# Matrix - Linear algebra problem

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

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

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 $$A$$.