Matrix orthogonal if and only if another matrix unitary

Question

For a positive integere $n$, suppose that $C$ and $D$ are real square matrices of size $n$. Let $F$ be defined as $F = C + iD$. And $G$ be defined as $$G=\left[\begin{matrix}C & -D\\D & C\end{matrix}\right]$$. Show that the matrix $G$ is orthogonal if and only if $F$ is unitary.

$F^{*} = C^T - iD^T$ and $FF^{*} = CC^T - iCD^T + iDC^T + DD^T = I = F^{*}F = C^TC + iC^TD - iD^TC + D^TD$ $GG^T = \left[\begin{matrix}C^2 + D^2 & CD -DC\\DC - CD & D^2 +C^2\end{matrix}\right] = G^TG = \left[\begin{matrix}C^2 + D^2 & DC - CD\\CD - DC & D^2 + C^2\end{matrix}\right] = \left[\begin{matrix}1 & 0\\0 & 1\end{matrix}\right]$ $\Rightarrow$ $C^2 + D^2 = 1$ and $CD - DC = DC - CD = 0$

But what should I do after? Suppose, that $FF^{*} = F^{*}F = I$ -- true. How we can imply from here that $G$ -- orthogonal?

And in the opposite direction..

Answer 1

Your computation of $G^T$ is false. You'll have to transpose the matrices inside too. Once you've the right computation for $GG^T$, the rest will be pretty clear to you.

Answer 2

For an alternative solution, notice that

$$\begin{bmatrix}C+iD & 0\\0 & C-iD\end{bmatrix} = \left(\frac1{\sqrt2}\begin{bmatrix} iI & -iI \\ I & I\end{bmatrix}\right)^*\begin{bmatrix}C & -D\\D & C\end{bmatrix}\left(\frac1{\sqrt2}\begin{bmatrix} iI & -iI \\ I & I\end{bmatrix}\right)$$

so $\begin{bmatrix}C & -D\\D & C\end{bmatrix}$ is unitary (i.e. orthogonal) if and only if $\begin{bmatrix}C+iD & 0\\0 & C-iD\end{bmatrix}$ is unitary, which holds if and only if $C+iD$ is unitary.