# How can I find the general form of an orthogonal matrix?

I know that the general form of orthogonal matrices is $$\begin{pmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{pmatrix}$$ since they are all rotation matrices but how do I prove it? I have done the reverse, i.e., for such a rotation matrix proven that it is orthogonal: $$\vec r^\prime . \vec r^\prime = \sum_{p=1}^n r_p r_p = \sum_{p=1}^n (\sum_{j=1}^n R_{pj} r_j) (\sum_{k=1}^n R_{pk} r_k) = \sum_{pjk} R_{pj} R_{pk} r_j r_k = \sum_{i} r_i r_i = \vec r . \vec r$$ which is only possible if $\sum_{p} R_{pj} R_{pk}$ = $\delta_{jk}$ which defines an orthogonal matrix. Does just reversing this process work for finding the general form? If not, what's the correct method?

• See wiki. – MisterRiemann Jul 4 '18 at 10:20
• You missed the matrices with determinant $-1.$ – gammatester Jul 4 '18 at 10:27

Actually, this is not the general form. It's the general form of the $2\times2$ orthogonal matrices with determinant $1$; there are also those with determinant $-1$.
Anyway, what you're after are those matrices $\left(\begin{smallmatrix}a&b\\c&d\end{smallmatrix}\right)$ such that $\left(\begin{smallmatrix}a&c\\b&d\end{smallmatrix}\right)\left(\begin{smallmatrix}a&b\\c&d\end{smallmatrix}\right)=\left(\begin{smallmatrix}1&0\\0&1\end{smallmatrix}\right)$. But this means that$$\left\{\begin{array}{l}a^2+b^2=1\\c^2+d^2=1\\ab+cd=0\end{array}\right.$$The first two equations mean that $(a,b)$ and $(c,d)$ have norm $1$, whereas the third one means that $(a,b)$ and $(c,d)$ are orthogonal. Since $\bigl\|(a,b)\bigr\|=1$, $(a,b)=(\cos\theta,\sin\theta)$, for some $\theta$. And, since $(c,d)$ is orthogonal to $(a,b)$ and since it also has norm $1$, it is equal to $\pm(-\sin\theta,\cos\theta)$. Therefore, the general form is$$\begin{pmatrix}\cos\theta&\mp\sin\theta\\\sin\theta&\pm\cos\theta\end{pmatrix}.$$Replacing $\theta$ by $-\theta$, one gets the form that you mentioned.