T/F: Subset of a spanning set always forms a basis

[Mr Davis 97]

Homework Statement

T/F: If a finite set of vectors spans a vector space, then some subset of the vectors is a basis.

Homework Equations

The Attempt at a Solution

It seems that the answer is true, due to the "Spanning Set Theorem," which says that we are allowed to remove vectors in a spanning set until we get a set that is linearly independent and thus forms a basis. However, what if our original set of vectors only spans the trivial vectors space $\{ 0 \}$? In that case we can't form a basis, so wouldn't the answer to this question be false?

 

[fresh_42]

However, what if our original set of vectors only spans the trivial vectors space ##\{ 0 \}?

This can only be the case, if all vectors in the set are $\vec{0}$. Then the subset $\{\}$ spans $\{0\}$.

 

[Mr Davis 97]

So in the problem statement, why can't we interpret that the finite set of vectors is a set of zero vectors?

 

[fresh_42]

So in the problem statement, why can't we interpret that the finite set of vectors is a set of zero vectors?

Who says we can't? A set $\{\vec{0},\vec{0},\vec{0},\vec{0},\ldots\}=\{\vec{0}\}$ spans the zero dimensional vector space $V=\{0\}$ with the basis $\{\}$. Of course this is on the edge of the definition and not very geometric, but it is allowed. And if you take away $\vec{0}$ from $\{\vec{0}\}$ you are left with the basis $\{\}$ of $V=\{0\}$.

One cannot have some non-zero vectors and their (linear) span will turn out to be $\{0\}$, because with every $0 \neq \vec{v}$ there will be at least all vectors $\mathbb{R}\cdot \vec{v}$ in the linear span of a set, that contains $\vec{v}$.
(I chose $\mathbb{R}$ as the corresponding field for simplicity, but any other will work as well.)