I have the group $GL(3,\mathbb R)$ (the group of invertible 3×3 matrices) acting on $M_{33}(\mathbb R)$ (the set of 3×3 real matrices) by $A\cdot M=AM$. Let
$$M_1 = \begin{pmatrix} 1 & 0 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \end{pmatrix}$$
$$M_2 = \begin{pmatrix} 1 & -1 & 0 \\ 0 & 2 & 2 \\ 1 & 1 & 2 \end{pmatrix}$$
I need to determine a matrix $A$ such that $$\text{Stab}(M_1)=A\text{Stab}(M_2)A^{-1}$$
I've shown that $M_1$ and $M_2$ lie in the same orbit because they are row equivalent and $M_1$ is basically the rref form of $M_2$.
I'm having trouble, however, understanding what the stabilizer of a matrix will be. For $M_1$, I believe it will contain, for example, the matrices corresponding to the row operations of multiplying row 3 (because row 3 is 0), but other than that, I have no idea what the Stab will be, let alone determine that matrix $A$.
The $A\text{Stab}(M_2)A^{-1}$ does look a bit like diagonalization, but I'm thinking $\text{Stab}(M_1)$ will contain non-diagonal matrices (for example, the matrix corresponding to adding 2 times (row 3) to (row 1)). Could anybody give me some hints as to how to approach this?
Best Answer
(1) let $G$ be a group acting on a set $X$. Two elemets $x,y$ of $X$ lying in the same orbit ($x=g\cdot y$) have conjugate stabilisers : $$Stab(x)=gStab(y)g^{-1},$$ where $Stab(x)=\{g\in G\vert g\cdot x=x\}$.
(2) In your case $G=GL(3,R)$ and $X=M_n(3,R)$ and the stabiliser of a matrix $M$ is the set of matrices $B$ such that $BM=M$.
(3) If $A\in Gl(3,R)$ satisfy $M_1=AM_2$ you get (using (1)) that $Stab(M_1)=AStab(M_2)A^{-1}$. So you should find such an $A$. Note that a row operation is given by a left multiplication.