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.

Given distinct proper rigid transformations $A, B \in \operatorname{SE}(n)$, what is the maximum dimension of the nullspace of $A - B$? That is, what is the maximum dimension of $\operatorname{Eq}(A, B) = \{x \in \mathbb{R}^n\ |\ A(x) = B(x)\}$?

I conjecture that the answer is $n - 2$ by observing simple cases:

When $n = 2$, proper rigid transformations are "defined" by $2$ affinely independent points, so $\operatorname{Eq}(A, B)$ can contain at most one point and has a dimension of $0$.

When $n = 3$, proper rigid transformations are "defined" by $3$ affinely independent points, so $\operatorname{Eq}(A, B)$ is at most the affine hull of two affinely independent points and has a dimension of $1$.

As this line of reasoning is not quite rigorous, I cannot tell if it easily extends to higher dimensions (also hampered by my lack of four-dimensional imagination!). Is my conjecture true for all integers $n \geq 2$?

share|improve this question
add comment

1 Answer 1

up vote 1 down vote accepted

Yes. WLOG, suppose $A$ and $B$ are linear (so there are no translations). Let $A,B$ also denote the matrix representations of the two transofrmations. As they belong to $SE(n)$, $A^TB\in SO(n)$. Hence $A^TB=UDU^\ast$ for some unitary matrix $U$ and some complex diagonal matrix $D$ whose diagonal entries have unit moduli. Therefore $$\operatorname{nullity}(A-B)=\operatorname{nullity}(I-D).$$ If $\operatorname{nullity}(A-B)=n-1$, $D$ must have $n-1$ eigenvalues equal to $1$. However, as $A^TB$ is real orthogonal and $A\not=B$, the remaining eigenvalue of $D$ must be equal to $-1$. But then $\det A^TB=-1$, which is a contradiction. Therefore the nullity of $A-B$ is at most $n-2$.

share|improve this answer
    
I'm sorry but i can't follow your argument well due to a inadequate exposure to linear algebra. Could you explain why there is no loss of generality in your assumption? –  Herng Yi Feb 7 '13 at 13:47
1  
@HerngYi Suppose $A(x)=Px+p$ and $B(x)=Qx+q$, where $P,Q$ are matrices and $p,q$ are translation vectors. The nullspace of $A-B$ is then the set $H=\{x: (P-Q)x+(p-q)=0\}$. Suppose $x_0\in H$. Then $H=x_0+\{u: (P-Q)u=0\}$. So the dimension of the affine subspace $H$ is equal to the dimension of the linear subspace $\{u: (P-Q)u=0\}$. In other words, as long as the dimension of the nullspace is concerned, $\dim H$ can be calculated as if $p=q=0$. –  user1551 Feb 8 '13 at 9:08
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.