Sum of vector subspaces generally I'm having trouble understanding the sum of two vector subspaces.
Let two subspaces: $G_{1}, T_{1} \subset\mathbb R^n$ be subspaces with bases
$ \left\{ \vec{g_{1}},...,\vec{g_{k}} \right\}$ and $\left\{ \vec{t_{1}},...,\vec{t_{m}} \right\}$, respectively.
The sum of vectors subspaces $G_{1} + T_{1}$ is defined as the vector subspace
$ \left\{ \vec{x} + \vec{y} | \vec{x} \in G_{1}, \vec{y} \in T_{1}\right\}$ of $\mathbb R^n$.
What will be the final result of the sum $G_{1} + T_{1}$?
Is it $ \left\{ \vec{g_{1}} + \vec{t_{1}},..., \vec{g_{k}} + \vec{t_{m}} \right\}$?
 A: The sum $G+T$ will be generated by $\{g_1,\dots,g_k,t_1,\dots t_m\}$, since $g_i=g_i+0,t_i=0+t_i$, and any sum $x+y$ with $x\in G$ and $y\in T$ can be written as a linear combination of these vectors.
Generally $\{g_1,\dots,g_k,t_1,\dots t_m\}$ is not a basis of $G+T$. Consider by example $\{(x,y,0)\}+\{(z,0,0)\}=\{(x,y,0)\}$ (i.e. if $V\subset W$, then $V+ W=W$).
$\{g_1,\dots,g_k,t_1,\dots t_m\}$ is a basis of $G+T$ if and only if $G\cap T=\{0\}$, in which case the sum is called a direct sum, and the notation is $\oplus$.
So if you're looking for a generating set, $\{g_1,\dots,g_k,t_1,\dots t_m\}$ is one. If you're looking for a basis, there is no fixed formula in general. The space generated by $\{g_i+t_j\}$ is a subspace of $G+T$, and in general is not $G+T$, as can be seen by $G=T=\mathbb{R}, g=1$ and $t=-1$.

Edit: 

Proposition: $\{(x,y,0)\}+\{(z,0,0)\}=\{(x,y,0)\}$

Proof: Since $(x,y,0)+(0,0,0)=(x,y,0)$, we have the inclusion RHS$\subset$LHS.
To prove the other inclusion, let $x\in$LHS. Then $x=u+v$ with $u\in\{(x,y,0)\}$ and $v\in\{(z,0,0)\}$. Write $u=(x_u,y_u,0)$ and $v=(z_v,0,0)$, then $x=u+v=(x_u+z_v,y_u,0)$, then $x\in$RHS.
