A set of vectors spans if they can be expressed as linear combinations. Say we have a set of vectors we can call S in some vector space we can call V. The subspace, we can call W, that consists of all linear combinations of the vectors in S is called the spanning space and we say the vectors span W.
Here is an example of vectors in R^3.
Say we have \(\displaystyle V_{1}=(1,1,2), \;\ v_{2}=(1,0,1), \;\ v_{3}=(2,1,3)\)
We want to see if they span or not.
We have to find whether an arbitrary vector, say, \(\displaystyle b=(b_{1},b_{2},b_{3})\) can be expressed as a linear combo \(\displaystyle b=k_{1}v_{1}+k_{2}v_{2}+k_{3}v_{3}\) of the vectors \(\displaystyle v_{1},v_{2},v_{3}\).
Set up a system of equations in terms of the components:
\(\displaystyle (b_{1},b_{2},b_{3})=k_{1}(1,1,2)+k_{2}(1,0,1)+k_{3}(2,1,3)\)
\(\displaystyle (b_{1},b_{2},b_{3})=(k_{1}+k_{2}+2k_{3}, \;\ k_{1}+k_{3}, \;\ 2k_{1}+k_{2}+3k_{3})\)
\(\displaystyle k_{1}+k_{2}+2k_{3}=b_{1}\)
\(\displaystyle k_{1} \;\ \;\ +k_{3}=b_{2}\)
\(\displaystyle 2k_{1}+k_{2}+3k_{3}=b_{3}\)
The system is consistent for all \(\displaystyle b_{1},b_{2},b_{3}\) iff the matrix of coefficients:
\(\displaystyle A=\begin{bmatrix}1&1&2\\1&0&1\\2&1&3\end{bmatrix}\)
has a determinant that is not equal to 0.
But this determinant does equal 0, so it DOES NOT span.
