In Heisenberg's group, why if $e^{tX}$ is upper triangular, then $X$ is strictly upper triangular? To find the Lie algebra of the Heisenberg group $H$, which we know to consist of upper triangular matrices, we see that exponentials of all strictly upper triangular matrices are in $H$. I do not get the following could you explain?
"On the other hand if $X$ is any matrix such that $e^{tX}$ is upper triangular, then all the entries of $X=\frac{d}{dt}|_{t=0}e^{tX}$ which are on or below diagonal must be zero so that $X$ is upper triangular"
Thus the Lie algebra of the Heisenberg group is the space of all $3\times 3$ real
matrices which are strictly upper triangular.
 A: The statement you cite is not true, but can be made so by adding "for all $t$" and "with entries $1$ on the diagonal" to the hypothesis. The examples of $\begin{pmatrix}0&1&0\\-1&0&0\\0&0&0\end{pmatrix}$ with $t=2\pi$, respectively of any diagonal matrix, show the necessity of these additions.
A: Suppose you know that $e^{tX}$ is in the Heisenberg group for all $t$. Then this means that
$$e^{tX} = \left(\begin{array}{ccc} 1 & a(t) & b(t) \\ 0 & 1 & c(t) \\ 0 & 0 & 1 \end{array}\right)$$
where $a(t),b(t),c(t)$ are smooth functions. Now if you take the derivative of $e^{tX}$ at $t = 0$, you are taking the derivative of the individual components of the matrix on the right. This means that
$$\frac{d}{dt}e^{tX}\bigg|_{t = 0}  = X =\left(\begin{array}{ccc} \frac{d}{dt}1 & a'(0) & b'(0) \\ 0 & \frac{d}{dt}1 & c'(0) \\ 0 & 0 & \frac{d}{dt}1 \end{array}\right) = \left(\begin{array}{ccc} 0 & a'(0) & b'(0) \\ 0 & 0 & c'(0) \\ 0 & 0 & 0\end{array}\right).$$
$a'(0), b'(0)$ and $c'(0)$ are just real numbers and the calculation above shows that $X$ is strictly upper triangular.
