Let \(V\) be a real vector space endowed with an inner product \(\langle \cdot, \cdot \rangle\).

It is known that a bijective map \( T : V \to V\) that preserves the inner product \(\langle \cdot, \cdot \rangle\) is linear.

That might not be the case if \(T\) is supposed to only preserve orthogonality. Let’s consider for \(V\) the real plane \(\mathbb R^2\) and the map \[
\begin{array}{l|rcll}
T : & \mathbb R^2 & \longrightarrow & \mathbb R^2 \\
& (x,y) & \longmapsto & (x,y) & \text{for } xy \neq 0\\
& (x,0) & \longmapsto & (0,x)\\
& (0,y) & \longmapsto & (y,0) \end{array}\]

The restriction of \(T\) to the plane less the x-axis and the y-axis is the identity and therefore is bijective on this set. Moreover \(T\) is a bijection from the x-axis onto the y-axis, and a bijection from the y-axis onto the x-axis. This proves that \(T\) is bijective on the real plane.

\(T\) preserves the orthogonality on the plane less x-axis and y-axis as it is the identity there. As \(T\) swaps the x-axis and the y-axis, it also preserves orthogonality of the coordinate axes. However, \(T\) is not linear as for non zero \(x \neq y\) we have: \[
\begin{cases}
T[(x,0) + (0,y)] = T[(x,y)] &= (x,y)\\
\text{while}\\
T[(x,0)] + T[(0,y)] = (0,x) + (y,0) &= (y,x)
\end{cases}\]