conditions under which real-matrix exponential are equivalent Consider $M_{1}$, $M_{2}\in\mathbb{R}^{2\times2}$, $k\in\mathbb{R}$, $M_{1}\neq M_{2}$. Under what conditions is $e^{M_{1}}=e^{kM_{2}}$?
Thanks!
 A: In essence, you are asking if $e^X=e^Y$ for some $X,Y\in M_2(\mathbb{R})$, how are $X$ and $Y$ related. The most obvious possibility, of course, is $X=Y$. Are there other possibilities?
This question can be answered using real Jordan decomposition. Every real matrix $A$ can be written as $PJP^{-1}$ for some $P\in GL_n(\mathbb{R})$ and some real matrix $J$ that is called the real Jordan form of $A$. When $A$ is 2-by-2, $J$ belongs to one of the following four classes of matrices:


*

*$J$ is a diagonal matrix whose diagonal entries are distinct and arranged in descending order, i.e. $J=\operatorname{diag}(\lambda,\mu)$ with $\lambda>\mu$. We have $e^J=\operatorname{diag}(e^\lambda,e^\mu)$ with $\lambda>\mu$ in this case.

*$J$ is a 2-by-2 Jordan block $J_2(\lambda)=\begin{pmatrix}\lambda&1\\0&\lambda\end{pmatrix}$. We have $e^J=e^\lambda J_2(1)=\begin{pmatrix}e^\lambda&e^\lambda\\0&e^\lambda\end{pmatrix}$ in this case.

*$J$ is a multiple of a rotation matrix but $J$ is not a multiple of $I$. That is, $J=\begin{pmatrix}p&-q\\q&p\end{pmatrix}$ with $q\neq0$. Such a matrix $J$ corresponds to the complex Jordan form $\operatorname{diag}(p+iq,p-iq)$. We have $e^J=e^p\begin{pmatrix}\cos q&-\sin q\\ \sin q&\cos q\end{pmatrix}$. So, $e^J$ belongs to class 3 if $q$ is not an integral multiple of $\pi$, or it belongs to class 4 otherwise.

*$J$ is a multiple of $I$.


Now suppose $e^X=e^Y=A$. Let $J_A$ be the real Jordan form of $A$, $X=PJ_XP^{-1}$ and $Y=QJ_YQ^{-1}$ be real Jordan decompostions of $X$ and $Y$, and $S=Q^{-1}P$.
Since $e^X=e^Y=A$, the matrices $e^{J_X}$ and $e^{J_Y}$ must be similar to $J_A$. When $J_A$ belongs to class 1, it is not hard to see that $J_X$ and $J_Y$ are identical class 1 matrices and $e^{J_X}=e^{J_Y}$ is also of class 1. Furthermore, since $e^X=e^Y$, we must have $Se^{J_X}=e^{J_X}S$. Hence $S$ commutes with a class 1 matrix. However, it is easy to show that every matrix $S$ that commutes with a class 1 matrix must also commute with all class 1 matrices (actually, $S$ must be a diagonal matrix). So, $X=PJP^{-1}=QSJS^{-1}Q^{-1}=QJQ^{-1}=Y$, i.e. $X$ must be equal to $Y$.
When $J_A$ belongs to class 2, by a similar argument, one can show that $J_X$ and $J_Y$ are identical class 2 matrices and $e^{J_X}=e^{J_Y}$ is a nonzero multiple of a class 2 matrix. As every matrix $S$ that commutes with a class 2 matrix must also commute with all class 2 matrices, we again have $X=PJP^{-1}=QSJS^{-1}Q^{-1}=QJQ^{-1}=Y$, i.e. $X$ must be equal to $Y$.
When $J_A$ belongs to class 3, things are different. One can show that $e^{J_X}=e^{J_Y}=J_A$ and hence both $J_X$ and $J_Y$ are class 3 matrices of the form
$$
\begin{pmatrix}a&-b-2m\pi\\b+2m\pi&a\end{pmatrix},
$$
with the same values of $a\in\mathbb{R}$ and $b\in(-\pi,\pi)\setminus0$ but with possibly different $m\in\mathbb{Z}$. So, it is no longer necessarily true that $J_X=J_Y$ or $X=Y$. However, $e^{J_X}$ still belongs to class 3. As every matrix that commutes with it must be a multiple of rotation matrix, and a rotation matrix commutes with all matrices of class 3, we have
$X=PJ_XP^{-1}=QSJ_XS^{-1}Q^{-1}=QJ_XQ^{-1}$. Therefore, while it is not necessarily true that $X=Y$ in this case, we may assume WLOG that $P=Q$.
When $J_A$ belongs to class 4, with $J_A=kI$ for some $k\neq0$, one can show that $e^{J_X}=e^{J_Y}=kI$ when both $J_X$ and $J_Y$ are of the form
$$
\begin{cases}
\begin{pmatrix}\log(k)&-2m\pi\\2m\pi&\log(k)\end{pmatrix}&k>0,\\
\begin{pmatrix}\log|k|&-(2m+1)\pi\\(2m+1)\pi&\log|k|\end{pmatrix}&k<0,
\end{cases}
$$
with possibly different values of $m\in\mathbb{Z}$ for $X$ and $Y$. Furthermore, as $e^{J_X}=e^{J_Y}=kI$ in this case, the matrices $P$ and $Q$ are arbitrary and hence $X$ can be different from $Y$.
A: If $M_1,M_2$ commute then $-M_1,kM_2$ likewise commute thus (I use this in the 2nd =)
$$ e^{M_1} = e^{kM_2} \qquad \Rightarrow \qquad I = e^{-M_1} e^{kM_2} =e^{kM_2-M_1} $$
Naturally, $kM_2-M_1=0$ solves the above equation. Moreover, this solution gives $kM_2 = M_1$ which is consistent with our supposition that the matrices commute.
Questions: is there any other matrix such that $e^A=I$? If $M_1,M_2$ do not commute then can we find additional solutions? I leave those questions for you.
