Find logarithm of matrix in diagonalform

Question

I'm trying to decide whether there is a matrix $A \in \text{Mat}(2, \mathbb{R})$ with

$$e^A = \begin{pmatrix}1 & -1 \\ -1 & 1\end{pmatrix}$$

I was already able to diagonalise the matrix into

$$\begin{pmatrix}1 & -1 \\ -1 & 1\end{pmatrix} = S^{-1} \cdot \text{diag}(0, 2) \cdot S \hspace{30pt} \text{with} \hspace{30pt} S = \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}$$

I'm not sure how to proceed from here. What could be the next step?

Answer 1

The answer is negative: $e^A$ cannot have a vanishing eigenvalue; in fact it is always invertible and the inverse is $e^{-A}$. So the matrix $A$ does not exists

Answer 2

There is no such matrix. If there was a matrix $B$ such that $e^B=A$, then\begin{align}0&=\det A\\&=\det e^B\\&=e^{\operatorname{tr} B}.\end{align}