When $e^A = e^B$ for matrices $A,B$?

Question

Let $A$ and $B$ be $n\times n$ complex matrices such that $e^A = e^B$. I would like to know relations between $A$ and $B$.

When $A,B\in\mathbb{C}$, we have a simple relation $e^A = e^B \Leftrightarrow A-B\in 2\pi i \mathbb{Z}$. Is there such simple relation for general matrices $A$ and $B$?

See https://math.stackexchange.com/q/660823 and in particular the answer by user loup blanc. Briefly speaking, the exponential map is "mostly" injective. — user1551, Dec 19 '18 at 04:00
See also here. My answer is basically that for any real number $x$ the exponential of $$\pmatrix{0&-x\cr4\pi^2/x&0\cr}$$ is equal to $I_2$ because the eigenvalues, $\pm2\pi i$ become $1$ when the exponential function is applied. — Jyrki Lahtonen, Dec 22 '18 at 10:28

score 3 · Answer 1 · answered Dec 19 '18 at 04:36

For any nonsingular $n \times n$ matrix $C$, there are matrices $A$ such that $\exp(A)=C$.

Suppose $C$ is in Jordan form. For each Jordan block of size $m$ with eigenvalue $\lambda$, $$\pmatrix{ \lambda & 1 & 0 & \ldots & 0 & 0\cr 0 & \lambda & 1 & \ldots & 0& 0\cr \ldots & \ldots & \ldots & \ldots & \ldots & \ldots\cr 0 & 0 & 0 & \ldots &\lambda & 1\cr 0 & 0 & 0 & \ldots & 0 &\lambda \cr}$$ $A$ can have a corresponding block of size $m$ where the diagonal elements are some branch of $\log(\lambda)$, and the elements in the $k$'th super-diagonal are $\dfrac{d^n}{d\lambda^n} \log(\lambda)/n!$. This is not a Jordan block, but its Jordan form will be a single Jordan block of size $m$.

If the blocks of $C$ are all different, the decomposition is unique, and the only choice in $A$ is which branches of logarithm to use. However, if there are identical blocks, there will be a continuum of choices. Thus since $$ C = \pmatrix{1 & 1 & 0 & 0\cr 0 & 1 & 0 & 0\cr 0 & 0 & 1 & 1\cr 0 & 0 & 0 & 1\cr}$$ commutes with $$ J = \pmatrix{\cos(t) & 0 & \sin(t) & 0\cr 0 & \cos(t) & 0 & \sin(t)\cr -\sin(t) & 0 & \cos(t) & 0\cr 0 & -\sin(t) & 0 & \cos(t)}$$ logarithms of $C$ will include $$ \eqalign{J &\pmatrix{ w_1 & 1 & 0 & 0\cr 0 & w_1 & 0 & 0\cr 0 & 0 & w_2 & 1\cr 0 & 0 & 0 & w_2\cr} J^{-1} \cr = &\pmatrix{ (w_1-w_2)\cos(t)^2+w_2 & 1 & -\cos(t)\sin(t)(w_1-w_2) & 0\cr 0 & (w_1-w_2)\cos(t)^2+w_2 & 0 & -\cos(t)\sin(t)(w_1-w_2)\cr -\cos(t)\sin(t)(w_1-w_2) & 0 & (-w_1+w_2)\cos(t)^2+w_1 & 1\cr 0 & -\cos(t)\sin(t)(w_1-w_2) & 0 & (-w_1+w_2)\cos(t)^2+w_1\cr}} $$ where $w_1$ and $w_2$ are any logarithms of $1$ (e.g. $0$ and $2\pi i$).

score 2 · Answer 2 · answered Jan 18 '19 at 22:36

$\textbf{Definition.}$ We say that the spectrum of $A\in M_n(\mathbb{C})$ is $2i\pi$ congruence free (denoted by $2i\pi$ CF) iff for every $u,v\in spectrum(A)$, $u-v\notin 2i\pi\mathbb{Z}^*$.

$\textbf{Theorem}.$ (for i), Hille https://link.springer.com/article/10.1007%2FBF01350286 ). Let $A,B\in M_n(\mathbb{C})$. If $spectrum(A)$ is $2i\pi$ CF and $e^A=e^B$, then

i) $AB=BA$ and, therefore,

ii) $A-B$ is similar to $diag(2i\pi\alpha_1,\cdots,2i\pi\alpha_n)$ where $\alpha_i\in\mathbb{Z}$.

$\textbf{Proof of ii)}$. From i), $e^{A-B}=I_n$ is diagonalizable and, therefore, $A-B$ is diagonalizable. Moreover $spectrum(A-B)\subset 2i\pi\mathbb{Z}$.

$\textbf{Remark}.$ i) Note that the matrices of Jyrki Lahtonen are not $2i\pi$ CF and don't pairwise commute.

ii) $A$ is $2i\pi$ CF iff $A$ is a polynomial in $e^A$ iff the function $X\mapsto e^X$ is one to one in a neighborhood of $A$.

When $e^A = e^B$ for matrices $A,B$?

2 Answers2