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I have been studying Rossmann's Lie Groups. In the context of this book, a linear group $G$ is a group of invertible real or complex matrices, and its Lie algebra $\mathfrak{g}$ consists of those matrices $X$ for which the exponential $e^{tX}$ belongs to $G$ for every $t\in\mathbb{R}$. As far as I understand, the matrices $X$ which are candidates to be an element of $\mathfrak{g}$ are taken a priori to be real if $G$ consists of real matrices. However, we can of course regard any such $G$ as a group of complex matrices as well; in any case, one should get the same $\mathfrak{g}$. This is what I want to prove.

It is clearly enough to show this for $G=\mathrm{GL}(n,\mathbb{R})$. Therefore, I want to show that any complex matrix $X$ for which $e^{tX}$ is real for every $t\in\mathbb{R}$ is itself real. This is easy to see if $X$ is diagonal, but what about general $X$? Any reference or sketch of proof is welcome.

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The derivative of $e^{tX}$ with respect to $t$ at $t = 0$ is $X$. –  Qiaochu Yuan Mar 10 '13 at 12:06
    
Jordan normal form? –  Berci Mar 10 '13 at 12:06
    
By the way, just so we're clear, there is a definition that doesn't have this potential ambiguity. –  Qiaochu Yuan Mar 10 '13 at 12:06
    
Lie Group,Lie Algebra and Representation Theory by Brian C Hall. –  Bunuelian Trick Mar 10 '13 at 13:05
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up vote 2 down vote accepted

As in Qiaochu Y's comment: if $f(t)=e^{tX}$ is real for every real $t$, then its derivative at $t=0$ is real, being a limit of a sequence or net of real numbers, and this derivative is (easily seen to be) $X$. There is no need to worry about "normal forms" or reduction to simpler cases.

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Thank you very much Professor. This was exactly the kind of answer I was looking for. –  Murat Güngör Mar 10 '13 at 15:47
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