Let \(e=(a,b,c)\) be a unit vector in \(\mathbb R^3\). Let \(T\) be rotation by \(180^\circ\) about the axis spanned by \(e\). Find the matrix of \(T\) in the standard basis.
Solution
1. Every \(v\in\RR^3\) has the orthogonal decomposition\[ v = \inner{v}{e}e + \left(v-\inner{v}{e}e\right), \]where the second summand is orthogonal to \(e\).
Since \(e\) is a unit vector, \[ \inner{v-\inner{v}{e}e}{e} = \inner{v}{e} -\inner{v}{e}\inner{e}{e} =0. \] Thus the first summand is the projection of \(v\) onto the rotation axis and the second lies in its orthogonal plane.
2. The half-turn satisfies\[ T(v)=2\inner{v}{e}e-v. \]
A rotation by \(180^\circ\) about the axis \(\RR e\) fixes every vector on that axis and sends every vector in the orthogonal plane to its negative. Applying this to the decomposition in step 1 gives \[ \begin{aligned} T(v) &= \inner{v}{e}e - \left(v-\inner{v}{e}e\right)\\ &= 2\inner{v}{e}e-v. \end{aligned} \]
3. In the standard basis, the matrix of \(T\) is\[ \boxed{ \begin{pmatrix} 2a^2-1&2ab&2ac\\ 2ab&2b^2-1&2bc\\ 2ac&2bc&2c^2-1 \end{pmatrix} }. \]
Regard \[ e= \begin{pmatrix} a\\ b\\ c \end{pmatrix}. \] For a column vector \(v\), one has \[ \inner{v}{e}=e^{\mathsf T}v. \] Hence step 2 becomes \[ T(v) = \left(2ee^{\mathsf T}-I_3\right)v. \] Now \[ ee^{\mathsf T} = \begin{pmatrix} a^2&ab&ac\\ ab&b^2&bc\\ ac&bc&c^2 \end{pmatrix}, \] so expanding \(2ee^{\mathsf T}-I_3\) gives the displayed matrix.
4. Q.E.D.
Step 3 is the requested standard-basis matrix.