Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Is the matrix exponential $exp:M_n(\mathbb C) \to GL_n(\mathbb C)$ injective? Can it be that $e^A=I$ where $A$ is not the zero matrix?

share|cite|improve this question
Consider $n=1$. Is $e^z$ injective on $\mathbb C$? – Jonas Meyer Jan 9 '12 at 7:50
Slightly less trivial is that it is not injective on $M_n({\mathbb R})$ for $n \ge 2$: consider $\exp\left( \pmatrix{0 & t\cr -t & 0\cr}\right)$ – Robert Israel Jan 9 '12 at 7:57
up vote 9 down vote accepted

Well, let me answer this in the obvious way. If $n=1$ then this is asking whether $\exp:\mathbb{C}\to\mathbb{C}^\times$ is injective. Once you see how to prove this isn't the case, you can generalized by using the fact that


share|cite|improve this answer

All the nice Lie Group elements look like $g=\exp(iA)$, for some "angle" $A$. Lets here think of finite dimensional and compact ones, which are connected to the unit. The $i$ is somewhat redundant, this is physicists notation to make $A$ have real eigenvalues if $g$ is unitary.

A good example would be the rotation group $SO(\mathbb{R},n)$ in any dimension. In three dimensions it's obvious that any rotation around $360°$ is represented by the unit element $I$.

In one dimension, you have $SO(2)$ or equivalently $U(1)$ with elements $e^{i\varphi}$ acting on the circle $S$ and for $\varphi=2\pi$ you get $e^{i\varphi}=1$. Alex Youcis's example acts on the cross product of circles $S^n$, with for example $S^2$ topologically being a torus.

If the group is compact as a manifold, then I think it's pretty reasonable that if you follow specific trajectories, you might come back to the unit.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.