<s>[INST] Let $A$ be a diagonal matrix and $B$ be a Hermitian and Idempotent matrix, then is $(I + ABA^H)^{-1} = (I - \frac{1}{2}ABA^H)$? Let $A$ be a diagonal matrix and $B$ be a Hermitian and Idempotent matrix, then is $(I + ABA^H)^{-1} = (I - \frac{1}{2}ABA^H)$, where $I$ is an identity matrix?
I am reading a paper, which states the above equality. I do not understand how this is true. Can someone please enlighten me?
EDIT: Constraint $A^H A = I$ should also be considered
 [/INST] This can be shown by a direct computation as such:
\begin{align}
(I + ABA^H)(I - \frac{1}{2}ABA^H) &= I - \frac{1}{2}ABA^H + \frac{1}{2}(ABA^H)^2 \\
&\stackrel{\spadesuit}{=} I - \frac{1}{2}ABA^* + \frac{1}{2}AB^2A^* \\
&= I - \frac{1}{2}ABA^H + \frac{1}{2}ABA^H = I
\end{align}
Assuming that $A^HA = I$ at $\spadesuit$, otherwise equality isn't assured. See this example:
\begin{align}
&A = \begin{bmatrix}1 & 0 \\ 0 & 2\end{bmatrix} &&B = \frac{1}{2}\begin{bmatrix}1 & 1 \\ 1 & 1\end{bmatrix}
\end{align}
Then:
$$(I + ABA^H)(I - \frac{1}{2}ABA^H) = \begin{bmatrix}\frac{5}{8} & -\frac{3}{4} \\ -\frac{3}{4} & -\frac{1}{2}\end{bmatrix} \neq I$$
</s>