\(\mathbb {R}^n\) and Subspaces of \(\mathbb {R}^n\)

We are familiar with two operations that can be applied to vectors in \(\RR ^n\), namely, addition and scalar multiplication. We learned that addition and scalar multiplication satisfy many nice properties (see Theorem th:vecproperties). These properties give \(\RR ^n\) an algebraic structure. We begin this section by introducing another property, called closure. Adding closure to the properties we studied earlier allows us to show that \(\RR ^n\) satisfies all of the properties of a vector space.

Closure

\(\RR ^n\) as a Vector Space

In Theorem ?? we learned that vector addition and scalar multiplication in \(\RR ^n\) satisfy the following eight properties:

For all vectors \(\vec {u}\), \(\vec {v}\), \(\vec {w}\in \RR ^n\), and scalars \(k, p\in \RR \),

1.
Commutative Property of Addition: \(\vec {u}+\vec {v}=\vec {v}+\vec {u}\)
2.
Associative Property of Addition: \((\vec {u}+\vec {v})+\vec {w}=\vec {u}+(\vec {v}+\vec {w})\)
3.
Existence of Additive Identity: \(\vec {u}+\vec {0}=\vec {u}\)
4.
Existence of Additive Inverse: \(\vec {u}+(-\vec {u})=\vec {0}\)
5.
Distributive Property over Vector Addition: \(k(\vec {u}+\vec {v})=k\vec {u}+k\vec {v}\)
6.
Distributive Property over Scalar Addition: \((k+p)\vec {u}=k\vec {u}+p\vec {u}\)
7.
Associative Property for Scalar Multiplication: \(k(p\vec {u})=(kp)\vec {u}\)
8.
Multiplication by \(1\): \(1\vec {u}=\vec {u}\)

In addition, observe that

  • \(\RR ^n\) is closed under addition (Why?)
  • \(\RR ^n\) is closed under scalar multiplication (Why?)

The eight properties of vector operations, together with closure, constitute the criteria for a set with two operations to be considered a vector space. So, \(\RR ^n\) is a vector space.

We will encounter other vector spaces later. Any vector space must be closed under both of its operations, and must satisfy the other eight properties in the list above. We will see (in Abstract Vector Spaces, for instance) that a wide variety of sets, with a wide variety of operations, are vector spaces (one important reason to study linear algebra). As we shall see, these sets and their operations may look very different, but the behavior of the elements under the two operations makes them vector spaces. For now, we simply focus on \(\RR ^n\).

Subspaces of \(\RR ^n\)

Now that we understand what it means for a set to be closed under addition and scalar multiplication, we are ready for the main definition.

We use the term subspace because it turns out that any subset of \(\RR ^n\) closed under both addition and scalar multiplication is also a vector space. In other words, by inheriting vector addition and scalar multiplication from \(\RR ^n\), and satisfying the properties of closure, a subset of \(\RR ^n\) will automatically satisfy all vector space properties. We will prove this in Theorem ?? of Abstract Vector Spaces.

Recall that the span of a set of vectors is the set of all linear combinations of those vectors (see Definition ??). It is easy to see from this definition, that the span of any set of vectors in \(\RR ^n\) must be closed under both addition and scalar multiplication, and therefore the span of those vectors is a subspace of \(\RR ^n\). This argument proves the following result, giving us an abundance of examples of subspaces:

In particular, if we take \(S\) to be the single vector \(\vec {v}\), we have that \(\text {span}(\vec {v})\) is a subspace of \(\RR ^n\). Geometrically, this subspace is a line with a direction vector \(\vec {v}\). Similarly, the span of two vectors is a subspace of \(\RR ^n\). If the two vectors are linearly independent, then the subspace is a plane in \(\RR ^n\).

Not every line or plane in \(\RR ^n\) is a subspace, however. The following important result provides us with a quick way to determine that some subsets are not subspaces.

Take any vector \(\vec {v}\) in \(V\), and note that \(0 \vec {v} = \vec {0}\) is in \(V\) because \(V\) is closed under scalar multiplication.

Theorem 12 shows that the only lines in \(\RR ^n\) that are subspaces are those that pass through the origin. The same holds true for planes and hyperplanes. For example, the plane \(z=3\) in \(\RR ^3\) is not a subspace of \(\RR ^3\), while any plane containing the origin is a subspace.

The proof is similar to what was done for the previous theorem and is left as an exercise.

Practice Problems

Let \(Y^+\) be the set of all vectors in \(\mathbb {R}^2\) whose \(y\) components are non-negative. Is \(Y^+\) closed under vector addition?
Yes No
Let \(Y^+\) be the set of all vectors in \(\mathbb {R}^2\) whose \(y\) components are non-negative. Is \(Y^+\) closed under scalar multiplication?
Yes No
Let \(X\) be the set of all vectors in \(\RR ^3\) that lie on either the \(x\)-axis, the \(y\)-axis, or the \(z\)-axis. Is \(X\) closed under vector addition?
Yes No
Let \(X\) be the set of all vectors in \(\RR ^3\) that lie on either the \(x\)-axis, the \(y\)-axis, or the \(z\)-axis. Is \(X\) closed under scalar multiplication?
Yes No
Determine whether the set \(V\) of vectors shown in the figure is closed under vector addition and scalar multiplication. Justify your responses. \(V\) consists of all vectors in \(\mathbb {R}^3\) in a slanted half-plane which has the \(x\)-axis as a boundary.

[Picture]

Is \(V\) closed under scalar multiplication?

Yes No

Is \(V\) closed under addition?

Yes No
Determine whether the set \(V\) of vectors shown in the figure is closed under vector addition and scalar multiplication. Justify your responses. \(V\) consists of all vectors along the line, as shown.

[Picture]

Is \(V\) closed under scalar multiplication?

Yes No

Is \(V\) closed under addition?

Yes No
Prove that if \(V\) is a subspace of \(\RR ^n\), then for any vector \(\vec {v} \in V\), the opposite vector, \(-\vec {v}\), is also in \(V\). (Theorem 14)
Let \(A\) be an \(m \times n\) matrix. Let \(V\) be the subset of \(\RR ^n\) consisting of all vectors \(\vec {x}\) such that \(A \vec {x} = \vec {0}\). Prove that \(V\) is a subspace of \(\RR ^n\). (This subspace is called the null space of the matrix \(A\). We will denote it \(\mbox {null}(A)\).)