Matrix equation $iA = AB$

Question

Let $A \in \mathrm{M}_{n, 2n}(\mathbb{C})$ be a matrix which rows are linearly independent over $\mathbb{R}$. Let $$ B := \begin{pmatrix} A \\ \overline{A} \end{pmatrix}^{-1} \begin{pmatrix} i 1_n & 0 \\ 0 & i 1_n \end{pmatrix} \begin{pmatrix} A \\ \overline{A} \end{pmatrix},$$ where $\overline{A}$ is the complex conjugate of $A$. I want show that $i A = AB$. Does anybody know how to do this?

I know that $B = \begin{pmatrix} A \\\ \overline{A} \end{pmatrix}^{-1} \begin{pmatrix} i A \\\ \overline{i A} \end{pmatrix}$, but I don´t know how to proceed.

Answer 1

$$ B = \begin{pmatrix} A \\\ \overline{A} \end{pmatrix}^{-1} \begin{pmatrix} i A \\\ \overline{i A} \end{pmatrix}\\ \begin{pmatrix} A \\\ \overline{A} \end{pmatrix} B = \begin{pmatrix} i A \\\ \overline{i A} \end{pmatrix}\\ \begin{pmatrix} A B \\\ \overline{A} B \end{pmatrix} = \begin{pmatrix} i A \\\ \overline{i A} \end{pmatrix}\\ $$

Reading the top $n$ lines shows $AB = iA$.