Theorem 1.1 – If $A$ is a nonempty subset of a vector space $V$, then the set $L(A)$ of all linear combination of the vectors in $A$ is a subspace, and it is the smallest subspace of $V$ which includes the set $A$.
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Is my answer rigorous enough? I am wondering if I am relying on the given theorem 1.1 too much.
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Do I need to show that $A = \{ \langle 1,1 \rangle, \langle 1,2 \rangle\}$ is a subset of $V$ or is that already obvious enough?
My answer:
Suppose $A$ is a nonempty subset of vector space $V$ in $\mathbb{R}^2$ and contains the vectors $\langle 1,1 \rangle$ and $\langle 1,2 \rangle$. Then by Theorem 1.1, $L(A)$, the set of all linear combinations of the vectors in $A$ is a subspace in $V$ and is the smallest subspace that contains the vectors in $A$.
Thus we can call $L(A)$ the linear span, or subspace generated by the two vectors in $A$.
Best Answer
Actually, your answer proves nothing. You are supposed to prove that every vector $(a,b)\in\mathbb R^2$ is a linear combination of $(1,1)$ and $(1,2)$. You can do that by proving that there are numbers $x,y\in\mathbb R$ such that $(a,b)=x(1,1)+y(1,2)=(x+y,x+2y)$. In other words, solve the system$$\left\{\begin{array}{l}x+y=a\\x+2y=b.\end{array}\right.$$You will get $x=2a-b$ and $y=b-a$. Since every vector of $\mathbb R^2$ is a linear combination of $(1,1)$ and $(1,2)$ then, indeed, $\operatorname{span}\bigl\{(1,1),(1,2)\bigr\}=\mathbb R^2$.