Does the inverse of a matrix with complex conjugate rows have complex conjugate columns?

Question

Let $A \in \mathbb{C}^{2n \times 2n}$ be an invertible matrix which consists of $n$ rows and their $n$ complex conjugates (in any random order). Is it true that $A^{-1}$ will consist of $n$ columns and their $n$ complex conjugates?

Answer 1

Yes. By assumption, the set of indices $\{1,2,\ldots,2n\}$ can be partitioned into two disjoint sets $\{i_1,i_2,\ldots,i_n\}$ and $\{j_1,j_2,\ldots,j_n\}$ such that for each $k$, the $i_k$-th row of $A$ is the complex conjugate of the $j_k$-th row. Therefore, if $P$ is the permutation matrix that exchange the $i_k$-th row with the $j_k$-th row for each $k$, then $PA=\overline{A}$. But then $A^{-1}P^T=(\overline{A})^{-1}=\overline{A^{-1}}$, i.e. the $i_k$-th column of $A^{-1}$ is the complex conjuage of the $j_k$-th column for each $k$.