Lie Group and a basis of tangent space Let $G \subset M(3\times 3, \mathbb{R})$ be the space of all matrices of the form 
$\left( \begin{array} 
&1 & a & b \\ 0 & 1 & c \\ 0 & 0 & 1\end{array}\right) $
where $a,b,c \in \mathbb{R}$.
Let  $I$ be the identity matrix. In particular, we can define:
$ t \mapsto \gamma (t)= I + tB $.
Note that $I \in G$, $\gamma(0)=I$ and $\gamma'(0)=B$.
What property has to satisfy $B$ to be in $T_{I}G$?
I guess that the basis can be the matrix with ones in the entries in the position of $a,b,c$ and the rest zero.
 A: Ok, here you go: you take a curve $\gamma: \mathbb{R} \to G$ and take the derivative $\gamma'(t)$ at $t$. This will give you a vector in $T_{\gamma(t)}M$ where $\gamma(t)$ is just a point on your manifold.
So you want a element in the tangent space of the $I$ element. What to you need? Well a curve with say $\gamma(0)=I$ then $\gamma'(0)$ will be a vector in $T_{I}M$. How about these ones:
\begin{eqnarray}
\gamma_1: \mathbb{R} &\to& G \\
t &\mapsto & \begin{bmatrix} 1 & t & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}\\
\gamma_2: \mathbb{R} &\to& G \\
t &\mapsto & \begin{bmatrix} 1 & 0 & t \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}\\
\gamma_3: \mathbb{R} &\to& G \\
t &\mapsto & \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & t \\ 0 & 0 & 1 \end{bmatrix}
\end{eqnarray} 
Now $\gamma_i(0)=I$ and 
$$\gamma_1'(0)=\begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}
 $$
and the others similarly. This will give yo a basis of your tangent space.
About your question regarding $B$. Well you need that $I+tB \in G$ for every $t$. So $B$ has to be a upper triangular matrix with zeros on the diagonal. But any such $B$ is fine and will give you a element in the tangent space as you see from the basis above. 
