We define closure under addition and scalar multiplication, and we demonstrate how to determine whether a subset of vectors in is a subspace of .

### VSP-0020: and Subspaces of

#### Closure

We are familiar with two operations that can be applied to vectors in , namely,
addition and scalar multiplication. We learned that addition and scalar multiplication
satisfy many nice properties. (See Theorem th:vecproperties of VEC-0030) These properties give an
algebraic structure, and make a *vector space*. Before we can give a formal
definition of a vector space we need to introduce the concept of *closure* under an
operation.

*closed under scalar multiplication*if for each element and for each scalar the product is also in .

is not closed under scalar multiplication because is not in .

The figure below helps us see that the sum of any two vectors in also lies in , but any negative scalar multiple of a vector in does not lie in .

#### as a Vector Space

In Theorem th:vecproperties of VEC-0030 we learned that vector addition and scalar multiplication in satisfy the following eight properties:

- (a)
- Commutative Property of Addition:
- (b)
- Associative Property of Addition:
- (c)
- Existence of Additive Identity:
- (d)
- Existence of Additive Inverse:
- (e)
- Distributive Property over Vector Addition:
- (f)
- Distributive Property over Scalar Addition:
- (g)
- Associative Property for Scalar Multiplication:
- (h)
- Multiplication by :

In addition, observe that

- is closed under addition (Why?)
- 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, is a vector
space.

We will encounter other vector spaces later. Any vector space must be closed under both operations, and must satisfy the other eight properties in the list above. We will see (in module VSP-0050, for instance) that there is a wide variety of sets, with a wide variety of operations, that are vector spaces (which is one of the reasons we 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 .

#### Subspaces of

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.

*subspace*of .

*subspace*because it turns out that any subset of closed under both addition and scalar multiplication is also a vector space in its own right. In other words, by satisfying the properties of closure, a subset of will automatically satisfy the eight vector space properties listed above. We will prove this in VSP-0050.

Recall that the *span* of a set of vectors is the set of all linear combinations of those
vectors. (See Definition def:span in VEC-0090) It is easy to see from this definition, that the
span of any set of vectors in must be closed under both addition and scalar
multiplication, and therefore the span of those vectors is a subspace of . This
argument proves the following result, giving us an abundance of examples of
subspaces:

In particular, if we take to be the single vector , we have that is a subspace of . Geometrically, this subspace is a line in the direction of the vector . Similarly, the span of two vectors is a subspace of . If the two vectors are linearly independent, then the subspace is a plane in .

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

- Proof
- Take any vector in , and note that is in because is closed under scalar multiplication.

Theorem th:zero_in_subspace shows that the only lines in that are subspaces are those that pass through the origin. The same holds true for planes and hyperplanes. For example, the plane in is not a subspace of , 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.