Let A be some $2×2$ matrix with real entries. Prove that $A^T$$A$ = $I$ if and only if $A$ is the rotation matrix or the reflection matrix.
My Progress: It can be shown that if $A$ is either the rotation or reflection matrix, then $A^T$$A$ = $I$ holds by matrix multiplication. Where I get suck is showing that if $A$ is a $2×2$ orthogonal matrix, then $A$ must either be the Rotation or Reflection Matrix. I suppose that since $A$ is orthogonal, it is distance preserving – and the only $2×2$ matrices that preserve distance are the Rotation and Reflection Matrices, but this isn't really a proof.
Any advice and help would be much appreciated.
Best Answer
With not so many variables running around, we may verify this claim algebraically. Write $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$. Then $A^TA = \begin{pmatrix} a^2+c^2 & ab+cd \\ ab+cd & b^2+d^2 \end{pmatrix}$. So $\begin{pmatrix} a \\ c\end{pmatrix}$ and $\begin{pmatrix} b \\ d\end{pmatrix}$ are unit vectors so that $\begin{pmatrix} a \\ c\end{pmatrix}\cdot \begin{pmatrix} b \\ d\end{pmatrix} = 0$. We can check that this implies $b = \pm c$ while $d = \mp a$. So $A = \begin{pmatrix} a & c \\ c & -a \end{pmatrix}$ or $A = \begin{pmatrix} a & -c \\ c & a \end{pmatrix}$, which are the forms of a reflection composed with a rotation and a rotation, respectively.