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I have some issues with a problem which is asking me to decompose the set of $2 \times 2$ complex matrices $\mathbb{C}^{2 \times 2}$ in orbits under the left multiplication operation on the group $GL_2 (\mathbb{C})$. I've proved that the subset of invertible matrices forms an orbit by using the fact that orbits are an equivalence relation and so are transitive. I've also shown (trivially) that the orbit of the zero matrix is of size one and contains only the zero matrix, but I'm not making any progress on decomposing the set of non-invertible, non-zero matrices - I don't understand what these orbits look like. Surely it isn't sufficient to say that any element of $g$ always carries a non-invertible matrix to a non-invertible matrix? There could be lots of smaller orbits contained in that subset, or am I misunderstanding/overthinking this?

Can someone help?

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Hint. Elementary matrices (i.e. the matrices that perform elementary row operations) form a subset of invertible matrices. Do you recall any special form of matrix that is obtained by applying a series of elementary row operations to a matrix?

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  • $\begingroup$ An echelon form row reduced matrix... depending on the rank of the starting matrix I'll end up with an rref matrix with up to $n$ zero rows... so that means there would be exactly $n$ orbits, one for each possible rank, and I found the full rank and zero rank cases already. That makes sense, I'll give it a try this afternoon and report back! Thanks! $\endgroup$ – Thomas May 9 '14 at 0:57
  • $\begingroup$ I mean $n + 1$ orbits. $\endgroup$ – Thomas May 9 '14 at 2:00
  • $\begingroup$ Ok I think I got it, there are exactly three orbits, the first one containing all matrices of rank 2 (invertible), the other one all matrices of rank 1, and the last one all matrices of rank 0 (= the zero matrix). Thanks for the hint! $\endgroup$ – Thomas May 9 '14 at 9:39
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    $\begingroup$ @Thomas Don't jump to conclusion so quickly. Consider $A=\pmatrix{1&0\\ 0&0}$ and $B=\pmatrix{1&3\\ 0&0}$. Both are row reduced echelon forms of the same rank. Is $B=PA$ for some invertible matrix $P$? What about matrices $\pmatrix{1&t\\ 0&0}$ with different $t$ in general? $\endgroup$ – user1551 May 9 '14 at 10:25
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    $\begingroup$ @Thomas Yes, you are right. There are actually infinitely many 2x2 row reduced echelon forms of rank 1. Shower is indeed helpful to research! $\endgroup$ – user1551 May 9 '14 at 10:52

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