Is $SU(3)$ generated by the exponentials of the Gell-Mann matrices? It is well known that the lie algebra of $SU(3)$ is spanned by the (half) Gell-Mann matrices $T_i=\lambda_i/2$, and that these generate $SU(3)$ via the map $\theta_iT_i \rightarrow e^{\theta_iT_i}$. 
However, the above map involves taking a linear combination of the generators and then taking the exponential. Do we generate $SU(3)$ if we take the exponential of this basis first? In other words, is it true that if we take the set $\{{e^{T_i}}\}, i=1,...8$, that these matrices generate $SU(3)$?   
 A: For SU$(3)$, the following set $F_m$ and its algebraic relations differ from that of the eight Gell Mann generators $g_m$. 
The $F_m$ have no simple relation to the $g_m$. For example, in the usual representation, $7$ of the $8$ generators $g_m$ have a diagonal element that is zero;  
Is the set of the eight finite SU(3) elements $F_m=e^{\pi \cdot i \cdot g_m / 2}$ $-$ defined by the eight SU(3) generators $g_m$.
Here are the Gell-Man matrices $g_n$: 
$g_1=\left ( \begin{array}{ccc} 0 & 1 & 0\\ 1 & 0 & 0 \\ 0 & 0 & 0 \end{array} \right )$, $g_2=\left ( \begin{array}{ccc} 0 & -i & 0\\ i & 0 & 0 \\ 0 & 0 & 0 \end{array} \right )$, $g_3=\left ( \begin{array}{ccc} -1 & 0 & 0\\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{array} \right )$, $g_4=\left ( \begin{array}{ccc} 0 & 0 & 1\\ 0 & 0 & 0 \\ 1 & 0 & 0 \end{array} \right )$, $g_5=\left ( \begin{array}{ccc} 0 & 0 & -i\\ 0 & 0 & 0 \\ i & 0 & 0 \end{array} \right )$, $g_6=\left ( \begin{array}{ccc} 0 & 0 & 0\\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{array} \right )$, $g_7=\left ( \begin{array}{ccc} 0 & 0 & 0\\ 0 & 0 & -i \\ 0 & i & 0 \end{array} \right )$, $g_8=\left ( \begin{array}{ccc} 1/\sqrt{3} & 0 & 0\\ 0 & 1/\sqrt{3} & 0 \\ 0 & 0 & -2/\sqrt{3} \end{array} \right )$, 
And these are the first matrices $F_n$: 
$F_1=\left ( \begin{array}{ccc} 0 & i & 0\\ i & 0 & 0 \\ 0 & 0 & 1\end{array} \right )$, $F_2=\left ( \begin{array}{ccc} 0 & 1 & 0\\ -1 & 0 & 0 \\ 0 & 0 & 1 \end{array} \right )$, $F_3=\left ( \begin{array}{ccc} i & 0 & 0\\ 0 & -i & 0 \\ 0 & 0 & 1 \end{array} \right )$, $F_4=\left ( \begin{array}{ccc} 0 & 0 & i\\ 0 & 1 & 0 \\ i & 0 & 0 \end{array} \right )$, $F_5=\left ( \begin{array}{ccc} 0 & 0 & 1\\ 0 & 1 & 0 \\ -1 & 0 & 0 \end{array} \right )$, $F_6=\left ( \begin{array}{ccc} 1 & 0 & 0\\ 0 & 0 & i \\ 0 & i & 0 \end{array} \right )$, $F_7=\left ( \begin{array}{ccc} 1 & 0 & 0\\ 0 & 0 & 1 \\ 0 & -1 & 0 \end{array} \right )$, Now $ F_8$ is less simple.
