# Linear Algebra/Span of a set

## Definition[edit | edit source]

Let V be a vector space over a field F. Choose n vectors **x _{1}**,

**x**,

_{2}**x**, ...,

_{3}**x**from the vector space V. The

_{n}**linear manifold**spanned by

**x**,

_{1}**x**,

_{2}**x**, ...,

_{3}**x**is defined to be all elements of V of the form a

_{n}_{1}

**x**+a

_{1}_{2}

**x**+a

_{2}_{3}

**x**+...+a

_{3}_{n}

**x**where a

_{n}_{1}, a

_{2}, a

_{3}, ..., a

_{n}are all elements of the field F, and shall be denoted S(

**x**,

_{1}**x**,

_{2}**x**, ...,

_{3}**x**). This is obviously a linear subspace of the vector space V. Since every linear subspace of V contains

_{n}**x**,

_{1}**x**,

_{2}**x**, ...,

_{3}**x**and their linear combinations, S(

_{n}**x**,

_{1}**x**,

_{2}**x**, ...,

_{3}**x**) is the smallest subspace containing

_{n}**x**,

_{1}**x**,

_{2}**x**, ...,

_{3}**x**.

_{n}## Theorem[edit | edit source]

If **y _{1}**,

**y**,

_{2}**x**, ...,

_{y}**y**are elements of S(

_{m}**x**,

_{1}**x**,

_{2}**x**, ...,

_{3}**x**), then S(

_{n}**y**,

_{1}**y**,

_{2}**y**, ...,

_{3}**y**) is contained within S(

_{m}**x**,

_{1}**x**,

_{2}**x**, ...,

_{3}**x**)

_{n}### Proof[edit | edit source]

All linear combinations of vectors which belong to a linear manifold also belong in the linear manifold (since a linear combination of linear combinations of vectors is also a linear combination of those vectors), and since any element of S(**y _{1}**,

**y**,

_{2}**y**, ...,

_{3}**y**) is a linear combinations of vectors within the manifold, it too is within the set, thus proving that S(

_{m}**y**,

_{1}**y**,

_{2}**y**, ...,

_{3}**y**) is contained within S(

_{m}**y**,

_{1}**y**,

_{2}**y**, ...,

_{3}**y**).

_{m}## Theorem[edit | edit source]

If **x** is linearly dependent on other vectors upon other vectors **x _{1}**,

**x**,

_{2}**x**, ...,

_{3}**x**, then it belongs to S(

_{n}**x**,

**x**,

_{1}**x**,

_{2}**x**, ...,

_{3}**x**) also

_{n}### Proof[edit | edit source]

**x**, **x _{1}**,

**x**,

_{2}**x**, ...,

_{3}**x**all belong to S(

_{n}**x**,

_{1}**x**,

_{2}**x**, ...,

_{3}**x**), then S(

_{n}**x**,

**x**,

_{1}**x**,

_{2}**x**, ...,

_{3}**x**) must be contained within S(

_{n}**x**,

**x**,

_{1}**x**,

_{2}**x**, ...,

_{3}**x**). Therefore, if x is linearly dependent upon

_{n}**x**,

_{1}**x**,

_{2}**x**, ...,

_{3}**x**, then S(

_{n}**x**,

**x**,

_{1}**x**,

_{2}**x**, ...,

_{3}**x**) is equal to S(

_{n}**x**,

_{1}**x**,

_{2}**x**, ...,

_{3}**x**).

_{n}## Theorem[edit | edit source]

The maximum number of linearly independent vectors of a set of vectors is equal to the dimension of the span of the set.

### Proof[edit | edit source]

Suppose that there are *d* linearly independent vectors among **x _{1}**,

**x**,

_{2}**x**, ...,

_{3}**x**with all other vectors being a linear combination of those

_{n}*d*linearly independent vectors. This number

*d*is the maximum number of linearly independent vectors among

**x**,

_{1}**x**,

_{2}**x**, ...,

_{3}**x**. Then any element of S(

_{n}**x**,

_{1}**x**,

_{2}**x**, ...,

_{3}**x**) must be a linear combination of those

_{n}*d*linearly independent vectors, so they form a basis, and so

*d*is the dimension of S(

**x**,

_{1}**x**,

_{2}**x**, ...,

_{3}**x**), which is equal to the maximum number of linearly independent vectors among

_{n}**x**,

_{1}**x**,

_{2}**x**, ...,

_{3}**x**.

_{n}