Exercise 19 — Linear Algebra

All of the subspaces that we’ve seen in some way use zero in their description. For example, the subspace in Example 2.3 consists of all the vectors from $\mathbb{R}^{2}$ with a second component of zero. In contrast, the collection of vectors from $\mathbb{R}^{2}$ with a second component of one does not form a subspace (it is not closed under scalar multiplication). Another example is Example 2.2, where the condition on the vectors is that the three components add to zero. If the condition there were that the three components add to one then it would not be a subspace (again, it would fail to be closed). However, a reliance on zero is not strictly necessary. Consider the set

\{\hskip 1.00006pt\begin{pmatrix}x\\ y\\ z\end{pmatrix}\mid x+y+z=1\hskip 1.00006pt\}

under these operations.

\begin{pmatrix}x_{1}\\ y_{1}\\ z_{1}\end{pmatrix}+\begin{pmatrix}x_{2}\\ y_{2}\\ z_{2}\end{pmatrix}=\begin{pmatrix}x_{1}+x_{2}-1\\ y_{1}+y_{2}\\ z_{1}+z_{2}\end{pmatrix}\qquad r\begin{pmatrix}x\\ y\\ z\end{pmatrix}=\begin{pmatrix}rx-r+1\\ ry\\ rz\end{pmatrix}

Show that it is not a subspace of $\mathbb{R}^{3}$ . (Hint. See Example 2.7).
Show that it is a vector space. Note that by the prior item, Lemma 2.9 can not apply.
Show that any subspace of $\mathbb{R}^{3}$ must pass through the origin, and so any subspace of $\mathbb{R}^{3}$ must involve zero in its description. Does the converse hold? Does any subset of $\mathbb{R}^{3}$ that contains the origin become a subspace when given the inherited operations?

Hefferon, Linear Algebra, Vector Spaces, Subspaces and Spanning Sets, Exercise 19

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