009 #5

    lanedance

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    maybe a little left & right muliplication
    [tex] D = P^{-1}(PDP^{-1})P = (P^{-1}P^{-1}^*)D(P^*P) [/tex]

    this will be satisfied if P is unitary
    [tex]P^*P = PP^* = I[/tex]
     
  7. Oct 23, 2009 #6

    lanedance

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    from before P is given by the eigenvectors (choose up to some multiplicative constant)
    [tex]P = \left| \begin{array}{ccc} u_1, u_2, u_3 \end{array} \right| [/tex]

    so in subscript form if j is the eigenvector and i is it's component, write
    [tex]P = P_{ij} = u_{j}^{(i)}[/tex]

    now for the transpose conjugate of a matrix
    [tex]P^* = (P^*)_{ij} = \bar{P}_{ji} = \bar{u}_{i}^{(j)} [/tex]

    matrix multipication is given by (with summation over the repeated indice)
    [tex]P^*P = (P^*)_{ij}P_{jk} = \bar{u}_{i}^{(j)} u_{k}^{(j)} [/tex]
    which looks just like a complex inner product between eigenvectors

    so to satisfy the unitary condition we were looking for
    [tex]PP^* = I \rightarrow \bar{u}_{i}^{(j)} u_{k}^{(j)} = \delta_{ik} [/tex]

    ie. all the columns of P must form an orthonormal basis, which is a property of unitary matricies in general. So the condition will be satisfied if you just normalise your eigenvectors to length 1.

    got a bit carried away there... but from this you should be able to write the general form of H no worries
     
  8. Oct 24, 2009 #7
    Ok that helps. I guess I got confused and thought he was looking for D, but I think you're right; I think he's looking for PDP-1

    Thanks!
     
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