Q
Show tha tif $Q= Q_1 + j Q_2$ is unitary where $Q_i\in \mathbb R^{m \times m}$, then the $2m \times 2m$ matrix
$$
Z = \begin{bmatrix}
Q_1 & -Q_2 \\
Q_2 & Q_1
\end{bmatrix}
$$
is orthogonal.
A
The distinction between orthogonal and unitary is usually that orthogonal implies that the transpose is the inverse while unitary implies that the hermitian is the inverse.
$$
\begin{align*}
Z^T Z &=
\begin{bmatrix}Q_1^T & Q_2^T \\ -Q_2^T & Q_1^T\end{bmatrix}
\begin{bmatrix}
Q_1 & -Q_2 \\
Q_2 & Q_1
\end{bmatrix} \\
&= \begin{bmatrix}
Q_1^TQ_1 + Q_2^TQ_2 & -Q_2^TQ_2 \\
-Q_2^TQ_1 + Q_1^T Q_2 & Q_2^T Q_2 + Q_1^T Q_1
\end{bmatrix}
\end{align*}
$$
My strategy is to look at what some of the terms evaluate to and then substitute those back in. So
$$
\begin{align*}
Q_1^TQ_1 &= (Q – jQ_2)^T(Q-jQ_2) \\
&= Q^TQ -j Q^TQ_2 – j Q_2^T Q- Q_2^TQ_2
\end{align*}
$$
And as I keep working it I don't really see things cleaning up…
If I assume that by orthogonal the authors meant unitary I get
$Z^H Z = 2I$
Which isn't what we're going for.
Best Answer
Note that $Q_1$ and $Q_2$ are real, so $Z$ is also real, and its conjugate transpose is the same as the usual transpose.
$Q$ being unitary means $$I = Q^H Q = (Q_1 - jQ_2)^\top(Q_1 + j Q_2) = Q_1^\top Q_1 + Q_2^\top Q_2 + j(Q_1^\top Q_2 - Q_2^\top Q_1).$$ Which in turn implies $$I = Q_1^\top Q_1 + Q_2^\top Q_2$$ $$0 = Q_1^\top Q_2 - Q_2^\top Q_1$$
This helps you reason about your block diagonal entries.
For the block off-diagonal entries, double check your matrix multiplication.