Solve matrix equation $e^A=e^B$ for nilpotent $A, B$.

Question

I need to solve equaton $e^A=e^B$ for nilpotent matrices A and B over field $\mathbb C$, where $B$ is fixed.

I solved equation $e^X=E$ for all matrices. The solution is any semisimple (in case $\mathbb C$ diagonalizable) matrix with eigenvalues $2\pi ik, k\in \mathbb Z$

So I solved my equation in commutative case $AB=BA$. $e^A=e^B \Leftrightarrow e^{A - B}=E \Leftrightarrow A - B$ is semisimple nilpotent matrix $\Leftrightarrow A-B=0 \Leftrightarrow A=B$

Sorry for poor results, but I have no ideas in non-commutative case.

Answer 1

Theorem: Let: $X\in M_n(\mathbb C)$ and: $||X||\lt \ln(2)$ which: $$||X||=(\sum_{i,j=1}^n |x_{ij}|^2)^{\frac12}$$ Then: $$Log(\mathbb e^X)=X$$

For the proof, See: Lie Groups, Lie Algebras, and Representations (page $50$) theorem: $2.7$.
Now, consider that for every $X\in M_n(\mathbb C)$, we can find positive real $t$, such that: $||tX||\lt \ln(2)$.
Lemma: if $X\in M_n(\mathbb C)$ and $X$ be nilpotent, then: $$Log(\mathbb e^X)=X$$ Proof: Since $X$ is nilpotent, so for every $z\in \mathbb C$, $zX$ and $\mathbb e^{zX}-I_n$ are nilpotent to. Chose $0\lt t\in \mathbb R$ such that, $||tX||\lt \ln(2)$. Now, by above theorem: $$Log(\mathbb e^{tX})=tX$$ Because of $\mathbb e^{tX}-I_n$ is nilpotent, $(\mathbb e^{tX}-I_n)^m=0$, for every $m\in \mathbb N$ which: $m\ge n$. So, $Log(\mathbb e^{tX})$ is a polynomial of $tX$ almost from degree $n$. But there are uncountably many $0\lt t\in \mathbb R$ which: $$Log(\mathbb e^{tX})=tX$$ But we know that both sides are polynomial functions of $tX$, so equality for all sufficiently small $t$ implies equality for all $t$. Thus for $t=1$ we get: $$Log(\mathbb e^X)=X$$ Now, consider that $A,B$ are nilpotent. So, $$Log(\mathbb e^A)=A$$
And: $$Log(\mathbb e^B)=B$$ So, from: $$\mathbb e^A=\mathbb e^B$$ we have: $$Log(\mathbb e^A)=Log(\mathbb e^B)$$ And by using the fact that $A,B$ are nilpotent, by above lemma, $A=B$.

Answer 2

Few remarks. 1. If $AB=BA$ and $e^A=e^B$, then $A-B$ is similar to $2i\pi diag((k_i)_i)$ where the $k_i$ are integers. If, moreover, $A,B$ are nilpotent, then $A-B$ is nilpotent and the $k_i$ are zero. 2. $AB\not= BA$ does not imply $e^{A-B}\not= e^Ae^{-B}$. 3. $e^A=e^B$ does not imply $e^{tA}=e^{tB}$ -as Save wrote-. (Correction) Yet the hamid's proof is correct; indeed $\log(e^{tX})$ is a polynomial of degree $<n$ in $e^{tX}-I$, then a polynomial of degree $<n$ in $tX$ and his reasoning works.

In fact, it suffices to use Taylor series. $e^A=I+A+A^2/2!+\cdots+A^{n-1}/(n-1)!=I+N$ where $N$ is nilpotent and $\log (I+N)=\sum_{i=1}^{n-1}(-1)^{i-1}N^i/i$. According to composition of Taylor series $\log(e^A)=A+\alpha_n A^n+\cdots=A$ and we are done.

EDIT. Answer to SaveMyLife . Let $A=2i\pi\ diag(1,2,0),B=2i\pi\;\begin{pmatrix}2&1&1\\1&3&-2\\1&1&0\end{pmatrix}$. Then $e^A=e^B=I,e^Ae^B=e^Be^A=e^{A+B}=I$ and the last, but not the least, for any positive integer $t$, $e^{tA}e^B=e^Be^{tA}=e^{tA+B}=I$. Yet $A,B$ are not simultaneously triangularizable (hence they don't commute !).