Originally Posted by

**oblixps** so what i'm trying to show is that the set of all 3x3 matrices with determinant = 1, such that for a fixed vector r, Ar = r is isomorphic to the set of all 2x2 matrices with determinant = 1.

since r is an eigenvector with eigenvalue 1 of the 3x3 matrix, the other 2 eigenvalues of the 3x3 matrix must multiply to 1 and i was thinking about how to map each 3x3 matrix with eigenvalues 1, $\displaystyle \lambda_1 $, $\displaystyle \lambda_2 $ to a 2x2 matrix with eigenvalues $\displaystyle \lambda_1 $ and $\displaystyle \lambda_2 $. however i am having trouble coming up with the explicit mapping. I tried taking a 3x3 matrix with those specified eigenvalues and thought about how to "trim the edges" (for example, remove the first column and first row) to get a 2x2 matrix with the same eigenvalues as the eigenvalues besides the 1 of the 3x3 matrix. but this process may "mess up" the determinant for some 3x3 matrices and end up creating a 2x2 matrix with different eigenvalues then the one i wanted. furthermore i don't even think that this is well defined.

can someone give me some hints in the right direction on how to define such a function from one set to the other? thanks.

let $\displaystyle k$ be the base field. of course we also need $\displaystyle r \neq 0.$ let $\displaystyle G=\{A \in SO(3): \ Ar=r \}.$ since $\displaystyle r \neq 0,$ there exists a basis $\displaystyle \{r,u,v\}$ of $\displaystyle k^3$ and, with respect to that basis, every element of $\displaystyle G$ is in the form

$\displaystyle A=\begin{pmatrix} 1 & a & b \\ 0 & x & y \\ 0 & z & t \end{pmatrix},$

for some $\displaystyle a,b,x,y,z,t \in k.$ now since $\displaystyle A \in SO(3),$ we have $\displaystyle B = \begin{pmatrix} x & y \\ z & t \end{pmatrix} \in SO(2)$ and $\displaystyle a =b=0.$ so the group isomorphism you're looking for is the map $\displaystyle A \mapsto B.$