Let $M_{n}(F)$, $n\times n$ matrices over $F$, and $G=GL(n,F)$, $n\times n$ invertible matrices. $G$ acts on $M_{n}(F)$ via left multiplication. Find the orbits of this group action.
Firstly, all the $G$ in $M$ is an orbit from the linear algebra fact. It's also clear that left multiplication can be seen as row operation. But when $G$ acts on the non-invertible matrices, things will become very complicated since there are so many possibilities of reduced row echelon forms. My question is how can we find a neater way to organize all those orbits and express them clearly ?
Best Answer
In the following I denote kernel and image of a matrix $X$ by $N(X)$ and $R(X)$, respectively. The orbit of a matrix $X$ is given by $$ \mathcal O(X) = \{Y\in M_n(F) : N(Y) = N(X)\}. $$ The inclusion "$\subset$" is obvious. Let us show the converse. For this, let $Y\in M_n(F)$ with $N :=N(Y) = N(X)$. Then, by the rank-nullity theorem, $\dim R(Y) = \dim R(X)$. Choose some subspace $M$ such that $F^n = N\oplus M$. Then both $\hat X := X|M$ and $\hat Y := Y|M$ are bijective, mapping $M$ to $R(X)$ and $R(Y)$, respectively. Hence $U := \hat Y\hat X^{-1} : R(X)\to R(Y)$ is a well-defined linear mapping. Let $x\in F^n$, $x = u + v$, where $u\in N$ and $v\in M$. Then $$ UXx = U\hat Xv = \hat Yv = Yx. $$ Hence, we have $UX = Y$. Now, let $L_1$ and $L_2$ be subspaces such that $$ F^n = L_1\oplus R(X) = L_2\oplus R(Y) $$ and choose some bijective $V : L_1\to L_2$. If $v\in L_1$ and $u\in R(X)$, define $$ G(v+u) := Vv + Uu. $$ Then $G\in\operatorname{GL}$ and $GX = UX = Y$. Thus, $Y\in\mathcal O(X)$.