Transformation that do not modify angles

Question

How can I prove for $R_2$ that the matrix $$T = \begin{bmatrix} a & b \\c & d \end{bmatrix} =\begin{bmatrix} a & b \\-b & a \end{bmatrix} $$ is the ony one that, when multiplied with 2 vectors, the angle between them keeps being the same, $$\frac {(Tv) \cdot (Tu)}{|Tv||Tu|} = \frac {v \cdot u}{|v||u|} = \cos \alpha.$$ If possible I would like to do it only by manipulating this formula

Answer 1

Hint:

Let us look at the action of the matrix on the basis vectors $e_1=\begin{bmatrix}1\\0\end{bmatrix}$ and $e_2=\begin{bmatrix}0\\1\end{bmatrix}$. $$Te_1=\begin{bmatrix}a\\c\end{bmatrix} \text{ and } Te_2=\begin{bmatrix}b\\d\end{bmatrix}.$$

Since $e_1 \perp e_2$, so for the angle to be preserved, the vectors $Te_1 \perp Te_2$. Thus $$ab+cd=0.$$ Moreover, the vectors $Te_1, Te_2$ should be linearly independent.

Hopefully you can take it from here.