A theorem in linear Algebra tells us
A matrix U is orthogonal iff $U^TU=I$ where I is the identity matrix.
I get the part that if U is orthogonal then $U^TU=I$, but why should $U^T$ also be orthogonal, consider the following case:
Let $$U =\begin{bmatrix}
a&b\\c&d
\end{bmatrix}
$$
let $$ a_1= \begin{bmatrix} a\\c \end{bmatrix} \ and\ a_2= \begin{bmatrix} b\\d \end{bmatrix}$$
we have,
$$a_1.a_2=0,\ |a_1|=1 \ and\ |a_2|=1$$
i.e. from the theorem
$$a^2+b^2=1,\ c^2+d^2=1,\ ac+bd=0 \ \ \ \ \ -> case1$$
Now, the real confusion is this site gives proof then $U^T$ is also orthogonal, but for that to happen this should hold true:
$$a^2+c^2=1,\ b^2+d^2=1,\ ab+cd=0\ \ \ \ \ ->case2$$
I don't see how is this possible in general cases where $U$ is an orthogonal matrix. How is $U^T$ orthongonal and how can both case1 and case2 hold at the same time in every case where $U$ is an orthogonal matrix?
Suppose the set of equalities in case 1 holds. We can write $a=\sin x,b=\cos x,c=\cos y,d=\pm\sin y$ for some angles $x$ and $y$, by the Pythagorean trigonometric identity (the sign in front of $d$ is important, and differentiates rotation and reflection matrices). Then $ac+bd=0$ is $\sin x\cos y\pm\cos x\sin y=\sin(x\pm y)=0$, so $x=-y$ or $x=y$ up to periodicity (modulo $2\pi$) if the sign in front of $d$ was $+$ or $-$ respectively. Hence $$a^2+c^2=\sin^2x+\cos^2x=1$$ $$b^2+d^2=\cos^2x+\sin^2x=1$$ $$ab+cd=\sin x\cos x-\cos x\sin x=0$$ and case 2 holds. A similar argument proves case 1 from case 2, so $U$ and $U^T$ are both orthogonal.