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

Regarding Diagonalization of Matrix by Spectral Theorem

  1. Mar 4, 2008 #1
    According to the spectral theorem for self-adjoint operators you can find a matrix P such that P[tex]^{-1}[/tex]AP is diagonal, i.e. P[tex]^{T}[/tex]AP (P can be shown to be orthogonal). I'm not sure what the result is if the same can be done for the following square (size n X n) and symmetric matrix of the form:
    A=
    [ U 0 U ]
    [ 0 0 0 ]
    [ U 0 U ]

    where U is square matrix and 0 is a matrix of zeros.

    If I am not mistaken the solution is that the columns of P are simply the eigenvectors of A??? can anyone confirm this?
     
  2. jcsd
  3. Mar 5, 2008 #2
    First if [itex]A[/itex] is Hermitian then [itex]U[/itex] is also Hermitian, then use the transformation,

    [tex]
    \left[ {\begin{array}{*{20}c}
    I & 0 & 0 \\
    0 & I & 0 \\
    { - I} & 0 & I \\
    \end{array}} \right]\left[ {\begin{array}{*{20}c}
    U & 0 & U \\
    0 & 0 & 0 \\
    U & 0 & U \\
    \end{array}} \right]\left[ {\begin{array}{*{20}c}
    I & 0 & { - I} \\
    0 & I & 0 \\
    0 & 0 & I \\
    \end{array}} \right] = \left[ {\begin{array}{*{20}c}
    U & 0 & 0 \\
    0 & 0 & 0 \\
    0 & 0 & 0 \\
    \end{array}} \right]

    [/tex]

    Now, we are back in the business because now it is the case where you start the argument, diagonalization of a self adjoint operator where we diagonalize [itex]U[/itex] now
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Regarding Diagonalization of Matrix by Spectral Theorem
  1. Diagonal of a Matrix (Replies: 2)

Loading...