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I am familiar with the matrix representation of the (lie) group of reals \begin{bmatrix}1&a\\0&1\end{bmatrix} with $a \in \mathbb{R}$. This is a really dumb thing I'm stuck on and I can't figure it out but what is the Lie algebra of this group? If I take the derivative and evaluate it at the identity I get \begin{bmatrix}0&1\\0&0\end{bmatrix} but when I exponentiate this to get back to the Lie group, it's clear that this is not the answer. I'm aware the the Lie Algebra representation is of the form \begin{bmatrix}0&a\\0&0\end{bmatrix}, but I can't for the life of me understand how you get that by differentiating near the identity? There's clearly something I'm missing.

Thanks for all your help!

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Why is that not the answer? $$ \exp \pmatrix{0 & a\cr 0 & 0\cr} = \pmatrix{1 & a\cr 0 & 1\cr}$$ as you should be able to check using whichever definition of the exponential you prefer.

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  • $\begingroup$ Right. I know that's the answer but I guess what I'm more confused about is how you could derive the Lie Algebra from the Lie group representation. Since the Lie Algebra is just the tangent space at the identity I figure all we have to do is just differentiate the representation, but that doesn't seem to be the case. Does that make sense? $\endgroup$ – hijasonno Aug 16 '16 at 0:15
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    $\begingroup$ It is the case. The Lie algebra is spanned by the matrix you got by differentiation, $\pmatrix{0 & 1\cr 0 & 0\cr}$. $\endgroup$ – Robert Israel Aug 16 '16 at 0:20
  • $\begingroup$ Oh jeez. Like I said, I knew it was something dumb I was forgetting. Thanks a ton! $\endgroup$ – hijasonno Aug 16 '16 at 0:21

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