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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.

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what you've written doesn't make sense. A fixed point of $M$ is a vector $v$ such that $Mv=v$, this is impossible if $M$ maps $R^3 \to R^2$. Could you rewrite the question to make it clear what you want? – m_t_ Jun 8 '11 at 9:01
@mt_ I hope it is clearer now. What I ment was that $M$ mapped $R^3$ onto a two dimensional subspace of $R^3$. – angerman Jun 8 '11 at 9:13
You only have eigenvalues of linear transformations from one vector space to itself. What you get when you go between two vector spaces, even if they are the same dimension, are not eigenvalues. You may want to look into something like singular value decomposition. – Aaron Jun 8 '11 at 9:29
up vote 1 down vote accepted

[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

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).

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Thanks. That got me thinking. Especially your edit. I've come to the conclusion that the condistion $Mv = 0$ is incorrect. I only know that $Mw = 0$. But I don't know anything about the relation of $w$ to $v$. You answer was very helpful though and helped to find some structure. – angerman Jun 8 '11 at 12:55

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