Every invertible square matrix with complex entries can be written as the exponential of a complex matrix. I wish to ask if it is true that
Every invertible real matrix with positive determinant can be written as the exponential of a real matrix. (We need +ve determinant condition because if $A=e^X$ then $\det A=e^{\operatorname{tr}(X)} > 0$.) If not is there a simple characterization of such real matrices (with +ve determinant) which are exponentials of other matrices ?
No, a real matrix has a real logarithm if and only if it is nonsingular and in its (complex) Jordan normal form, every Jordan block corresponding to a negative eigenvalue occurs an even number of times. So, it is possible that a matrix with positive determinant is not the matrix exponential of a real matrix. Here are two counterexamples: $\pmatrix{-1&1\\ 0&-1}$ and $\operatorname{diag}(-2,-\frac12,1,\ldots,1)$. For more details, see
Walter J. Culver, On the existence and uniqueness of the real logarithm of a matrix, Proceedings of the American Mathematical Society, 17(5): 1146-1151, 1966.
Another characterization is as follows: $A$ is the exponential of a real matrix iff $A$ is the square of a real invertible matrix. In particular, remark that if $A=e^X$, then $A=(e^{X/2})^2$.
Concerning the Lee's post, if $A$ is in a neighborhood of $I$, then $A=I+B$ with $||B||<1$ and we can take $X=B-B^2/2+B^3/3+\cdots$.
EDIT: An outline of the proof. Let $A=B^2$ where $B$ is invertible real; we may assume that $B$ is in Jordan form. For each Jordan block $B_k=\lambda I_k+J_k$ of $B$ s.t. $\lambda<0$, change $B_k$ with $-B_k$. Finally you obtain a matrix $C$ s.t. $C^2=A$. It is not difficult to show that such a matrix $C$ which has no $<0$ eigenvalues is the exp of a real matrix.
