Let $V$ be a vector space and $S=${$v_1,…,v_p$} be a set in in $V$. Prove $SPAN(S)$is the smallest subspace of $V$ that contains the vectors $v_1,…,v_p$.
Proof: let $B=${$v_1,…,v_n$} with $n≤p$ be a subset of $S$ and also a basis for $SPAN(S)$, then $SPAN(S)=SPAN(B)$. Now, assume that there is a smaller basis {$v_1,…,v_{n-1}$} s.t. $SPAN(v_1,…,v_{n-1})=SPAN(B)$. Because $B$ is linearaly independent $v_n$ is not a linear combination of {$v_1,…,v_{n-1}$}, so $v_n∉SPAN(v_1,…,v_{n-1})=SPAN(B)$, but this is a contradiction as $v_n∈SPAN(B)=SPAN(S)$. Therefore $SPAN(S)$is the smallest subspace of $V$ that contains the vectors $v_1,…,v_p$.
Please comment and provide feedback on the validity of my proof. Thank you.
Best Answer
Your proof assumes that, given a set $S=\{v_1,\ldots,v_p\}$, then there is a $n\leqslant p$ such that $\{v_1,\ldots,v_n\}$ is a basis of $\operatorname{span}(S)$. That is not true. For instance, if $v_1=0$ and $v_2\ne0$, then there is no such $n$.
You can do it as follows. Let $W$ be a subspace of $V$ such that $W\supset S$. Then, since $W\supset S$ and since $W$ is a vector space, $W$ contains all linear combinations of elements of $S$. In other words, $W\supset\operatorname{span}(S)$. This proves that $\operatorname{span}(S)$ is contained in every subspace of $V$ containing $S$. Therefore, since $\operatorname{span}(S)$ is a subspace containing $S$, it is the smallest such subspace.