# Decomposing the set of $2 \times 2$ complex matrices into orbits under left multiplication

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?

• 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! – Thomas May 9 '14 at 0:57
• I mean $n + 1$ orbits. – Thomas May 9 '14 at 2:00
• @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? – user1551 May 9 '14 at 10:25