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Let $ A = \mathbb{R}^{3}$, let $R$ be an equivalence relation such that $\left(v_{1},v_{2}\right) \in R$ if and only if $\exists P \in M_{3 \times 3}\left(\mathbb{R}\right)$ such that $P$ is orthogonal and $Pv_{1} = v_{2}$

Describe and sketch the equivalence classes of $R$ in $\mathbb{R}^{3}$

I am having difficulty attempting this question. I attempted to find the equivalence class, $C$, of $\left(1,0,0\right)^{T}$ for which I used: $$P = \begin{bmatrix}1 & 0 & 0 \\ 0 & \cos\left(\theta\right) & \sin\left(\theta\right)\\ 0 & -\sin\left(\theta\right) & \cos\left(\theta\right)\\ \end{bmatrix}$$ and said that $\displaystyle v = \left(\alpha,\beta,\gamma\right)^{T} \in C \implies \alpha = 1, \tan\left(\theta\right) = \frac{\gamma - \beta}{\gamma + \beta}$

But I feel that I am not following the right route with this approach. Thank you in advance.

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Euclidean
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2 Answers2

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Geometrically, the orthogonal matrices are the "rigid motions", they preserve length of every vector and they preserve the angle between any two vectors. In particular, rotations are all (special) orthogonal transformations.

The equivalence classes are concentric spheres, centred at the origin. Thinking of a vector as a radius of a sphere, an orthogonal transformation will take that radius to another radius of the same sphere.

Fly by Night
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An orthogonal matrix is an isometry when viewed as an operator. So for any orthogonal matrix $P$ this holds: $|v|=|Pv|$. In particular if $(v_1, v_2)\in R$ then $|v_1|=|v_2|$.

The other implication also holds, i.e. if $|v_1|=|v_2|$ then you can find an orthogonal matrix $P$ such that $Pv_1=v_2$ (for explicit construction in dimension 3 see this).

All in all, an equivalence class of $v$ in $R$ is a set of all vectors with the same norm, i.e. a sphere.

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