If we consider the equation : $$e^x=1+x, $$ where $x\in\mathbb{R}$, then the only solution of the equation is $x=0$, and it's a consequence of the convexity of the exponential function.
Now,if we consider the equation : $$e^X=I_n+X, $$ where $X$ is a $n\times n$ real matrix, $X=0$ is a solution.
Is it the only solution to the problem ? If yes, how to prove it ?
On
We can obtain all complex solutions.
$\textbf{Proposition}$. Let $X\in M_n(\mathbb{C})$ and $U=\{z\in\mathbb{C};e^z=1+z\}$. Then $e^X=I+X$ iff $X$ is similar to a matrix in the form
$diag(K_1,\cdots,K_p,\lambda_1,\cdots,\lambda_{n-2p})$, where
$K_i=\begin{pmatrix}0&1\\0&0\end{pmatrix}$, $\lambda_j\in U$.
$\textbf{Proof}$. We may assume that $X=diag(\mu_1I_{r_1}+J_{r_1},\cdots,\mu_kI_{r_k}+J_{r_k})$ where $J_r$ is the nilpotent Jordan block of dimension $r$. Then it suffices to assume that $X=\mu I_n+J_n$, where $n\geq 2$.
We obtain $e^{\mu}e^J=(\mu+1)I+J$, that is
$\mu\in U$ and $e^{\mu}(J+J^2/2+J^3/6+\cdots)=J$, that is
$\mu\in U,\mu\in 2i\pi\mathbb{Z}$ and $J^2=0$, that is
$\mu=0,J^2=0$, and we are done.$\square$
$\textbf{Remark}$. If we consider only the real matrices $X$, then $X$ is similar, over $\mathbb{R}$, to
$diag(K_1,\cdots,K_p,L_1,\cdots,L_q,0,\cdots,0)$, where $L_j=\begin{pmatrix} Re(\lambda_j)&Im(\lambda_j)\\-Im(\lambda_j)&Re(\lambda_j)\end{pmatrix}$ and $\lambda_j\in U\setminus \{0\}$ that is, the set
$user1551 \oplus Wojowu$.
No. E.g. for any $X$ such that $X^2=0$, we have $e^X=I+X$.