The exponential map is neither injective nor surjective when $n\ge 2$? Consider the exponential map
$$\exp: M_n(\mathbb R) \to \{g:\det(g)>0 \}.$$
When $n=1$ this is bijective. But what happens when $n>1$? I was trying to come up with examples that fails the injectivity/surjectivity but I don't have some good examples in mind. It would be great if there is a series of examples that fails all $n\ge 2$.
Thanks in advance!
 A: Hint: Consider $\mathbb{C}$ as $\mathbb{R}^2$ and $exp(it)$ it is not injective where $i$ is represented in the basis $(1,i)$ by $\pmatrix{0& -1\cr 1&0}$.
A: The other answer gives an example where injectivity fails. For surjectivity, the comment on your answer gives the example
$$
A = \pmatrix{-1&1\\0&-1},
$$
and indeed there is no matrix $X$ such that $A = \exp(X)$. To see that this is the case, suppose for the purpose of contradiction that such an $X$ exists. Note that for any eigenvalue $\lambda$ of $X$, $e^\lambda$ is an eigenvalue of $e^X$. It follows that in order for $e^X$ to have an eigenvalue of $-1$, its eigenvalues must be of the form  $\pi k i$ for some odd integer $k$.
Because $X$ is a real matrix, its eigenvalues come in conjugate pairs, which is to say that both $\pm \pi ki$ must be eigenvalues of $X$ for some odd $k$.  Thus, $X$ is diagonalizable over $\Bbb C$, which means that $e^X$ is diagonalizable over $\Bbb C$ with $-1$ as its only eigenvalue, which means that $e^X = -I$, so $e^{X}\neq A$.
