Linear transformation , proof existence of angle between two vectors

Question

Let $J:\mathbb R^2\to \mathbb R^2$ be the linear transformation defined by $J(p_1,p_2)=(−p_2,p_1),\forall (p_1,p_2)\in \mathbb R^2$. Let $p,q\in \mathbb R^2\setminus\{(0,0)\}$. Show that there exists a unique number $\theta$ satisfying $\cos\theta=\dfrac{\langle p,q\rangle}{\lVert p\rVert\lVert q\rVert}$ and $\sin\theta=\dfrac{\langle p,J(q)\rangle}{\lVert p\rVert\lVert q\rVert}, 0≤\theta<2\pi$. The oriented angle from $q$ to $p$ is $\theta$.

We can prove $J$ is inner product and norm preserving because $J(J(x))=-x$ for $x\in\mathbb R^2$. But I don't understand how to proof the existence of the angle above.

Answer 1

By the Cauchy-Schwarz inequality $$ -1\le \frac{\langle p,q\rangle}{\|p\|\,\|q\|}\le1\ , $$ and \begin{align} \langle p, J(q)\rangle^2+ \langle p,q \rangle^2&=p_1^2q_2^2+p_2^2q_1^2+p_1^2q_1^2+p_2q_2^2\\ &=\left(p_1^2+p_2^2\right)\left(q_1^2+q_2^2\right)\\ &=\left\|p\right\|^2 \left\|q\right\|^2\ . \end{align}

• If $\ \frac{\langle p,q\rangle}{\|p\|\,\|q\|}=-1\ $, then $\ \langle p, J(q)\rangle=0\ $, and the equation $\ \cos\theta= \frac{\langle p,q\rangle}{\|p\|\,\|q\|}\ $ has the unique solution $\ \theta=\pi\ $ in the interval $\ 0\le\theta<2\pi\ $. Since $\ \sin\pi=0= \langle p, J(q)\rangle=0\ $, then the two equations have the unique solution $\ \theta=\pi\ $ in this case.
• Likewise, if $\ \frac{\langle p,q\rangle}{\|p\|\,\|q\|}=1\ $, then $\ \langle p, J(q)\rangle\ $, and the equation $\ \cos\theta= \frac{\langle p,q\rangle}{\|p\|\,\|q\|}\ $ has the unique solution $\ \theta=0\ $ in the interval $\ 0\le\theta<2\pi\ $. Since $\ \sin0=0= \langle p, J(q)\rangle\ $, then the two equations have the unique solution $\ \theta=0\ $ in this case.
• On the other hand, if $\ -1<\frac{\langle p,q\rangle}{\|p\|\,\|q\|}<1\ $, then then the equation $\ \cos\theta= \frac{\langle p,q\rangle}{\|p\|\,\|q\|}\ $ has exactly two solutions in the interval $\ 0\le\theta<2\pi\ $. It has a unique solution, $\ \theta_0\ $, say, in the interval $\ 0<\theta<\pi\ $, and a second solution $\ \theta_1=2\pi-\theta_0\ $ in the interval $\ \pi<\theta<2\pi\ $. Now $\ \langle p, J(q)\rangle=\pm\sqrt{\|p\|^2\|q\|^2 -\langle p,q\rangle^2}\ne0\ $. If $\ \langle p, J(q)\rangle>0\ $, then $\ \sin\theta_0=$$\sqrt{1-\cos^2\theta_0}=$$ \langle p, J(q)\rangle\ $, whereas $\ 0>\sin\theta_1\ne \langle p, J(q)\rangle\ $, so $\ \theta=\theta_0\ $ is the only solution of the two equations in this case. Similarly, if $\ \langle p, J(q)\rangle<0\ $, then $\ \theta=\theta_1\ $ is the only solution.