Show an orthonormal basis is a set of eigenvectors

Question

Consider the orthonormal basis $B= \{ {\bf u}, {\bf v}, {\bf w} \}$ for $\mathbb{R}^3$ and the transformation $T: \mathbb{R}^3 \to \mathbb{R}^3$ defined by

$$T({\bf x}) = {\bf x} - 2 ({\bf x} \cdot {\bf u}) {\bf u} .$$

Show that $B$ is a set of eigenvectors for $T$ and find the eigenvalues corresponding to the eigenvectors ${\bf u}$ and ${\bf v}$. Describe $T$ geometrically.

Answer 1

Consider

$$T({\bf u}) = {\bf u} - 2 ({\bf u} \cdot {\bf u}) {\bf u} = - {\bf u}$$

This is an eigenvalue equation with eigenvalue $\lambda = -1$.

Now plug in to find $T({\bf v})$ and use your knowledge of orthogonality.

Can you continue?

As for geometry: Note what $T$ does to each eigenvector. Imagine what it does to points on a sphere. Think about inversions...