<s>[INST] Complex Matrix is Orthogonal if and only if... Let D be a 2x2 matrix with entries in the complex numbers. Prove that D is orthogonal if and only if, it is of the form:
\begin{pmatrix}
a & -b\\ 
b & a
\end{pmatrix}
or
\begin{pmatrix}
a & b\\ 
b & -a
\end{pmatrix}
Proof. I've already proved that if D is equal to those forms then, it implies that D is an orthogonal matrix. But how can I prove this? If D is orthogonal, then it must be of the form:
\begin{pmatrix}
a & -b\\ 
b & a
\end{pmatrix}
or
\begin{pmatrix}
a & b\\ 
b & -a
\end{pmatrix}
Update: a and b must satisfy that $a^2+b^2=1$
 [/INST] Let a 2x2 matrix $A$ be orthogonal i.e $AA^t = A^tA = E$. Notice that
$$ E= \left (\begin{array}{cc}a & b \\ c & d\end{array} \right )\left (\begin{array}{cc}a & c \\ b & d\end{array} \right ) \Rightarrow \|(a,b)\| = \|(c,d)\| = 1,\ (a,b)\perp (c,d) $$
Conversely, suppose a 2x2 matrix $A$ has orthonormal columns $(a,b)^t$ and $(c,d)^t$. Then
$$ A^tA = \left (\begin{array}{cc}a & b \\ c & d\end{array} \right )\left (\begin{array}{cc}a & c \\ b & d\end{array} \right ) = \left ( \begin{array}{cc}a^2 + b^2 & 0 \\ 0 & c^2+d^2 \end{array} \right ) = E $$
Likewise $AA^t = E$.

For the complex case we say a square matrix $A$ is unitary, if $A^HA=AA^H = E$, where $A^H := \overline{A}^t$. Orthonormality of column vectors remains an equivalent condition.
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