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If $k$ is a field of characteristic $p$, we can define a map $\exp:\mathfrak{gl}_n(k)\to GL_n(k)$ by:

$$\exp(A)=\sum_{i=0}^{p-1}\frac{A^i}{i!}$$

In the answer to this question, we see that if $A^p=B^p=0$, and if $\exp(A)=\exp(B)$, then $A=B$. So if $p>n$, $\exp$ is injective when restricted to nilpotents. I'd like to know whether or not $\exp$ is injective on all of $\mathfrak{gl}_n(k)$.

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Jared
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1 Answers1

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I am not sure if there is any nice characterisation of injectiveness, but the truncated matrix exponential is not injective in all cases. Let $p=3>n=2$ and $A=\pmatrix{1&\ast\\ 0&0}$. Then $A^m=A$ for every $m\ge1$ and $$\exp(A) = I + A + \frac12 A^2 = I + A + 2A = I = \exp(0).$$

user1551
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