Let $A$ and $B$ be $n\times n$ hermitean matrices. When do we have $e^{iA}=e^{iB}$? Can we somehow classify those pairs of matrices that have the same exponential?

Here are some observations that I made:

  • If $A$ and $B$ commute, then the condition is satisfied if and only if the spectrum of $A-B$ is contained in $2\pi\mathbb Z$. (Both directions can fail if $A$ and $B$ don't commute, as the following two points show.)
  • The condition $e^{iA}=e^{iB}$ can also be satisfied if $A$ and $B$ do not commute. Take, for example, $A=2\pi\begin{pmatrix}1&0\\0&-1\end{pmatrix}$ and $B=2\pi\begin{pmatrix}0&1\\1&0\end{pmatrix}$. They do not commute but $e^{iA}=e^{iB}=I$. The spectrum of their difference is not in $2\pi\mathbb Z$.
  • If $A=\sqrt2\pi\begin{pmatrix}1&0\\0&-1\end{pmatrix}$ and $B=\sqrt2\pi\begin{pmatrix}0&1\\1&0\end{pmatrix}$, then the eigenvalues of $A-B$ are $\pm2\pi$ but $e^{iA}\neq e^{iB}$.
  • The exponential map is not a homomorphism so finding the kernel is not enough; cf. the Baker–Campbell–Hausdorff formula.
  • Having the same exponential is an equivalence relation.

Here is a rather cheap answer to this question. Let $E_\lambda$ denote the eigenspace of $\lambda \in \mathrm{spec}(A)$. Let $F_\lambda$ denote the eigenspace of $\lambda \in \mathrm{spec}(B)$. Then $\exp(iA) =\exp(iB)$ if and only if, for every $\lambda_0 \in \mathbb{R}$, we have $$\bigoplus_{\lambda \in \mathrm{spec}(A) \cap (\lambda_0 + 2 \pi \mathbb{Z})} E_\lambda = \bigoplus_{\lambda \in \mathrm{spec}(B) \cap (\lambda_0 + 2 \pi \mathbb{Z})} F_\lambda.$$

Roughly speaking, the condition is that $A$ and $B$ should have the same eigenspaces, after identifying the eigenvalues whose exponentials are equal.

  • $\begingroup$ Excellent, thank you! A similar solution occurred to me last night, but I think I won't write it up since we already have this one. $\endgroup$ – Joonas Ilmavirta May 1 '15 at 10:27
  • $\begingroup$ No problem! Of course, this isn't a particularly deep characterization. I guess a similar thing works if $A$ and $B$ are just normal and $f : \mathbb{C} \to \mathbb{C}$ is any continuous function (instead of $f(t) = e^{it}$). $\endgroup$ – Mike F May 1 '15 at 16:27

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