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Inner product propety with Scalar Matrix (Proof)

  1. Dec 1, 2015 #1


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    1. The problem statement, all variables and given/known data
    Let A be an nxn matrix, and let |v>, |w> ∈ℂ. Prove that (A|v>)*|w> = |v>*(A†|w>)

    † = hermitian conjugate

    2. Relevant equations

    3. The attempt at a solution
    Struggling to start this one. I'm sure this one is likely relatively quick and painless, but I need to identify the trick behind it.

    Are there any tips?

    My only current thought is to use the properties of inner products. Should I look at each element of the matrix A one by one?
  2. jcsd
  3. Dec 1, 2015 #2


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    How is this for a solution:

    Writing in sum notation, we note that matrix A has corresponding eigenvalue such that A|v_i> = t|v_i>

    We replace it in the left hand side so that
    t|v_i>|w_i> = |v_i>A†|w_i>

    Now, since the hermitian conjugate takes the transpose and the complex conjugate, we see that

    A^T^*|w_i> = t*|w_i>

    Applying properties of inner product spaces, to take the scalar t* out of the right hand side, we take its complex conjugate.

    So we are left with
    t|v_i>|w_i> on both sides
  4. Dec 1, 2015 #3


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    The problem that I am having right now is this:

    A^T^* (|w_i>) = t*|w_i>
    where T is transpose and * is complex conjugate

    I know that the eigenvalues of a matrix and it's transpose are equal. But that does mean that the eigenvalues of a complex matrix and its conjugate are off by conjugation?

    If so, this proof is concluded.
  5. Dec 2, 2015 #4


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    I don't understand why you are assuming that |v> is an eigenvector of A.

    I would do the proof by considering the elements of the matrix and the vectors. For instance, defining |u> = A|v>, we have
    u_i = \sum_j A_{ij} v_j
    and so on.
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