Orthogonal Rotation Matrix

Question

Consider an arbitrary rotation of the coordinate frame. How would you go about proving a Rotation Matrix $R$ would have to be orthogonal if $r^2$=$r'^2$

Given: $(x', y', z')$= $R$$(x, y, z)$

Answer 1

With $r^\prime=Rr$, $r^{\prime2}=(r^{\prime T}r^\prime)_{11}=(r^TR^TRr)_{11}=r^2$ provided $R^TR=I$.

Answer 2

I assume that we are working in the vector space $\Bbb R^n$ for some $n \in \Bbb N$.

Note that

$r^2 = r'^2 \Longleftrightarrow r = r', \tag 0$

since $r, r' \ge 0$.

I take it that

$r = r' \tag 1$

is a way of saying that $R$ preserves lengths of vectors, that is,

$\Vert \vec x' \Vert = \Vert R \vec x \Vert = r' = r = \Vert \vec x \Vert, \tag 2$

for all vectors $\vec x$; this in turn implies

$\langle R \vec x, R \vec x \rangle = \langle \vec x', \vec x' \rangle = \Vert \vec x' \Vert^2 = \Vert \vec x \Vert^2 = \langle \vec x, \vec x \rangle, \; \forall x; \tag 3$

now with

$\vec x = \vec y + \vec z \tag 4$

we have

$\langle R \vec y + R\vec z, R\vec y + R\vec z \rangle = \langle R(\vec y + \vec z), R(\vec y + \vec z) \rangle = \langle \vec y + \vec z, \vec y + \vec z \rangle; \tag 5$

by the bilinearity of $\langle \cdot, \cdot \rangle$ this yields

$\langle R \vec y, R \vec y \rangle + \langle R \vec y, R\vec z \rangle + \langle R \vec z, R \vec y \rangle + \langle R \vec z, R \vec z \rangle$ $= \langle \vec y, \vec y \rangle + \langle \vec y, \vec z \rangle + \langle \vec z, \vec y \rangle + \langle \vec z, \vec z \rangle; \tag 6$

in light of (3) this becomes

$\langle R \vec y, R\vec z \rangle + \langle R \vec z, R \vec y \rangle = \langle \vec y, \vec z \rangle + \langle \vec z, \vec y \rangle, \tag 7$

or

$2\langle R \vec y, R \vec z \rangle = 2 \langle \vec y, \vec z \rangle \tag 8$

by the symmetry of $\langle \cdot, \cdot \rangle$. Thus,

$\langle R \vec y, R \vec z \rangle = \langle \vec y, \vec z \rangle, \tag 9$

or

$\langle \vec y, R^TR \vec z \rangle = \langle \vec y, \vec z \rangle, \; \forall \vec y, \vec z, \tag{10}$

which yields

$R^TR \vec z = \vec z, \; \forall z, \tag{11}$

and finally

$R^TR = I; \tag{12}$

that is, $R$ is an orthogonal matrix.