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The displayed orthogonal matrix transforms the coordinates of a fixed vector, rather than actively moving that vector. Write $\theta =\arctan (1/2)$. Then $P=R(-\theta )$, so the new unit basis vectors, expressed in the old orthonormal basis, are the columns of $P^T$:The axes have rotated counterclockwise through $\theta$, while the coordinate column transforms by the opposite rotation. For the example point $\mathbf{r}=(1,0)$, its new coordinates are $(2,-1)/\sqrt{5}$. Reconstruction gives $(2{\mathbf{e}}_1^{\prime }-{\mathbf{e}}_2^{\prime })/\sqrt{5}=(1,0)$, confirming that the point is unchanged. The middle panel of the figure in part 1 shows these axes and the signed projections. Since $P^{T}P=I$ and $\det P=1$, lengths and orientation of the basis are preserved.

Let $\mathbf{u}=(2,1)/\sqrt{5}$ be a unit vector along the fixed line. The orthogonal projection parallel to the line is $\mathbf{u}{\mathbf{u}}^T$; a reflection keeps that component and reverses its orthogonal complement. ThereforeIts eigenvalues are $+1$ along $(2,1)$ and $-1$ along $(-1,2)$. It preserves lengths and unsigned areas but reverses orientation. For example, $(1,0)$ maps to $(3/5,4/5)$, with their midpoint on the reflection line and their separation perpendicular to it; the right panel of the figure in 1 illustrates this.