Let $v_1, \dots , v_n$ be vectors which span a vector space $V,$ and let $u$ be a vector in $V.$ Prove
that $\operatorname{Span}\{v_1, \dots, v_n, u\} = V .$
So far I have:
Let v1, …, vn be vectors which span a vector space V and let u be a vector in V. I wish to prove that span{u, v1, …, vn} = V. We know that span{v1,…,vn}=V and thus for some arbitrary vector z in V, then k1v1+…+knvn=z, where k1,…,kn are scalars. Let span{v1,…,vn,u} be the set of all linear combinations of the form k1v1+…+knvn+cu, where k1,…,kn,c are scalars. Therefore k1v1+…+knvn+cu=z+cu.
I'm stuck here. I know I need to show that z=k1v1+…+knvn+cu, but I don't know how to get there.
Best Answer
You are thinking too complicated. The span of a set is either defined as the set of all finite linear combinations from the set or equivalently as the smallest subspace of $V$ containing the set.
Anyway, it is trivial to see that the span has some monotonous property: If $A\subseteq B$ any finite linear combination of elements of $A$ is also one of elements of $B$ (or any subspace that contains $B$ also contains $A$). Thus $\mathrm{span} A\subseteq\mathrm{span} B$.
Thus you get $$ V = \mathrm{span}\{v_1,\ldots,v_n\} \subseteq \mathrm{span}\{v_1,\ldots,v_n,u\}\subseteq V$$ This implies that actually $$ V = \mathrm{span}\{v_1,\ldots,v_n\} = \mathrm{span}\{v_1,\ldots,v_n,u\}= V$$