# Linear algebra: distinguishing between Vector Subspace and more general sub-set of vectors

Though this is not a specific homework problem, I'm trying to make sure I'm getting (all) the concepts in the section on Vector Spaces and Subspaces (LA, Lay, Sect. 4.1, 4th Ed).

The concept of a vector space seems clear. And the fact that a subspace to a vector space is a specific subset which contains the zero vector and is closed under addition and (scalar) multiplication also seems clear.

By the above, a sub-set of R3 defined by all vectors defined by $(a+b, 0, 2a-3b)$ (where $a , b \in R$) WOULD be a subspace, whereas a sub-set defined by $(a+b, -1, 2a-3b)$ WOULD NOT be a subspace (that is, the second sub-set does NOT include the zero vector).

My trouble is reconciling the wording of "Theorem 1" in that section (all the lower-case "$v$" variables are vectors): If $v_1, . . . ,v_n$ are in vector space $V$, then $Span (v_1, . . ., v_n)$ is a subspace of $V$.

I'm trying to make sure that I'm right in the following: one could make up a sub-set of vectors drawn from any vector space by any means one wanted; any such set would be just that: a set of vectors with no guarantee of special properties. However, if a set has the additional properties listed above (has zero vector, closed), it is ALSO a sub-SPACE of that vector space (and therefor a vector space in-and-of itself). Is that correct?

And as an aside, the reason a "Span" operation between one (or more) vectors from a vector space will always result in a sub-space (another vector space) is that the definition of a "vector" is the path/line from the origin to the indicated point in some n-space (some $\mathbb{R}^n$); thus, when creating a 'Span', by definition each of the vectors is originating from the origin (equivalent to the zero vector), and thus include it in the Span. (And this is why a set which does NOT include the zero vector is not a sub-space, though it may be a sub-set of vectors in the n-space ($\mathbb{R}^n$)). Is that also correct?

Thanks for taking the time to answer. (Studying LA and VC on my own, and live in a town with no 4-year college, so no ready access to professorial or grad student help!)

• This seems to be a pure math question, it belongs on Mathematics instead. Commented Nov 19, 2015 at 20:50

From the definition of $$Span(v_1, v_2, ..., v_r) = \{\sum_{i=1}^r \alpha_iv_i : \alpha_i \in \mathbb{K}, i =1, 2, ..., r\}$$
for vectors $v_1, v_2, ..., v_r$ in a vector space $V$ over a field $\mathbb{K}$, it follows that $Span(v_1, v_2, ..., v_r)$ it's a vector space itself (it's easy to see that $0$ is in $Span(v_1, v_2, ..., v_r)$ and to prove the closure).