The question is
Show that the length of a free vector is not changed by rotation, that is, that $\lVert v \rVert = \lVert Rv \rVert$
My attempt is let $R=\begin{bmatrix} c_\theta & -s_\theta \\ s_\theta & c_\theta \end{bmatrix}$ and $v=[v_x \ v_y]^T$ where $c_\theta = \cos\theta,s_\theta=\sin\theta$. $$ \begin{align} \lVert Rv \rVert &= \lVert \begin{bmatrix} c_\theta & -s_\theta \\ s_\theta & c_\theta \end{bmatrix} \begin{bmatrix} v_x \\ v_y \end{bmatrix} \rVert = \lVert \begin{bmatrix} v_xc_\theta - v_ys_\theta \\ v_xs_\theta + v_yc_\theta \end{bmatrix} \rVert \\ &= \sqrt{ (v_xc_\theta - v_ys_\theta )^2 + (v_xs_\theta + v_yc_\theta)^2} \\ &= \sqrt{ v^2_xc^2_\theta + v^2_ys^2_\theta + 2v_xv_yc_\theta s_\theta + v^2_xs^2_\theta + v^2_yc^2_\theta - 2v_xv_yc_\theta s_\theta} \\ &= \sqrt{v^2_x(c^2_\theta+s^2_\theta) + v^2_y(c^2_\theta+s^2_\theta) } \\ &= \sqrt{v^2_x + v^2_y } \\ &= \lVert v \rVert \end{align} $$ Obviously this works for 2D but I can't guarantee my proof to work in higher dimensions. Any suggestions for alternatives to prove this? Also, is my proof valid for higher dimensions?
Since you seem to know that any rotation is orthogonal, it is pretty easy: An orthogonal matrix satisfies $R^TR=I$ or equivalently $R^T = R^{-1}$ which is either your definition of an orthogonal matrix or a consequence of your definition (it might be the case that you defined orthogonal matrices in a different way).
Then we can calculate $$||v||^2 = \langle v, v \rangle = \langle v, R^TRv \rangle = \langle Rv, Rv \rangle = ||Rv||^2$$ and by taking the square root we have what we wanted to show.