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Let A be some $2 \times 2$ 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 $2 \times 2$ 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 $2 \times 2$ matrices that preserve distance are the Rotation and Reflection Matrices, but this isn't really a proof.

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    $\begingroup$ Observe that $A$ also preserves orthogonality of vectors: if $x^T y = 0$, then $(Ax)^T(Ay) = x^T (A^T A) y = x^TIy = x^Ty = 0$. Similarly, $A$ maps unit vectors to unit vectors. Now consider how this constrains $Ae_1$ relative to $Ae_2$, where $e_1, e_2$ is the standard basis of $\mathbb R^2$. $\endgroup$
    – user169852
    Commented Sep 20, 2018 at 17:03

2 Answers 2

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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.

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    $\begingroup$ To clarify why $b=\pm c$: notice that $ab=-cd$ from orthogonality, so $a^2b^2=c^2d^2$; also notice that $a^2=1-c^2$ and $d^2=1-b^2$ from unit magnitude. So $(1-c^2)b^2=c^2(1-b^2)$, so $b^2=c^2$ and $b=\pm c$. We can use analogous methods to derive $a^2=d^2$, then use $ab=-cd$ to show $d=\mp a$. $\endgroup$
    – jskattt797
    Commented May 9, 2020 at 1:04
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Write $a_{11}$ as $\cos \theta$ so $ a_{12} = \pm \sin \theta $ because the first row must have norm 1. Likewise, $ a_{21} = \pm \sin \theta $ as well. It follows that $a_{22} = \pm \cos \theta$; with the sign chosen so as to make the rows (or equivalently, columns) orthogonal. This idenitifies all the possible forms of the matrix.

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