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If we let $S$ be the the set of real $m \times n$ matrices and take G to be the direct product $G=GL(m,\mathbb{R})\times GL(n,\mathbb{R})$ and consider the action of G on S as follows

$(P,Q) \star A=PAQ^{-1}$

how is S decomposed into G orbits?

Having verified that this is indeed an action on the set, I think I'm looking to find the equivalence classes of $\sim$ where $A \sim B$ if $B$ can be expressed in the form $PAQ^{-1}=B$ for some matrices $P$, $Q$ of the required dimension.

Intuitively, I imagine we call always find invertible matrices $P$, $Q$ that satisfy the condition, though I'm not sure if this is the case. Any help would be much appreciated. Best.

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    @MiamiMath: the B_r in Calle's answer is called RREF. For 1x1s you can just exclude 0, but for 2x2s you get three orbits, [0,0;0,0] is one very small orbit, but [1,0;0,0] is a larger orbit, and [1,0;0,1] is even larger (and disjoint). Invertible matrices don't change the rank, just like they don't change a nonzero matrix into a zero one.2012-02-22

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Let $S_r$ be the set of real matrices of rank $r$. Then $S_r$ is an orbit of this action. I.e., if $A, B$ are matrices with rank $r$, there exists a $(P, Q) \in G$ such that $(P, Q) \star A = PAQ^{-1} = B.$

The easiest way to prove this is probably by choosing the matrix $B_r$ to be the $m \times n$ matrix with $r$ ones on it's "diagonal" and the rest of the elements being zero, then showing that every $A \in S_r$ can be transformed into $B_r$ using the action.

Moreover, no orbit can contain more than $S_r$ since $\operatorname{rank}(PAQ^{-1}) = \operatorname{rank}(AQ^{-1}) = \operatorname{rank} A.$

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    This is of course the "correct" answer. It is the same over any field, not just the real numbers.2012-02-22