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I need some assistance with the proof for part (b) of the following problem statement:

Problem Statement: Decompose the set $\mathbb{C}^{2\times2}$ of $2\times2$ complex matrices into orbits for the following operations of $GL_{2}(\mathbb{C})$:

(a) left multiplication

(b) conjugation

So, I was able to decompose $\mathbb{C}^{2\times2}$ into four orbits for part (a) fairly simply, but I am a little unsure how to go about part (b).

So I know that in the general case, the orbit of $A\in\mathbb{C}^{2\times2}$ is defined as: $$O_{A}=\left\{B\in\mathbb{C}^{2\times2} \mid B=P^{-1}AP \text{ for } P\in GL_{2}(\mathbb{C})\right\}$$

I was thinking using the fact that for any $A\in\mathbb{C}^{2\times2}$ there is some $P\in GL_{2}(\mathbb{C})$ such that $B=P^{-1}AP$ has Jordan block form, being either a diagonal or triangular matrix? But this would not necessarily be true $\forall A\in\mathbb{C}^{2\times2}, P\in GL_{2}(\mathbb{C})$? Because any $A\in\mathbb{C}^{2\times2}$ would have a distinct $P\in GL_{2}(\mathbb{C})$ such that conjugation by $P$ results in the Jordan form of $A$?

Would the use of an equivalence relation defined as $A\sim B$ iff $B=P^{-1}AP$ for some $P\in GL_{2}(\mathbb{C})$ be helpful?

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    $\begingroup$ Jordan normal form is more or less the answer to this question; you just need to figure out when two Jordan normal forms are conjugate which is not so bad. $\endgroup$ – Qiaochu Yuan Nov 15 '15 at 22:50
  • $\begingroup$ So should I consider cases of $A$ having two distinct eigenvalues, one distinct eigenvalue of multiplicity 2, and no eigenvalues? $\endgroup$ – yung_Pabs Nov 15 '15 at 22:52
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    $\begingroup$ There's no such thing as "no eigenvalues" over $\mathbb{C}$. We only have the first two cases. $\endgroup$ – Qiaochu Yuan Nov 15 '15 at 23:01
  • $\begingroup$ Oh yes that's right. So for any matrix $A$ if it has two-distinct eigenvalues, it's Jordan form is a diagonal. If it has one-distinct eigenval, it is either diagonal or triangular (depending on the dimension of the eigenspace). But since we must consider conjugation by all invertible matrices, doesn't that mean that there will be some matrices $P^{-1}AP$ that are not Jordan form? $\endgroup$ – yung_Pabs Nov 15 '15 at 23:21
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    $\begingroup$ Those are two different collections of orbits, yes (although the notation you're using is not great because it looks like you're talking about double cosets, which you aren't). $\endgroup$ – Qiaochu Yuan Nov 16 '15 at 23:33

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