Let $\textbf{A}:\Re^n \rightarrow \Re^n$ be given by $\textbf{A} \vec{x}=\vec{x}+l(\vec{a},\vec{x})\vec{a}$ where $||\vec{a}|| = 1, \vec{a} \in \Re^n, l \in \Re.$ And let $\textbf{A}$ be orthogonal for $l=0 \vee l=-2$. Find an orthonormal basis z of $\Re^n$ with respect to which the matrix $A_z$ is diagonal.
What I've got so far: for the first part of the basis we can take $\vec{a}$ as it's on $\Re^n$ and has length 1. Next we take $\vec{a}^\bot$ which forms a basis of $\Re^n$ together with $\vec{a}$. It can be normalized by dividing it by its length. Is this right so far? And I don't know how to show if it's diagonal to $A_z$.
The matrix $A_z$ will be diagonal if and only if each basis element of $z$ is an eigenvector of $\bf A$.
To include $a$ into the basis $z$ is a good start, since ${\bf A}(a)=(1+l)a$.
Note, however, that $a^\perp=\{x\in\Bbb R^n:(a,x)=0\}$ is a whole subspace of dimension $n-1$.
But, luckily any vector $x\in a^\perp$ is an eigenvector of $\bf A$, because then ${\bf A}(x)=x$.
So, we can extend the single vector $a$ by an arbitrary orthonormal basis of $a^\perp$, and this will work.