Let $v \in \mathbb R^3$. Given a matrix $M : \mathbb R^3 \mapsto \{v\}^\perp$, that is, there is one vector $v$ such that $\forall x \in \mathbb R^3\setminus\{v\}$: $\langle Mx,v \rangle = 0$. Furthermore I know that a vector $w$ exists, such that $Mw = 0$.
What is known about the motions and the domain of the eigenvalues (and -vectors?) of $M$, if it has
- $0$ fixed points (e.g. a translation)
- $1$ fixed point (e.g. a rotation (+scaling))
- $≥2$ fixed points (e.g. mirroring)
- $>2$ fixed points (lin. indep.) (e.g. identity)?
Any hint or pointers to relevant literature or search term are greatly appreciated as well.
[Edit: Removed most of the original version, as it was ill-formed and actually contained errors. A downvote is due!]
$Mw=0$, so $w$ is an eigenvector belonging to eigenvalue $\lambda=0$. As $M$ is a 3x3 matrix, it has at most two other eigenvalues (counted with multiplicity).
Any eigenvector belonging to a non-zero eigenvalue must lie in $V=<v>^\perp$, because its image lies in that plane.
A fixed point ($\neq0$) is an eigenvector belonging to eigenvalue $\lambda=1$, and by the previous point $\in V$.
The restriction $M|_V$ of $M$ onto the plan $V$ is a mapping $V\rightarrow V,$ $\lambda=1$ may be a double root of the characteristic equation of $M|_V$, but the corresponding eigenspace may have dimension one only. In that case we can say that geometrically $M|_V$ is a shearing (see http://en.wikipedia.org/wiki/Transformation_matrix#Shearing).
There is no reason to think that $M|_V$ would have any real eigenvalues, but the fixed points come from there, so for that reason $M|_V$ and its potential for having eigenvalue $\lambda=1$ received the special attention in the previous paragraph.
It is, of course, possible that $w\in V$. There is no reason to think that $w$ would be ortohogonal to $V$, unless we can choose $w=v$. (therefore we cannot conclude that $M$ would be an orthogonal projection at this point as I errorneously stated in the erased version).