Regarding Diagonalization of Matrix by Spectral Theorem

JustYouAsk

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?

trambolin

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