It is well known that a map $T: {\bf R}^2 \rightarrow {\bf R}^2$ that preserves the standard inner product is linear. What about transformations preserving the determinant?
Assume $det(T(v_1), T(v_2)) = det(v_1, v_2)$ for all $v_1$, $v_2 \in {\bf R}^2$. Is $T$ a linear transformation of ${\bf R}^2$?
Maybe some additional assumption is needed: $T$ bijective, or $T$ continuous... Any hint appreciated.
On
The determinant is a bilinear function, hence the transformation $T$ must be linear to preserve bilinearity.
Then to preserve the value of determinants, $\color{green}{\det(T)=1}$ must hold because
$$\det(T(v_1),T(v_2))=\det(TV)=\det(T)\det(V)=\det(T)\det(v_1,v_2).$$
It is not required that $T$ be orthogonal.
$T$ is linear. First note that for the standard basis, $\det(T(e_1), T(e_2)) = \det(e_1, e_2) = 1 \neq 0$, hence $T(e_1)$ and $T(e_2)$ are independent vectors. So are $f_1= R(T(e_1))$ and $f_2 = R(T(e_2))$, where $R$ is the clockwise rotation of angle 90 degrees. But for every vector $u$, one has \begin{equation} T(u)\cdot f_i = \det(T(u), T(e_i)) = \det(u,e_i) \end{equation} Hence $T(u)\cdot f_i$ depends linearly on $u$. It remains to say that there exist real numbers $a,b,c,d$ such that for every vector $w$ \begin{equation} w=(a w\cdot f_1 + b w\cdot f_2) f_1 + (c w\cdot f_1 + d w\cdot f_2) f_2 \end{equation} Applying this to $w=T(u)$, we see that it depends linearly on $u$
Edit: One can go a little step further and prove that for a map $T:{\mathbb R}^2\to {\mathbb R}^2$, the two following properties are equivalent
Indeed, 2. implies 1. by the above proof, and 1. implies 2. because if $T$ is a linear operator, the expression $\det(T(u), T(v))$ is a alternating bilinear form of $u$ and $v$, thus a multiple of the determinant.
The constant $C$ is the determinant of $T$, in particular the set of all functions $T$ satisfying the initial problem is exactly the special linear group $SL_2({\mathbb R})$.