Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Is it true that $Ax+b$ where $A^\dagger A=I=AA^\dagger$ and $A$ is an $n\times n$ real matrix, $x,b\in \mathbb R^n$ must have either a fixed point or a fixed $n-1$ hyperplane? If not, is it true for a smaller $n$? Thanks.

[Edit:] Perhaps if we only consider the small $n$ cases such as $n=2,3$, the question might be more friendly and hopefully I will get a response? Thanks.

share|improve this question
    
Rotate and translate along the axis of rotation. –  Louis Feb 3 '12 at 23:46
    
@Louis: Is there an algebraic proof? And why is the translation necessarily along the axis of rotation? $A$ could be a reflection too, no? –  Charlie Feb 4 '12 at 0:06
add comment

1 Answer 1

up vote 1 down vote accepted

If $A$ is an $n$ by $n$ orthogonal matrix and $b\in {\Bbb R}^n$, then the map $\phi: x \mapsto Ax+b$ need not have any fixed points in ${\Bbb R}^n$. For example, $A$ could be the identity and $b\ne 0$.

In general, by a change of coordinate system, $A$ can be made block-diagonal, where each block is either the size-1 block $1$, the size-1 block $-1$, or a size-2 block which is a rotation matrix $R_\theta=\left(\cos \theta\ \ \sin \theta\atop-\sin \theta\ \ \cos \theta\right)$, $\theta\in(0,\pi)$. Suppose that there are $k$ blocks $1$, $\ell$ blocks $-1$, and $m$ blocks $R_{\theta_1}$, $\ldots$, $R_{\theta_m}$, where $k+\ell+2m=n$. Breaking up ${\Bbb R}^n$ according to these block types, we can write ${\Bbb R}^n=U\oplus V\oplus W$, where $\dim U=k$, $\dim V=\ell$, $\dim W=2m$. We also have projections $\pi_U$, $\pi_V$, and $\pi_W$ from ${\Bbb R}^n$ to $U$, $V$, and $W$. We can then take the equation $$x=\phi(x)=Ax+b \qquad (*)$$ and project it, getting $$\pi_U(x)=\pi_U(x)+\pi_U(b), \qquad (1)$$ $$\pi_V(x)=-\pi_V(x)+\pi_V(b), \qquad (2) $$ $$\pi_W(x)={\rm diag}(R_{\theta_1},\ldots,R_{\theta_m}) \pi_W(x)+\pi_W(b). \qquad (3)$$ (1) will never be satisfied if $\pi_U(b)\ne 0$ but is always satisfied if $\pi_U(b)=0$. (2) and (3) always have unique solutions for $\pi_V(x)$ and $\pi_W(x)$. So, (*) will have no solutions if $\pi_U(b)\ne 0$, but will have an affine space of solutions of dimension $k$ if $\pi_U(b)=0$.

In dimensions 2 and 3, $\phi$ has a special name depending on $k$, $l$, $m$, and $\pi_U(b)$:

Dimension 2:

  • $k=2$, $\ell=0$, $m=0$, $b=0$: identity (the whole plane fixed)

  • $k=2$, $\ell=0$, $m=0$, $b\ne 0$: translation (no fixed points)

  • $k=1$, $\ell=1$, $m=0$, $\pi_U(b)=0$: reflection (line of fixed points)

  • $k=1$, $\ell=1$, $m=0$, $\pi_U(b)\ne 0$: glide reflection (no fixed points)

  • $k=0$, $\ell=2$, $m=0$: inversion in a point (rotation by 180°; one fixed point)

  • $k=0$, $\ell=0$, $m=1$: rotation (one fixed point)

Dimension 3:

  • $k=3$, $\ell=0$, $m=0$, $b=0$: identity (all of space fixed)

  • $k=3$, $\ell=0$, $m=0$, $b\ne 0$: translation (no fixed points)

  • $k=2$, $\ell=1$, $m=0$, $\pi_U(b)=0$: reflection (a plane of fixed points)

  • $k=2$, $\ell=1$, $m=0$, $\pi_U(b)\ne 0$: glide reflection (no fixed points)

  • $k=1$, $\ell=2$, $m=0$, $\pi_U(b)=0$: 180° rotation (reflection in a line; an axis of fixed points)

  • $k=1$, $\ell=0$, $m=1$, $\pi_U(b)=0$: rotation (an axis of fixed points)

  • $k=1$, $\ell=2$, $m=0$, $\pi_U(b)\ne 0$: 180° rotation plus translation along rotation axis (screw motion; no fixed points)

  • $k=1$, $\ell=0$, $m=1$, $\pi_U(b)\ne 0$: rotation plus translation along rotation axis (screw motion; no fixed points)

  • $k=0$, $\ell=3$, $m=0$: inversion in a point (180° rotation in an axis plus reflection in a plane normal to this axis; one fixed point)

  • $k=0$, $\ell=1$, $m=1$: rotation in an axis plus reflection in a plane normal to this axis (rotatory reflection or rotoreflection; one fixed point)

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.