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Math Help - Hermitian Eigenvalues

  1. #1
    Senior Member sfspitfire23's Avatar
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    Hermitian Eigenvalues

    Let v be the column vector [1-i ; -3 ; -2i ; 0] and suppose v is an eigenvector of a Hermitian matrix M with eigenvalue -3. Calculate v^H M

    What I did-

    We know by definition Mv = -3v

    Now, M = M^H because M is Hermitian. Then,

    M^Hv = (v^H M)^H = conj(v^H M).

    So, v^H M = conj(-3v)

    Thoughts?

    Thanks!
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  2. #2
    MHF Contributor

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    Re: Hermitian Eigenvalues

    i get:

    vHM = vHMH = (Mv)H = (-3v)H = (-3)(1+i,-3,2i,0) = (-3-3i,9,-6i,0) (row-vector).

    (since we're already using _H for the conjugate-transpose, why not keep using it?).
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  3. #3
    Senior Member sfspitfire23's Avatar
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    Re: Hermitian Eigenvalues

    Ah. Does my logic work going the other way?

    So you started from v^H M ....does what I did starting from Mv make sense too?
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