##### 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 also not a subspace.

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