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Show that there is no matrix $X\in \operatorname{M}_{2\times2}(\mathbb{C})$ with $\operatorname{Tr}(X)=0$ and $e^X=A$, where

$$ A=\begin{pmatrix} -1 & 1 \\ 0 & -1 \\ \end{pmatrix}.$$

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Note that since ${\rm tr}X=0$, the eigenvalues of $X$ are given by

$$\lambda_{1,2}=\pm\sqrt{-{\rm det}X}$$

If ${\rm det}X\neq 0$ then $X$ has two different eigenvalues and it can be diagonalized. Thus $e^{X}$ can also be diagonalized in contradiction with $A=e^{X}$ which cannot.

If ${\rm det}X=0$ then $X\sim\left(\begin{matrix}0&x\\0&0\end{matrix}\right)$. This matrix is nilpotent of order two and therefore $e^{X}=I+X=\left(\begin{matrix}1&x\\0&1\end{matrix}\right)$ which again cannot be equal to $A$.

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