##### Example3.5.2Subsets that are not subspaces

It will be helpful to first look at some examples of subsets of \(\real^2\) that are not subspaces. First, consider the set of vectors in the first quadrant of \(\real^2\text{;}\) that is, vectors of the form \(\twovec{x}{y}\) where both \(x,y \geq 0\text{.}\) This subset is illustrated on the left of Figure 3.

If this subset were a subspace of \(\real^2\text{,}\) any linear combination of vectors in the first quadrant must also be in the first quadrant. If we consider the vector \(\vvec=\twovec{3}{2}\text{,}\) however, we can form the linear combination \(-\vvec=\twovec{-3}{-2}\text{,}\) which is not in the first quadrant, as seen on the right of Figure 3. Therefore, the set of vectors in the first quadrant is not a subspace.

This shows something important, however. Suppose that \(S\) is a subspace and \(\vvec\) is a vector in \(S\text{.}\) Any scalar multiple of \(\vvec\) is a linear combination of \(\vvec\) and so must be in \(S\) as well. This means that the line containing \(\vvec\) must be in \(S\text{.}\)

With this in mind, let's consider another example where we look at vectors that are in either the first or third quadrant; that is, we will consider vectors of the form \(\twovec{x}{y}\) where either \(x,y\geq 0\) or \(x,y\leq 0\text{,}\) as seen on the left of Figure 4.

If \(\vvec\) is a vector in this set, then the line containing \(\vvec\) is in the set. However, if we consider the vectors \(\vvec = \twovec{0}{3}\) and \(\wvec=\twovec{-2}{0}\text{,}\) then their sum \(\vvec+\wvec = \twovec{-2}{3}\) is not in the subset, as seen on the right of Figure 4. This subset is therefore not a subspace.