In texts on linear algebra, a standard example for a $2 \times 2$ orthogonal matrix is the rotation matrix given by
$$
R = \left[ \begin{array}{cc}
\cos \theta & – \sin \theta \\
\sin \theta & \cos \theta
\end{array} \right]
$$
which also has the property that $\mbox{det}(R) = \cos^2 \theta + \sin^2 \theta = 1$.
It is interesting to know that the converse of this result is true for $2 \times 2$ orthogonal matrices.
In other words, it can be shown :
If $A$ is a $2 \times 2$ real orthogonal matrix with $\mbox{det}(A) = 1$, then
$$
A = \left[ \begin{array}{cc}
\cos \theta & – \sin \theta \\
\sin \theta & \cos \theta
\end{array} \right]
$$
for some angle $\theta$. In other words, $A$ is a rotation matrix.
I saw this result in a book but without proof. (I assume that the proof is simple.)
Suppose that
$$
A = \left[ \begin{array}{cc}
a & b \\
c & d \\
\end{array} \right]
$$
To proceed, we note that the rows of $A$ form an orthonormal basis of $\mathbf{R}^2$.
This gives the conditions
$$
a^2 + b^2 = 1. \tag{1}
$$
$$
c^2 + d^2 = 1. \tag{2}
$$
$$
a c + b d = 0. \tag{3}
$$
Since $\mbox{det}(A) = 1$, we also know that
$$
a d – b c = 1. \tag{4}
$$
How to proceed next?
Best Answer
You need to use the orthogonality condition on $A$, namely, $$ A^T = A^{-1} \tag{1} $$
Noting that $\mbox{det}(A) = 1$, we can simplify as $$ \left[ \begin{array}{cc} a & c \\ b & d \end{array} \right] = \left[ \begin{array}{cc} d & -b \\ -c & c \\ \end{array} \right] \tag{2} $$
Equating like terms in (2), we get $$ c = -b, \ \ d = a $$
Thus, we rewrite $A$ as $$ A = \left[ \begin{array}{cc} a & b \\ -b & a \\ \end{array} \right] $$
Since the rows of $A$ form an orthonormal basis of $R^2$, we must have $$ a^2 + b^2 = 1 $$
If we define $$a = \cos \phi$$ for some $\phi$, then $$ b = \sqrt{1 - \cos^2 \phi} = \sin \phi $$
Hence, a $2 \times 2$ orthogonal matrix $A$ with $|A|=1$ has the representation as a rotation matrix given by $$ A = \left[ \begin{array}{cc} \cos \phi & \sin \phi \\ -\sin \phi & \cos \phi \end{array} \right] $$
If we define $\phi = - \theta$, then we can write $A$ equivalently as $$ A = \left[ \begin{array}{cc} \cos \theta & - \sin \theta \\ \sin \theta & \cos \theta \end{array} \right] $$