<s>[INST] If $P^{-1}(D + A)P$ is a diagonal matrix, with $D$ a diagonal matrix and $D+A$ hermitian, is $P^{-1}AP$ also diagonal? Suppose $D+A$ is hermitian matrix, and $D$ is a diagonal matrix. Then $A$ is a hermitian matrix as well.
Since $D+A$ is a hermitian matrix, there exists $P$ such that $P^{-1}(D+A)P$ is a diagonal matrix.
The question is, how do I prove that same $P$ also diagonalizes $A$ such that $P^{-1}AP$ is a diagonal matrix? Or is there a counterexample?
 [/INST] Here is an counter-example. Put \begin{align*} A = \begin{bmatrix} 3 &  i \\ -i & 1 \end{bmatrix},  D= \begin{bmatrix} -2 &  0 \\ 0 & 0 \end{bmatrix} \end{align*}
We can find an unitary matrix $U = \frac{1}{\sqrt{2}}  \begin{bmatrix} 1 &  1 \\ i & -i \end{bmatrix} $ such that $$U^{-1}(D+A)U = \frac{1}{2}
\begin{bmatrix} 1 &  -i \\ 1 & i \end{bmatrix} \begin{bmatrix} 1 &  i \\ -i & 1 \end{bmatrix}\begin{bmatrix} 1 &  1 \\ i & -i \end{bmatrix}=  \begin{bmatrix} 0 & 0 \\ 0 & 2 \end{bmatrix}$$
However, $$U^{-1}AU = \frac{1}{2}
\begin{bmatrix} 1 &  -i \\ 1 & i \end{bmatrix} \begin{bmatrix} 3 &  i \\ -i & 1 \end{bmatrix}\begin{bmatrix} 1 &  1 \\ i & -i \end{bmatrix}=  \begin{bmatrix} 1 & 1 \\ 1 & 3 \end{bmatrix}$$
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