The proof in the textbook linear algebra 4th edition by Stephen H. Friedberg Arnold J. Insel Lawrence E. Spence
In the last part:
" Let v $\in$ S. If v $\in$ $\beta$, then clearly v $\in$ span($\beta$). Otherwise, if v $\notin$ $\beta$, then the preceding
construction shows that $\beta$ $\cup$ {v} is linearly dependent. So v $\in$ span($\beta$) by
Theorem 1.7 (p. 39). Thus S $\subseteq$ span($\beta$)"
＂$\beta$ $\cup$ {v} is linearly dependent" is also correct when v $\in$ V-$\beta$ besides v $\in$ S-$\beta$, right?
So can we say it alternatively :
Let v $\in$ "V". If v $\in$ $\beta$, then clearly v $\in$ span($\beta$). Otherwise, if v $\notin$ $\beta$, then the preceding
construction shows that $\beta$ $\cup$ {v} is linearly dependent. So v $\in$ span($\beta$) by Theorem 1.7 (p. 39). Thus "V" $\subseteq$ span($\beta$)."?
Is this correct?
Best Answer
No, it is not correct. The “preceding construction” only shows that if you add an element $v$ of $S$ to $\beta$, then the set $\beta\cup\{v\}$ is linearly dependent (since $\beta$ is a maximal linearly independent subset of $S$). But the construction of $S$ doesn't allow you to deduce directly that $\beta\cup\{v\}$ is linearly dependent for an arbitrary element of $V$.