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$$^{-1}$$AP is diagonal, i.e. P$$^{T}$$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 $A$ is Hermitian then $U$ is also Hermitian, then use the transformation,

$$\left[ {\begin{array}{*{20}c}<br /> I & 0 & 0 \\<br /> 0 & I & 0 \\<br /> { - I} & 0 & I \\<br /> \end{array}} \right]\left[ {\begin{array}{*{20}c}<br /> U & 0 & U \\<br /> 0 & 0 & 0 \\<br /> U & 0 & U \\<br /> \end{array}} \right]\left[ {\begin{array}{*{20}c}<br /> I & 0 & { - I} \\<br /> 0 & I & 0 \\<br /> 0 & 0 & I \\<br /> \end{array}} \right] = \left[ {\begin{array}{*{20}c}<br /> U & 0 & 0 \\<br /> 0 & 0 & 0 \\<br /> 0 & 0 & 0 \\<br /> \end{array}} \right]<br />$$

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 $U$ now