Suppose we have a matrix $g \in M_n(\mathbb{C})$ of the following form $$g = \begin{pmatrix} 0 & 0 & \cdots & 0 & -g_n\\ 1 & 0 & \cdots & 0 & -g_{n-1}\\ 0 & 1 & \cdots & 0 & -g_{n-2}\\ \vdots & \vdots & & \vdots & \vdots\\ 0 & 0 & \cdots & 1 & -g_1\\ \end{pmatrix}$$ with $g_i \in \mathbb{C}$, $g_n \neq 0$. Furthermore, let $X \in M_n(\mathbb{C})$ be a solution of $g^t X \overline{g}=X$. Then i have the following two questions:
- How can i show, that the entries $X=(X_{i,j})$ depend only on $i-j$?
- Assuming 1. is true. Does the same statement hold true for the real case, i.e. $g \in M_n(\mathbb{R})$ of said form with $g_n \neq 0$ and with $X \in M_n(\mathbb{R})$ satisfying $g^t X g=X$?
Regarding the first question, i tried an approach via direct computation. Assuming $g^t X \overline{g}=X$ we have $$ X_{i,j} = (g^t X \overline{g})_{i,j} = \sum_{k=1}^n \left(\sum_{p=1}^n g^t_{i,p} X_{p,k} \overline{g}_{k,j} \right)$$ Then one also verifies, that $$(g^t)_{i,p} = \left\{\begin{array}{lll} -g_{i-p+1}&:& i=n \\ -(i-p)&: & p=i+1 \wedge i<n\\ 0&: & else \end{array}\right.$$ $$(\overline{g})_{k,j} = \left\{\begin{array}{lll} -\overline{g}_{-(k-j)+1}&:& j=n \\ k-j&: & k=j+1 \wedge j<n\\ 0&: & else \end{array}\right.$$ So we can see, that $g^t$ and $\overline{g}$ depends on $i-p$ or $k-j$ respectively, but why does $X$ depend on $i-j$? Help would be really appreciated.
Yes. Since every real matrix is a complex matrix, it suffices to consider the complex case. Let $\{e_1,e_2,\ldots,e_n\}$ be the standard basis of $\mathbb C^n$. When $i<n$, we have $ge_i=\overline{g}e_i=e_{i+1}$. Therefore, when $i,j<n$, $$ x_{ij}=e_i^tXe_j=e_i^t(g^tX\overline{g})e_j=(ge_i)^tX(\overline{g}e_j)=e_{i+1}^tXe_{j+1}=x_{i+1,j+1}. $$ Hence $x_{ij}$ depends only on $i-j$.