# 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!

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:

1. $J$ 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.
2. $J$ 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.
3. $J$ 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.
4. $J$ 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 $Y$.

• If $P_{X}\neq P_{Y}$, it isn't immediately clear to me why $e^{a}P_{X}\begin{pmatrix}\cos b & -\sin b\\ \sin b & \cos b \end{pmatrix}P_{X}^{-1}=e^{a}P_{Y}\begin{pmatrix}\cos b & -\sin b\\ \sin b & \cos b \end{pmatrix}P_{Y}^{-1}$. Could you elaborate? Thanks! – chris Mar 25 '13 at 0:37
• @chris This is because $P_X=P_YR$ for some $R$ that is a scalar multiple of rotation matrix. As rotation matrices commute, we have $R\begin{pmatrix}\cos b & -\sin b\\ \sin b & \cos b\end{pmatrix}R^{-1}=\begin{pmatrix}\cos b & -\sin b\\ \sin b & \cos b\end{pmatrix}$. – user1551 Mar 25 '13 at 11:01
• Thanks for your great response! – chris Mar 26 '13 at 14:42

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.

I'll admit right now, I used wolfram alpha to get the first step of what I'm about to say (by making substitutions to simplify the result given when you ask it for the matrix exponential of [[a,b],[c,d]])...

If your matrix is of the form

$$M=\left(\begin{matrix}H-f&b\\\frac{g^2-f^2}{b}&H+f\end{matrix}\right)$$ Then the matrix exponential takes the form

$$e^M = \frac{e^{H-g}}{2g}\left(\begin{matrix}g+f+e^{2g}(g-f)&b(e^{2g}-1)\\\frac{g^2-f^2}{b}(e^{2g}-1)&g-f+e^{2g}(g+f)\end{matrix}\right)$$

So, we can use this to try to find non-trivial solutions to $e^{M_1}=e^{M_2}$. Using those subscripts, we have $$K_1(2g_1+L_1(g_1-f_1))=K_2(2g_2+L_2(g_2-f_2))\\ K_1b_1L_1 = K_2b_2L_2\\ K_1 \frac{g_1^2-f_1^2}{b_1}L_1 = K_2\frac{g_2^2-f_2^2}{b_2}L_2\\ K_1(2g_1+L_1(g_1+f_1)) = K_2(2g_2+L_2(g_2+f_2))$$ Where, for notational compression (and hiding of the exponentials), we have

$$K_n=\frac{e^{H_n-g_n}}{2g_n}\\ L_n=e^{2g_n}-1$$ Taking the difference between the first and fourth equations and dividing by 2, we get three equations with the same form:

$$K_1L_1f_1=K_2L_2f_2\\ K_1L_1b_1 = K_2L_2b_2\\ K_1L_1 \frac{g_1^2-f_1^2}{b_1} = K_2L_2\frac{g_2^2-f_2^2}{b_2}$$ From these, we can immediately say

$$\frac{b_1}{f_1}=\frac{b_2}{f_2}\\ \frac{g_1^2-f_1^2}{b_1^2}=\frac{g_2^2-f_2^2}{b_2^2}$$ The second of these can be rearranged, using the first of them ($b_2=b_1f_2/f_1$), to give

$$\frac{g_1^2}{f_1^2}=\frac{g_2^2}{f_2^2}$$ So

$$\frac{g_1}{f_1} = \pm \frac{g_2}{f_2}$$ We can convert these into regular a,b,c,d form by noting that $a=H-f$ and $d=H+f$, so $f=(d-a)/2$. Similarly, $bc=g^2-f^2=g^2-(d-a)^2/4$, so $g^2=bc+(d-a)^2/4$. And so, we know that

$$\frac{d_1-a_1}{b_1}=\frac{d_2-a_2}{b_2}\\ \frac{b_1}{c_1}=\frac{b_2}{c_2}$$ These two conditions immediately tell us that $M_1$ and $M_2$ must commute. Proof:

$$\left(\begin{matrix}a_1&b_1\\c_1&d_1\end{matrix}\right)\left(\begin{matrix}a_2&b_2\\c_1\!\!\frac{b_2}{b_1}&a_2+\frac{b_2}{b_1}(d_1-a_1)\end{matrix}\right) = \left(\begin{matrix}a_1a_2+c_1b_2&a_2b_1+b_2d_1\\a_2c_1+c_1d_1\frac{b_2}{b_1}&c_1b_2+d_1a_2+d_1(d_1-a_1)\frac{b_2}{b_2}\end{matrix}\right)=\left(\begin{matrix}a_2&b_2\\c_1\!\!\frac{b_2}{b_1}&a_2+\frac{b_2}{b_1}(d_1-a_1)\end{matrix}\right)\left(\begin{matrix}a_1&b_1\\c_1&d_1\end{matrix}\right)$$ Since our matrices commute, we can immediately know that $e^{M_1-M_2}=I$.

So all we have to do is find a non-zero solution to $e^M=I$. If we let $g=n\pi i$, with $n$ being an integer other than zero (for $n=0$, the following doesn't apply), then we can see that for

$$M = \left(\begin{matrix}H-f&b\\-\frac{f^2+n^2\pi^2}{b}&H+f\end{matrix}\right)$$ we can write

$$e^M = (-1)^ne^H\left(\begin{matrix}1&0\\0&1\end{matrix}\right)$$

Therefore, as long as $H=i\pi(2m-n)$ for some integer $m$, and integer $n\neq 0$, then we have a matrix that satisfies $e^M=I$. For the case of $n=2k$ (integer $k\neq 0$) and $m=0$, we have

$$M = \left(\begin{matrix}-f&b\\-\frac{4k^2\pi^2+f^2}{b}&f\end{matrix}\right)$$

As an example that is more fun to test, consider $f=b=\pi$ and $k=1$. Now we have

$$M = \left(\begin{matrix}-\pi&\pi\\-5\pi&\pi\end{matrix}\right)$$ Which satisfies $e^M=I$.