Given the set of unipotent matrices:
$S = \left\{ A\in GL_{2}(\mathbb{R}) \;:\; A=\left( \begin{matrix} 1 & a \\ 0 & 1 \end{matrix} \right) , \; a\in\mathbb{R} \right\}$
I can show this is a subgroup of the general linear group $GL_{2}(\mathbb{R})$ under matrix multiplication. I can also define a curve:
$\begin{array}{rl} \gamma: \mathbb{R} &\longrightarrow& S \\ t &\longmapsto& g = \gamma(t) = I_{2} + tA \end{array}$
with $A = \left( \begin{matrix} 1 & a \\ 0 & 1 \end{matrix} \right) $ which gives:
$\gamma(t) = I_{2}+tA = \left( \begin{matrix} 1+t & at \\ 0 & 1+t \end{matrix} \right)$
with $\gamma(0)=I_{2}$ and $\dot{\gamma}(0)=A$. I can also show this defines a group homomorphism so that the image of $\gamma(t)$ defines a one-parameter subgroup of $GL_{2}(\mathbb{R})$. From this, I can compute the tangent space at the identity $T_{1}S$:
$T_{1}S = \left\{ \dot{\gamma}(0) \;:\; \gamma(0)=I_{2}\right\} = \left\{ \left( \begin{matrix} 1 & a \\ 0 & 1 \end{matrix} \right) , \; a\in\mathbb{R} \right\} = S$.
Is this correct? Is the tangent space at the identity $T_{1}S=S$ ? Further to this, I am completely clueless how to show it is a Lie subalgebra of $gl_{2}(\mathbb{R})$ (I know in principle how to show it but can;t seem to do it). Indeed, I am struggling to show it is even a subspace of $2\times 2$ matrices!
Any help would be much appreciated.
You've made a mistake in the definition of your curve. You should take $$ \gamma(t) = \pmatrix{1&at\\0&1} $$ You should find then that $$ T_1(S) = \left\{\pmatrix{0&a\\0&0} : a \in \Bbb R\right\} $$