Consider the semi-direct product $GL_2(2,\mathbb{R})\times \mathbb{R}^2$ acting on $\mathbb{R}^2\ni z$ by: $$ (A,a)\cdot z = Az+a $$ and where the Lie bracket rule is $[(A,a),(B,b)]=([A,B],Ab-Ba)$.
I have several questions:
- What's the point to precise that we have a semidirect product ?
- How to show that the Lie albegra generated by $(A,a),(B,0)$ where $A= \begin{pmatrix} u & 0 \\ 0 & v \end{pmatrix}$ $a=\begin{pmatrix} 0 \\ 1 \end{pmatrix}$ and $B= \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$ is $gl(2,\mathbb{R})\oplus \mathbb{R}^2$ if $u\neq v$ ?
I am not familiar with your course, but perhaps you are meant to recognize the affine group on $\mathbb{R}^2$. As you see in WP, you may represent it by 3x3 matrices involving 2x2 matrices and 2-vectors in its entries, and you may directly check the semidirect structure: the matrix entries know nothing of the vectors, but the vectors reflect the matrices in the group composition law, and, ipso facto, in the Lie algebra. (By the way, it is unfortunate you use the same symbols, A,a in the group action expression and the Lie algebraic considerations. I would use $(M,v)\cdot z = Mz+v$ in your first equation and keep the A,B,a,b for the subsequent Lie-algebraic generators' part.)
You may then directly check from the group composition law that the Lie-algebra $[(A,a),(B,b)]=([A,B],Ab-Ba)$ follows.
Apply this bracket rule then to your given two generators, (A,a) $$A= \begin{pmatrix} u & 0 \\ 0 & v \end{pmatrix} , \qquad a=\begin{pmatrix} 0 \\ 1 \end{pmatrix}$$
and (B,0), $$B= \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}.$$
The resulting two-vector is $$\begin{pmatrix} 1 \\ 0 \end{pmatrix}$$ spanning $\mathbb{R}^2$ together with a, and the commutator of A with B is proportional to $\sigma_1$, if you were familiar with the Pauli matrices, $$ (v-u) \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}, $$ linearly independent from A and B, so its spans $gl(2,\mathbb{R})$ provided $u\neq v$. I fear this looks like a homework problem, so your assignment is to derive the Lie bracket rule out of the group composition law, also to be derived!