Jordan form in $SL_2(\Bbb C)$

Question

Is there a matrix $M \in \mathrm{GL}_2(\Bbb C)$ such that $\mathrm{tr}(M)=0$ but $$ M \neq S^{-1} \begin{pmatrix}a & 0\\ 0 & -a\end{pmatrix} S \quad\text{and}\quad M \neq S^{-1} \begin{pmatrix}0 & 1\\ 0 & 0\end{pmatrix} S $$ for any $S \in \mathrm{SL}_2(\Bbb C)$ and $a \in \Bbb C \setminus \{0\}$ ?

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My motivation comes from the fact that this is true for some $S \in \mathrm{GL}_2(\Bbb C)$ : the Jordan normal form of $M$ is either $$ \begin{pmatrix}a & 0\\ 0 & -a\end{pmatrix} \quad\text{or}\quad \begin{pmatrix}0 & 1\\ 0 & 0\end{pmatrix} $$ since the trace of $M$ is $0$. But my question is: can we still do this with some $S \in \mathrm{SL}_2(\Bbb C)$ ?

Answer 1

If $M=S^{-1}NS$ for some matrix $S\in \operatorname{GL}_2(\Bbb C)$, then $$M=S^{-1}NS=(\det(S))^{\frac{1}{2}}S^{-1}NS (\det(S))^{-\frac{1}{2}}=\left(\frac{S}{\sqrt{\det(S)}}\right)^{-1}N\left(\frac{S}{\sqrt{\det(S)}}\right)=T^{-1}NT,$$ with $T=\frac{S}{\sqrt{\det(S)}}\in \operatorname{SL}_2(\Bbb C)$, since $$\det(T)=\det\left(\frac{S}{\sqrt{\det(S)}}\right)=\frac{\det(S)}{\sqrt{\det(S)}^2}=1.$$

Answer 2

Conjugacy in $\mathrm{SL}_2(\Bbb C)$ is the same as conjugacy in $\mathrm{GL}_2(\Bbb C)$. You can replace your conjugating matrix $S\in\mathrm{GL}_2(\Bbb C)$ by $t S$ where $t^{-2}=\det S$.