Subsection TS Testing Subspaces

In Example SC3 we proceeded through all ten of the vector space properties before believing that a subset was a subspace. But six of the properties were easy to prove, and we can lean on some of the properties of the vector space (the superset) to make the other four easier. Here is a theorem that will make it easier to test if a subset is a vector space. A shortcut if there ever was one.

Theorem TSS Testing Subsets for Subspaces

Suppose that $V$ is a vector space and $W$ is a subset of $V$, $W\subseteq V$. Endow $W$ with the same operations as $V$. Then $W$ is a subspace if and only if three conditions are met

$W$ is nonempty, $W\neq\emptyset$.
If $\vect{x}\in W$ and $\vect{y}\in W$, then $\vect{x}+\vect{y}\in W$.
If $\alpha\in\complex{\null}$ and $\vect{x}\in W$, then $\alpha\vect{x}\in W$.

Proof

So just three conditions, plus being a subset of a known vector space, gets us all ten properties. Fabulous! This theorem can be paraphrased by saying that a subspace is “a nonempty subset (of a vector space) that is closed under vector addition and scalar multiplication.”

You might want to go back and rework Example SC3 in light of this result, perhaps seeing where we can now economize or where the work done in the example mirrored the proof and where it did not. We will press on and apply this theorem in a slightly more abstract setting.

Example SP4 A subspace of $P_4$

Much of the power of Theorem TSS is that we can easily establish new vector spaces if we can locate them as subsets of other vector spaces, such as the vector spaces presented in Subsection VS.EVS.

It can be as instructive to consider some subsets that are not subspaces. Since Theorem TSS is an equivalence (see Proof Technique E) we can be assured that a subset is not a subspace if it violates one of the three conditions, and in any example of interest this will not be the “nonempty” condition. However, since a subspace has to be a vector space in its own right, we can also search for a violation of any one of the ten defining properties in Definition VS or any inherent property of a vector space, such as those given by the basic theorems of Subsection VS.VSP. Notice also that a violation need only be for a specific vector or pair of vectors.

Example NSC2Z A non-subspace in $\complex{2}$, zero vector

Example NSC2A A non-subspace in $\complex{2}$, additive closure

Example NSC2S A non-subspace in $\complex{2}$, scalar multiplication closure

There are two examples of subspaces that are trivial. Suppose that $V$ is any vector space. Then $V$ is a subset of itself and is a vector space. By Definition S, $V$ qualifies as a subspace of itself. The set containing just the zero vector $Z=\set{\zerovector}$ is also a subspace as can be seen by applying Theorem TSS or by simple modifications of the techniques hinted at in Example VSS. Since these subspaces are so obvious (and therefore not too interesting) we will refer to them as being trivial.

Definition TS Trivial Subspaces

Given the vector space $V$, the subspaces $V$ and $\set{\zerovector}$ are each called a trivial subspace.

We can also use Theorem TSS to prove more general statements about subspaces, as illustrated in the next theorem.

Theorem NSMS Null Space of a Matrix is a Subspace

Suppose that $A$ is an $m\times n$ matrix. Then the null space of $A$, $\nsp{A}$, is a subspace of $\complex{n}$.