### Vector spaces: Spans

### Basis and dimension

If a subspace of a vector space is the span of two independent sets of vectors, then the number of vectors of the two sets must be the same. This number is called the dimension of the space. This property is a consequence of the following result.

Number of spanning vectors

Let #m# and #n# be natural numbers, #V# a vector space, and #\vec{a}_1 ,\ldots ,\vec{a}_n#, #\vec{b}_1,\ldots ,\vec{b}_m# vectors of #V# such that

- #V= \linspan{\vec{a}_1 ,\ldots ,\vec{a}_n }# and
- #\vec{b}_1 ,\ldots ,\vec{b}_m# is an independent set of vectors.

Then:

- #m\leq n#.
- If #V# is also spanned by #\vec{b}_1 ,\ldots ,\vec{b}_m#, then #m=n#.

This result justifies the following definition of dimension:

Basis and dimension

An independent spanning subset of a vector space #V# is called a **basis** of #V# (the plural is **bases**).

The number of elements of a basis is called the **dimension** of #V# and denoted as #\dim{V}#.

If #V# is the null space, then the dimension is #0#. If there is no finite number of spanning vectors of #V#, then the dimension is infinite, and we write #\dim{V}=\infty#.

For every natural number #n# the set of vectors in #\mathbb{R}^n# \[\vec{e}_1 =\rv{1,0,0,\ldots ,0},\phantom{x}

\vec{e}_2 =\rv{0,1,0,\ldots ,0},\phantom{x}\ldots,\phantom{x}\vec{e}_n =\rv{0,0,0,\ldots ,1}\] is a basis of #\mathbb{R}^n#; it is also called the **standard basis** of #\mathbb{R}^n#. Accordingly, #\dim{\mathbb{R}^n}=n#. Similarly, #\dim{\mathbb{C}^n}=n#. Here, too, we use the name **standard basis**.

The following properties are useful for determining the dimension.

Properties of the dimension

If #V# is a vector space with #\dim{V} =n\lt\infty#, then each basis of #V# consists of exactly #n# vectors. Let #\vec{b}_1 ,\ldots ,\vec{b}_m# be a set of vectors in #V#.

- If #m\lt n#, then #\linspan{\vec{b}_1 ,\ldots ,\vec{b}_m }# is a proper subspace of #V#.
- If #m\gt n#, then the set #\vec{b}_1 ,\ldots ,\vec{b}_m# is dependent.
- If #m=n#, then the set #\vec{b}_1 ,\ldots ,\vec{b}_m# is a basis of #V# if and only if the sequence is independent.

Select two vectors in #\mathbb{E}^2# which are both distinct from the zero vector and which are not scalar multiples of each other. Then, any other vector in the plane can be written as a linear combination of these two (this is especially well known for the two vectors that have been given the coordinates #\rv{1,0}# and #\rv{0,1}#, respectively, in a coordinate system). Moreover, the two are linearly independent. From these two facts, we conclude that they form a basis.

The reasoning for #\mathbb{E}^3# is similar.

**Pass Your Math**independent of your university. See pricing and more.

Or visit omptest.org if jou are taking an OMPT exam.