Find $A$ such that $e^A = B$ for a diagonal-matrix $B$ I'm trying to decide where're there exists a matrix $A \in \text{Mat}(2, \mathbb{R})$ such that
$$e^A = \begin{pmatrix}1 & 0 \\ 0 & -1\end{pmatrix} =: B$$
In particular I don't want to use the identity $\det(B) = e^{\text{trace}(A)}$. Is there another way to approach this?
 A: We shall prove that, for any $B\in\text{Mat}_{n\times n}(\mathbb{R})$ with $\det(B)\leq 0$, the equation $\exp(A)=B$ has no solution $A\in\text{Mat}_{n\times n}(\mathbb{R})$.  In particular, the matrix $B$ in question has $\det(B)=-1<0$, which fits our hypothesis.  
Suppose on the contrary that $A$ exists.  Let $f:[0,1]\to\text{Mat}_{n\times n}(\mathbb{R})$ be the map $$f(t):=\exp(tA)$$
for all $t\in[0,1]$.  Therefore, $f$ forms a path from the identity matrix $I$ to $\exp(A)=B$.  (Note that $f$ is continuous as it is differentiable: $f'(t)=A\,f(t)$ for all $t\in[0,1]$.)  
Now, $g:[0,1]\to\mathbb{R}$ given by
$$g(t):=\det\big(f(t)\big)$$
for each $t\in[0,1]$ is a path in $\mathbb{R}$ from $\det(I)=+1$ to $\det(B)\leq 0$.  (Since $\det$ is a polynomial function, it is continuous.  Therefore, a composition of $\det$ with the continuous map $f$ is another continuous map $g$.)  Therefore, $0$ lies in the image of $g$.  Consequently, $g(\tau)=0$ for some $\tau\in(0,1]$.  Ergo, $\exp(\tau A)$ has determinant $0$, and whence noninvertible.  However, we know that $\exp(-\tau A)$ is the inverse of $\exp(\tau A)$, leading to a contradiction.
A: $\exp(\mathcal{M}_n(\mathbb{R}))=\{M\in\text{GL}_n(\mathbb{R}), \exists A\in\text{GL}_n(\mathbb{R}),M=A^2\}$. In your case, there doesn't exist $A\in\text{GL}_n(\mathbb{R})$ such that $A^2=B$ because $-1$ is an eigenvalue of $B$.
Proof of the first equality: Notice first that $\exp(\mathcal{M}_n(\mathbb{C}))=\text{GL}_n(\mathbb{C})$ and that for all $A\in\mathcal{M}_n(\mathbb{R}),\exp(A)=\exp\left(\frac{A}{2}\right)^2$. Let $A\in\text{GL}_n(\mathbb{R})$ such that there exists $B\in\mathcal{M}_n(\mathbb{R})$ such that $A=B^2$. There exists $P\in\mathbb{C}[X]$ such that $B=\exp(P(B))$ (it results from the proof of $\exp(\mathcal{M}_n(\mathbb{C}))=\text{GL}_n(\mathbb{C})$), then $\overline{B}=\exp(\overline{P}(B))$ and since $B\overline{B}=\overline{B}B=A$ we have $P(B)\overline{P}(B)=\overline{P}(B)P(B)$ and thus $A=\exp(P(B)+\overline{P}(B))=\exp(Q(B))$ where $Q=P+\overline{P}\in\mathbb{R}[X]$, and since $B\in\mathcal{M}_n(\mathbb{R})$, $Q(B)\in\mathcal{M}_n(\mathbb{R})$.
