Find matrix of linear transformation and show it's diagonalizable

[schniefen]

TL;DR

Let $V$ be an $n$-dimensional inner product space, where $n>0$, and let $F$ be the linear transformation on $V$ defined by $F(\textbf{u})=\langle \textbf{u},\textbf{c} \rangle \textbf{b}-\langle \textbf{b},\textbf{c} \rangle \textbf{u}$, where $\textbf{b},\textbf{c} \in V$ and $\langle \textbf{b},\textbf{c} \rangle \neq 0$. Show that $V$ has a basis consisting of eigenvectors of $F$ and find the matrix of $F$ with respect to some such basis.

The strategy here would probably be to find the matrix of $F$. How would one go about doing that? Since $V$ is finite dimensional, it must have a basis...

 

[tnich]

I think it would be simpler to just figure out what an eigenvector of $F(.)$ looks like. Under what conditions does $F(\mathbf u) = \lambda \mathbf u$ for some $\lambda$ and $\mathbf u$?

 

[WWGD]

This is possible when the arithmetic and geometric multiplicities of each eigenvalue agree.