A question related to pole of a matrix( Pole defined in the question)

Question

For a $3\times 3$ matrix A, say that a point p on the on the unit sphere centred at the origin in $\mathbb{R}^3$ is a pole of A if Ap=p. Denote by $SO_3$ , the subgroup of $GL_{3}(\mathbb{R})$ consisting of all the orthogonal matrices with determinant 1.

(A) Show that if $A\in SO_3$ , then A has a pole.

(B) Let G be a subgroup of $SO_3$. Show that G acts on the set { $p\in \mathbb{S}^2$ | p is a pole for some matrix $A\in G$}

This question is from my abstract algebra assignment and I was unable to solve it.

For (A), I am sorry to say I don't have any clue. I have been given that for any matrix $AA^T= I = A^T A$ and det A=1. The problem is how to find such a p that satisfies Ap=p.

For (B), if G acts on the given set (say X) then I X =X and $(G_1 G_2) X = (G_1 )(G_2 X)$.

then I p = p, but ${G_1}{ G_2 } p $ , but I can't write $G_2 p = G_2$ because p is pole of some matrix $A \in G$ not necessarily $G_2$. So, I got struck here.

Can you please guide?

Answer 1

For part a, it suffices to show that $\det(A - I) = 0$. Since you have already showed that $\det(A) = 1$, it suffices to note that $$ \begin{align} \det(A - I) &= \det(A^T) \det(A - I) \\ & = \det(A^TA - A) \\ &= \det(I - A) = \det(-(A - I)) = - \det(A - I). \end{align} $$ For part b, it suffices to note that every $p$ is a pole of the identity matrix $I \in G$.

Answer 2

if you are really clueless on a groups problem, a common technique is to step back and look at the generators of a group. In particular consider $O_n(\mathbb R)$. The generators of this group are Householder (reflection) matrices:
$H:= I - 2 \mathbf {xx}^T$ for some $\big \Vert \mathbf x\big \Vert_2=1$. You can verify that this reflection is in fact an involution i.e. $H^2=I$

By examining the eigenvalues or trace of an involution or using matrix determinant lemma (or the geometry) you can see
$\det\big(H\big)=-1$.

Every $Q \in O_n(\mathbb R)$ may be written as a product of at most $n$ Householder matrices (exercise).

Since you are interested in $SO_n(\mathbb R) \subset O_n(\mathbb R)$, this means every matrix may be written as the product of an even number of Householder matrices, which turns into: for $A \in SO_3(\mathbb R)$

$A = H_1\cdot H_2 = \big(I - 2 \mathbf x_1\mathbf x_1^T\big)\big(I - 2 \mathbf x_2\mathbf x_2^T\big)$
There must exist some unit length $\mathbf x_3$such that $A\mathbf x_3 = \mathbf x_3$ -- i.e. up to re-scaling, select non-zero

$\mathbf x_3 \in \ker \left( \begin{bmatrix}\mathbf x_1^T \\\mathbf x_2^T \\ \end{bmatrix}\right)$

This also implies the desired answer for $B$ -- consider the order 2 group $G:=\big\{H, I\big\}$ for some $H$ -- arbitrary $p$ isn't a fixed point for $H$ but is for $I$ and $I$ is in every subgroup, giving the answer.