# Regarding Diagonalization of Matrix by Spectral Theorem

1. Mar 4, 2008

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?

2. Mar 5, 2008

### trambolin

First if $A$ is Hermitian then $U$ is also Hermitian, then use the transformation,

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

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