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I was reading about orthogonal matricies and noticed that the $2 \times 2$ matrix $$\begin{pmatrix} \cos(\theta) & \sin(\theta) \\ -\sin(\theta) & \cos(\theta) \end{pmatrix} $$ is orthogonal for every value of $\theta$ and that every $2\times 2$ orthogonal matrix can be expressed in this form. I then wondered if this can be generalized to any smooth paramtrization of the unit circle. More precisely, let $x(t)$ and $y(t)$ be a smooth parametrization of the unit circle for all $t$ in some interval $I \subseteq \Bbb R$ such that $|\langle x(t),y(t) \rangle| = 1$ for all $t \in I$. Is the matrix $$A= \begin{pmatrix}x(t)&y(t)\\x'(t) & y'(t) \end{pmatrix} $$ necessarily orthogonal? At first I thought yes, but I'm having trouble proving it. Letting $v = \langle x(t), y(t) \rangle$, it suffices to show three things: 1) $|v| = 1$, 2) $v \cdot v' = 0$, and 3) $|v'| = 1$. (1) follows straight from how $x(t)$ and $y(t)$ were defined. (2) can be obtained by differentiating the equation $x(t)^2 + y(t)^2 = 1$: \begin{align*} &\frac{d}{dt} \Big[ x(t)^2 + y(t)^2 \Big]= 0 \\ &\implies 2x(t)x'(t) + 2y(t)y'(t)= 0 \\ &\implies v \cdot v' = 0. \end{align*} But I couldn't find a way to prove (3). Now I am unsure whether (3) is true at all. Is $A$ even orthogonal in the first place? Any hints would be much appreciated.

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  • $\begingroup$ For the sine/cosine case that matrix is orthogonal since it is the wronskian matrix of the DE $$y''+y=0$$, but in general two functions $x(t),y(t)$ are not two independent solutions of the same differential equation. $\endgroup$ Jun 1, 2016 at 13:37
  • $\begingroup$ In fact, the matrix defined by $x(t) = 1, y(t) = 0$ is not even nondegenerate. $\endgroup$ Jun 1, 2016 at 13:59

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First, not every orthogonal matrix is of the stated form, only rotation matrices. There are orthogonal reflection matrices, with determinant $-1$: $$ \left[ \begin{array}{@{}rr@{}} \cos\theta & \sin\theta \\ \sin\theta & -\cos\theta \end{array}\right]. $$

Second, your conjecture is not true: Only unit speed parametrizations have the stated property, because the rows of an orthogonal matrix form an orthonormal set (in the obvious sense), and particularly must be unit vectors.

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The third point implies that the parametrization goes over the circle at a constant speed $1$, which is, of course, false in general.

Take for example $v(t)=\langle \sin t^2,\cos t^2\rangle $ for $t\in[0,\sqrt{2\pi})$.

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In general, (3) is not true. Since $\left\vert v \right\vert^2 = x(t)^2 + y(t)^2 = 1$ , we can write $$x(t) = \cos \theta(t), \quad y(t) = \sin \theta(t)$$ for some function $\theta(t)$. But computing gives $\left\vert\dot v\right\vert^2 = \dot\theta(t)^2$, so the condition $|\dot v|^2 = 1$ (which just says that $v$ is a unit speed parameterization) forces $\dot\theta(t) = \pm 1$, and this condition yields just the matrices of the given form (the rotations). Note that a composition of any rotation with a reflection is another reflection and hence orthogonal, but these reflections cannot be of the form in the question: Reflections have determinant $-1$ but matrices of the given form have determinant $(\cos \theta)(\cos \theta) - (\sin \theta)(-\sin \theta) = +1$.

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No, that is not the case. Ad you show, the rows in your $A$ will necessarily be orthogonal to each other, and the first row has norm $1$ -- but the norm of the second row is not necessarily $1$.

If you scale each of $x'(t)$ and $y'(t)$ in the second row by $\dfrac{1}{\sqrt{x'(t)^2+y'(t)^2}}$, you do get all of $SO(2)$, assuming that your $x'(t)$ and $y'(t)$ never vanish at the same time.

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(3) is only sometimes true, and it is at only the times that (3) is true that A is orthogonal.

More particularly, the reason A is orthogonal when A is some multiple of the rotation matrix is that (x(t), y(t)) describes a circle, so position is orthogonal to velocity.

For A where position is not orthogonal to velocity, the conjecture is false.

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