Complex Matrix is Orthogonal if and only if...

Question

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 $a^2+b^2=1$

Answer 1

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 $AA^t = E$.

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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.

Answer 2

If $A= \begin{pmatrix}a & b \\ c & d\end{pmatrix}$ then we get from $A^TA= I$ that $a^2+c^2=1= b^2+d^2$ and also $ab+cd = 0$. Then $ab = -cd$ so squaring it gives $a^2b^2 = c^2d^2$. I'll let you finish the rest.

Answer 3

To Prove: If A is a 2x2 orthogonal matrix, A is of the form given in the question.

Proof: For any general non-singular matrix, $A= \begin{pmatrix}a & b \\ c & d\end{pmatrix}$, $\begin{equation}\det(A) = ad-bc \end{equation}$ and $A^T= \begin{pmatrix}a & c \\ b & d\end{pmatrix}$.

Using $ \begin{align}A A^T = I \end{align} $, we get $\begin{equation}a^2 + b^2 = c^2 + d^2 = 1\end{equation}$ and $\begin{equation}ac=-bd\end{equation}$.

Using $\begin{align} A^T = A^{-1} \end{align}$, we get $\begin{equation}ad-bc = \frac{d}{a} = \frac{a}{d} = \frac{c}{-b} = \frac{-b}{c}\end{equation}$.

On solving $\begin{equation}d^2 = a^2 \end{equation}$ and $\begin{equation} b^2 = c^2 \end{equation}$ (obtained from above equation) and satisfying in $\begin{equation}ac=-bd\end{equation}$. You get the required matrices.