When we talk about a vector space over a field , the word scalar refers to an element of .
A subspace of a vector space is a subset of which
contains the zero vector ,
is closed under addition, meaning that for all we have , and
is closed under scalar multiplication, meaning that for all scalars and all we have .
We write to mean that is a subspace of .
The idea this definition captures is that a subspace of is a nonempty subset which is itself a vector space under the same addition and scalar multiplication as .
If and and are scalars then . This follows by using closure under scalar multiplication and closure under addition lots of times.
If is any vector space, . This is because, as a vector space, contains the zero vector, is closed under addition, and is closed under scalar multiplication.
A subspace of other than is called a proper subspace.
For any vector space we have . Certainly this set contains the zero vector. It is closed under addition because , and it is closed under scalar multiplication by Lemma 4.3.3. This is called the zero subspace.
Let be the set of vectors in whose first entry is zero. Then . We check the three conditions in the definition of subspace.
The zero vector in is . This has first coordinate 0, so it is an element of .
Let , so that and for some real numbers and . Then has first coordinate 0, so it is an element of .
Let v be as above and . Then which has first coordinate 0, so .
All three conditions hold, so . Of course, a similar argument shows the vectors in with first entry 0 are a subspace of for any field and any .
To every matrix we associate two important subspaces. The nullspace (Definition 3.6.5) is the set of all vectors such that , and the column space is the set of all linear combinations of the columns of .
Let be an matrix with entries from the field . The nullspace contains the zero vector as . It is closed under addition as if then and so
and therefore . It is closed under scalar multiplication because if is any scalar then so . Therefore .
The column space , defined to be the set of all linear combinations of the columns of , is a subspace of . We won’t prove that here, because it is a special case of Proposition 4.7.1 which we prove later.
The set of all vectors in with first entry 1 is not a subspace of . It doesn’t contain the zero vector (and it doesn’t meet the other two conditions either).
is not a subspace of . It contains the zero vector 0, it is closed under addition because if you add two integers you get another integer. But it is not closed under scalar multiplication: is a scalar, , but is not in .
Let be the set of all functions with . This is a subspace of the vector space of all functions . The zero vector in is the constant function that always takes the value zero, so certainly it belongs to . If then , so . If and then so .
. The transpose operation satisfies and , which you should check. This makes the three conditions straightforward to check.
is not a subspace of . For example, contains and but you can check that .