Proof of Theorem G

Let $B$ be a basis of $V$. Since $\dimension{V}=t$, Definition B and Theorem BIS imply that $B$ is a linearly independent set of $t$ vectors that spans $V$.

1. Suppose to the contrary that $S$ is linearly independent. Then $B$ is a smaller set of vectors that spans $V$. This contradicts Theorem SSLD.
2. Suppose to the contrary that $S$ does span $V$. Then $B$ is a larger set of vectors that is linearly independent. This contradicts Theorem SSLD.
3. Suppose to the contrary that $S$ does not span $V$. Then we can choose a vector $\vect{w}$ such that $\vect{w}\in V$ and $\vect{w}\not\in\spn{S}$. By Theorem ELIS, the set $S^\prime=S\cup\set{\vect{w}}$ is again linearly independent. Then $S^\prime$ is a set of $m+1=t+1$ vectors that are linearly independent, while $B$ is a set of $t$ vectors that span $V$. This contradicts Theorem SSLD.
4. Suppose to the contrary that $S$ is linearly dependent. Then by Theorem DLDS (which can be upgraded, with no changes in the proof, to the setting of a general vector space), there is a vector in $S$, say $\vect{v}_k$ that is equal to a linear combination of the other vectors in $S$. Let $S^\prime=S\setminus\set{\vect{v}_k}$, the set of “other” vectors in $S$. Then it is easy to show that $V=\spn{S}=\spn{S^\prime}$. So $S^\prime$ is a set of $m-1=t-1$ vectors that spans $V$, while $B$ is a set of $t$ linearly independent vectors in $V$. This contradicts Theorem SSLD.