Discover Linear Algebra: A first course in linear algebra

Suppose we have a subcollection of vectors in a vector space that satisfies the three conditions of the Subspace Test. We would like to verify that this subcollection is a vector space all on its own, using the same vector addition and scalar multiplication operations as the larger space. But as explored in Discovery 16.1 and discussed in Subsection 16.3.1, we don’t need to verify all ten vector space axioms — we get the six algebra axioms for free from knowing that is how vector algebra works in the larger space. The remaining four axioms are Axiom A 1, Axiom A 4, Axiom A 5, and Axiom S 1, so we really only need to verify that the subcollection contains the zero vector, and is closed under vector addition, scalar multiplication, and taking negatives. Furthermore, from Condition ii and Condition iii of the test, we already know that the subcollection is closed under addition and scalar multiplication. So we are down to checking the zero vector and negatives.

To show that the subcollection must contain the zero vector, consider that Condition i of the test guarantees that the subcollection contains some vector $\mathbf{v}\text{.}$ But then Condition iii of the test tells us that the subcollection must also contain every scalar multiple of $\mathbf{v}\text{.}$ In particular, by applying Rule 1.b of Proposition 15.6.2 we may say that the subcollection must contain $0\mathbf{v}=\mathbf{0}\text{,}$ as desired.

To show that the subcollection must be closed under taking negatives, consider a vector $\mathbf{w}$ in the subcollection. Again, Condition i of the test says that the subcollection must also contain every scalar multiple of $\mathbf{w}\text{.}$ In particular, by applying Rule 2.e of Proposition 15.6.2 we may say that the subcollection must contain $(-1)\mathbf{w}=-\mathbf{w}\text{,}$ as desired.

Finally, we will consider the “only if” part of the statement. Suppose we have a subcollection of vectors in a vector space that we already know is a subspace. A subspace is itself a vector space, so it must be nonempty (since it at least contains some zero vector by Axiom A 4), it must be closed under vector addition (Axiom A 1), and it must be closed under scalar multiplication (Axiom S 1). In other words, it must pass the Subspace Test.

We have already established that a subspace is closed under the vector operations. Verifying that it also contains the zero vector and is closed under taking negatives is exactly as in the proof of Proposition 16.5.1 above, since we know that a subspace always passes the Subspace Test.

It remains to show that a subspace is closed under linear combinations. So suppose that ${\mathbf{v}}_1,{\mathbf{v}}_2,\dots ,{\mathbf{v}}_{\ell }$ are vectors in the subspace. Since the subspace is closed under scalar multiplication, the vectors $k_1{\mathbf{v}}_1,k_2{\mathbf{v}}_2,\dots ,k_{\ell }{\mathbf{v}}_{\ell }$ are all also in the subspace. And then, since the subspace is also closed under addition, the linear combination $k_1{\mathbf{v}}_1+k_2{\mathbf{v}}_2+\cdots +k_{\ell }{\mathbf{v}}_{\ell }$ is also in the subspace.

We verified these examples of subspaces in Example 16.4.6 and Example 16.4.7 of Subsection 16.4.2.

Recall that $\mathrm{Span}S$ is the collection of all possible linear combinations of vectors in $S\text{.}$ First we verify that $\mathrm{Span}S$ contains every vector in $S\text{.}$ Indeed, if $\mathbf{v}$ is a vector in $S\text{,}$ then it is trivially a linear combination of vectors in $S$ by $\mathbf{v}=1\mathbf{v}\text{.}$

Let $V$ represent the vector space from which the collection of vectors $S$ is taken. First, we know that every vector in $\mathrm{Span}S$ is a vector in $V\text{,}$ because the vectors in $\mathrm{Span}S$ are linear combinations of the vectors in $S\text{,}$ and $V$ is closed under taking linear combinations (Proposition 16.5.2, where $V$ is considered as a subspace of itself using Proposition 16.5.4). So $\mathrm{Span}S$ is a subcollection of $V\text{.}$

Now let’s apply the Subspace Test to $\mathrm{Span}S\text{.}$

We wish to show that every other subspace that contains the vectors of $S$ must also contain $\mathrm{Span}S$ as a subspace. So suppose we have another subspace that contains the vectors of $S\text{.}$ Then it must contain every linear combination of the vectors in $S\text{,}$ since subspaces are closed under taking linear combinations (Proposition 16.5.2). That is, if a subspace contains all of the vectors in $S\text{,}$ then it must also contain all of the vectors in $\mathrm{Span}S\text{.}$

Note: There is no need to use the Subspace Test to prove that $\mathrm{Span}S$ is a subspace of this other subspace — we already know from Statement 1 that $\mathrm{Span}S$ is a subspace of $V\text{,}$ the vector space from which the vectors $S$ are taken. So $\mathrm{Span}S$ is a vector space all on its own, hence will be a subspace of any space that contains all of its vectors. (See the definition of subspace in Section 16.2.)

A vector space $V$ always has an obvious spanning set — itself! That is, we claim that $V=\mathrm{Span}V$ is always true. To verify this, we must demonstrate that each vector in the collection $V$ is also in the collection $\mathrm{Span}V\text{,}$ and vice versa, so that they are exactly the same collection of vectors. However, by applying Statement 1 we can immediately say that $\mathrm{Span}V$ is a subspace of $V$ (implying every vector in $\mathrm{Span}V$ is in $V$) that contains every vector of the spanning set $V$ (i.e. every vector in $V$ is in $\mathrm{Span}V$).

Recall that $\mathrm{Span}{S}^{\prime }$ is the collection of all possible linear combinations of the vectors in $S^{\prime }\text{.}$ So assuming that each vector in $S$ can be expressed as a linear combination of the vectors in $S^{\prime }$ is the same as assuming that each vector in $S$ is in $\mathrm{Span}{S}^{\prime }\text{.}$ But Statement 2 of Proposition 16.5.5 tells us that $\mathrm{Span}S$ is the smallest subspace that contains all the vectors in $S\text{,}$ and that $\mathrm{Span}S$ must therefore be a subspace of $\mathrm{Span}{S}^{\prime }\text{.}$

Now we assume both that each vector in $S$ can be expressed as a linear combination of the vectors in $S^{\prime }$ and that each vector in $S^{\prime }$ can be expressed as a linear combination of the vectors in $S\text{.}$ Then we can apply Statement 1 of this proposition twice, first to conclude that $\mathrm{Span}S$ is a subspace of $\mathrm{Span}{S}^{\prime }\text{,}$ and second to conclude that $\mathrm{Span}{S}^{\prime }$ is a subspace of $\mathrm{Span}S\text{.}$ But then $\mathrm{Span}S$ is a subcollection of the vectors in $\mathrm{Span}{S}^{\prime }\text{,}$ and also vice versa. This can only happen if they are in fact the same collection of vectors.