I know how to show that multiplying by an orthogonal matrix preserves the angle and distance between two vectors. I have seen everywhere that Orthogonal matrices are kind of related to rotations and reflections.


What is a good definitive proof that reflection/rotation $\iff$ orthogonal matrix. Both for the $2$x$2$ case and the $3$ x $3$ case. (both directions)


What is a proof that Given an isometry $T$, there exists an $n$ × $n$ orthogonal matrix $A$ and a vector $u$ such that the associated map $T$ on position vectors is given by $T (p) = Ap + u$. Is this bit when there is a combination of rotation and reflection?


Here is a sketch of these proofs. The details of this are described in great detail in Michael Artin's Algebra - a book well-worth reading.

1a) For the $2\times 2$ case: If $A$ is orthogonal, then multiply by a suitable reflection to ensure that $\det(A) = 1$. Then note that $$ A = \begin{pmatrix} a & b \\ c & d \end{pmatrix} $$ where the vector $(a,c)$ and $(b,d)$ are orthonormal. Check that this must happen iff $\exists \theta \in [0,2\pi]$ such that $$ a = \cos(\theta) = d, \text{ and } b = \sin(\theta) = -c $$ 1b) For the $3\times 3$: Again, assume $\det(A) = 1$. Now one needs to prove that $1$ is an eigen-value of $A$. This gives a one-dimensional subspace $W$ which is invariant under $A$. One can then prove that $W^{\perp}$ is also invariant under $A$, so $A$ will now restrict to a $2\times 2$ orthogonal matrix on $W^{\perp}$. Use part (a) to conclude that $A$ must be a rotation.

2) First note that if $T$ is an isometry, so is $T - T(0)$, so you may assume that $T(0) = 0$. Now since $\|T(x)\|^2 = \|x\|^2$ for all $x$, you can use the Polar identity to conclude that $$ \langle Tx,Ty\rangle = \langle x,y\rangle \quad\forall x,y $$ and this gives that $T$ must be an orthogonal matrix.


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