Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Something about hermitian matrixes

  1. Jun 19, 2005 #1
    Hi all,

    I don't understand to one part of proof of this theorem:

    All eigenvalues of each hermitian matrix A are real numbers and, moreover, there exists unitary matrix R such, that

    [tex]
    R^{-1}AR
    [/tex]

    is diagonal


    Proof: By induction with respect to n (order of matrix A)

    For n = 1 it's obvious.
    Suppose that the theorem holds for 1, 2, ..., n-1
    We know that [itex]\exists[/itex] eigenvalue [itex]\lambda[/itex] and appropriate eigenvector [itex]x \in \mathbb{C}[/itex].
    Using Steinitz's theorem, we can extend [itex]x[/itex] to orthonormal base of [itex]\mathbb{C}^{n}[/itex].
    Suppose that [itex]||x|| = 1[/itex] and construct matrix [itex]P_n[/itex] from vectors of this base ([itex]P_n[/itex] will have these vectors in its columns).

    [itex]P_n[/itex] is unitary [itex]\Leftarrow P_{n}^{H}P_n = I[/itex], because standard inner product of two different vectors in the orthonormal base is zero and inner product of two identical vectors is 1.

    This holds:

    [tex]
    \left(P_{n}^{H}A_{n}P_{n}\right)^{H} = P_{n}^{H}A_{n}^{H}\left(P_{n}^{H}\right)^{H} = P_{n}^{H}A_{n}P_{n}
    [/tex]

    Last line is what I don't understand, probably it's trivial but I can't see that

    [tex]
    \left(P_{n}^{H}A_{n}P_{n}\right)^{H} = \left(P_{n}^{H}\right)^{H}A_{n}^{H}P_{n}^{H} = P_{n}^{H}A_{n}^{H}\left(P_{n}^{H}\right)^{H}
    [/tex]

    (the second equality)

    Thank you for the explanation.
     
  2. jcsd
  3. Jun 19, 2005 #2

    matt grime

    User Avatar
    Science Advisor
    Homework Helper

    because you're forgetting that taking the daggeer reverses the order of the matrices i'l; use start instead, but (AB)*=B*A*

    the second equality as you have it is wrong, but then it isn't supposed to be true.
     
  4. Jun 19, 2005 #3
    Thank you a lot Matt, I was looking at it for ten minutes and it's as simple as normal transposition :rolleyes:
     
  5. Jun 19, 2005 #4

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

    Well,the first part (the real values of eigenvalues of hermitean operators) can be proven quite easily for a hermitean linear operator defined on a dense everywhere subset of a separable Hilbert space.

    Daniel.
     
  6. Jun 19, 2005 #5

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

    The key to the second part is to remark that that matrix

    [tex] M^{\dagger}AM \ ,\ M\in U(n,\mathbb{C})[/tex]

    is hermitean,which means that the linear operator associated is hermitean.A hermitean linear operator in a finite dimensional complex Hilbert space admits a spectral decomposition (moreover,the spectrum is purely discrete),which means that the operator [itex] M^{\dagger}AM [/itex] has zero off-diagonal matrix elements.

    Daniel.
     
  7. Jun 19, 2005 #6
    Thank you Daniel for this explanation, but I don't have a clue what Hilbert space is (I only heard of it) and what hermitean linear operator is.

    However, I've been studying the proof on and I again encountered place I don't understand to.

    If I continue from where I finished my initial post:

    ...

    And thus [itex]P_{n}^{H}A_{n}P_{n}[/itex] is hermitian matrix.

    Next,

    [tex]
    \left( \begin{array}{cc} \lambda & 0 ..... 0 \\ 0 & A_{n-1} \\ 0 \end{array} \right)
    [/tex]

    Because this matrix is equal to its hermitian transposition, [itex]\lambda \in \mathbb{R}[/itex]

    // I'm not sure why this matrix is here and whether it should mean that it is the matrix [itex]P_{n}^{H}A_{n}P_{n}[/itex], I really don't know...anyway, let's continue

    From induction presumption [itex]\exists[/itex] unitary matrix [itex]R_{n-1}[/itex] such, that

    [tex]
    R_{n-1}^{-1}A_{n-1}R_{n-1} = D_{n-1}
    [/tex]

    Let's take

    [tex]
    S = \left(\begin{array}{cc} 1 & 0 ..... 0 \\ 0 & R_{n-1} \\ 0 \end{array} \right)
    [/tex]

    [tex]
    R_n = P_{n}S
    [/tex]

    S is unitary, as well as [itex]P_{n}[/itex]. Is also their product unitary? (In another words, is product of two unitary matrixes unitary matrix?) Let's see.

    [tex]
    R_{n}^{H}R_{n} = \left(P_{n}S\right)^{H}P_{n}S = S^{H}P_{n}^{H}P_{n}S = I
    [/tex]

    So, it holds that [itex]R_{n}[/itex] is also unitary. Is [itex]R_{n}[/itex] the matrix we're looking for?

    [tex]
    R_{n}^{-1}A_{n}R_{n} = \left(P_{n}S\right)^{H}AP_{n}S = S^{H}P_{n}^{H}AP_{n}S = \left(\begin{array}{cc} 1 & 0 ..... 0 \\ 0 & R_{n-1}^{H} \\ 0 \end{array} \right)
    \left(\begin{array}{cc} \lambda & 0 ..... 0 \\ 0 & A_{n-1} \\ 0 \end{array} \right)
    \left(\begin{array}{cc} 1 & 0 ..... 0 \\ 0 & R_{n-1} \\ 0 \end{array} \right) =
    \left(\begin{array}{cc} \lambda & 0 ..... 0 \\ 0 & D_{n-1} \\ 0 \end{array} \right) = D
    [/tex]

    Q.E.D

    What I don't understand is that according to this,

    [tex]
    P_{n}^{H}AP_{n} = \left(\begin{array}{cc} \lambda & 0 ..... 0 \\ 0 & A_{n-1} \\ 0 \end{array} \right)
    [/tex]

    Why?

    Thank you.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Something about hermitian matrixes
  1. Hermitian Matrix (Replies: 1)

Loading...