Does $\exp(At)=\exp(Bt)$ for infinitely many $t$ imply $A=B$ where $A,B$ are square matrices?

Question

Suppose $A$ and $B$ are two square matrices so that $e^{At}=e^{Bt}$ for infinite (countable or uncountable) values of $t$ where $t$ is positive.

Do you think that $A$ has to be equal to $B$?

Thanks, Trung Dung.

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Maybe I do not state clearly or correctly.

I mean that the equality holds for all $t\in (0, T)$ where $T>0$ or $T=+\infty$, i.e. for uncountable $t$. In this case I think some of the counter-examples above do not work because it is correct for countable $t$.

Answer 1

Take $A=\pmatrix{0&1\\-1&0}$. Then $\exp(tA)=I$ for $t=2n\pi$ ($n$ integer). That is $\exp(tA)=\exp(tB)$ infinitely often for $B$ the zero matrix.

Answer 2

No. Let $A$ be the null matrix and let $B=2\pi i\operatorname{Id}$. Then$$(\forall t\in\mathbb{Z}):e^{tA}=e^{tB}.$$

Answer 3

$\textbf{Proposition 1.}$ Let $(t_k)$ be a positive sequence that converges to $l$ s.t., for every $k$, $t_k\not= l$.

If, (*) for every $k$, $e^{t_kA}=e^{t_kB}$, then $A=B$.

$\textbf{Proof.}$ There is $k$ s.t. $t_kA$ is $2i\pi$ congruence free (for every $\lambda,\mu\in spectrum(A)$, $t_k(\lambda-\mu)\notin 2i\pi\mathbb{Z}^*$). From a result by Hille, $t_kA,t_kB$ commute, that implies $AB=BA$.

In the sequel, we may assume that $A=\lambda I+N$ where $N$ is nilpotent and (*). Let $\mu\in spectrum(B)$; then, for every $k$, $e^{t_k\mu}=e^{t_k\lambda}$, that is $t_k(\lambda-\mu)\in 2i\pi\mathbb{Z}$; thus $\lambda=\mu$ and $B=\lambda I+M$ where $M$ is nilpotent and $e^{t_kM}=e^{t_kN}$. Note that the exponential map is injective on the nilpotent matrices; then $t_kM=t_kN$ and $A=B$. $\square$

EDIT. In the same way as above, we can prove that follows.

$\textbf{Proposition 2.}$ Let $(t_k)$ be a generic real sequence; for instance, the $(t_k)$ are iid and each follows the normal law $N(0,1)$.

If, (*) for every $k$, $e^{t_kA}=e^{t_kB}$, then $A=B$ with probability $1$.

In other words, that the OP says is true except if we make an ad-hoc sequence so that it does not work.

Answer 4

Note that $t\mapsto e^{tA}-e^{tB}$ is a matrix of power series $c_{ij}(t)=\sum_{l=0}^\infty c_{ij}^l t^{(l)}$ with $\infty$ as its radius of convergence. By Identity theorem for power series these series are identical zero if the set $N=\{t\,\vert \,c_{ij}(t)=0\}$ has a limit point.

Thus all $c_{ij}^{(l)}=0$ provided for instance that $c_{ij}(\frac1n)=0$ for all $n\in\mathbb{N}$. But then $A-B=0$ since $A-B=(c_{ij}^{(1)})_{1\leq i,j\leq n}$.