Before discussing about linear Span we have to learn Linear Combination of vectors.

# Linear Combination Definition

Let V be vector space over a field F. Let α_{1}, α_{2}, …, α_{n} ∈ V and c_{1}, c_{2}, …, c_{n} ∈ F, then

\[{{c}_{1}}{{\alpha }_{1}}+{{c}_{2}}{{\alpha }_{2}}+…+{{c}_{n}}{{\alpha }_{n}}\]

is called a linear combination of the vectors α_{1}, α_{2}, …, α_{n}

## Linear Span or Generator

Let S ⊆ V. The linear span of a subset of S of the vector space V, denoted by L(S), is the set of all linear combinations of the vectors in S.

In other words, if S is a subset of V, then the linear span of S is the set

\[\left\{ {{c}_{1}}{{\alpha }_{1}}+{{c}_{2}}{{\alpha }_{2}}+…+{{c}_{n}}{{\alpha }_{n}}~~|~~{{\text{c}}_{\text{1}}},{{\text{c}}_{\text{2}}},\text{ }\ldots ,{{\text{c}}_{\text{n}}}\in \text{ F}~~and~~{{\alpha }_{\text{1}}},{{\alpha }_{\text{2}}},\text{ }\ldots ,{{\alpha }_{\text{n}}}\in \text{ S} \right\}\]

Theorem |

**Let S be a non-empty subset of a vector space V. Then the linear span L(S) of S is a subspace of V and it is the smallest subspace containing the set S.**

**Proof:**

Let α = c_{1} α_{1} + c_{2} α_{2} + … + c_{n} α_{n} α_{i} ∈ S, c_{i} ∈ F i=1, 2, …, n

and β = d_{1} β_{1} + d_{2} β_{ 2} + … + d_{m} β_{ m} β_{ i} ∈ S, d_{i} ∈ F j=1, 2, …, m

be any two elements of L(S), to prove the theorem we have to prove that α + β ∈ L(S) and for any c ∈ F, cα ∈ L(S)

We have, α + β = c_{1} α_{1} + c_{2} α_{2} + … + c_{n} α_{n} + d_{1} β_{1} + d_{2} β_{ 2} + … + d_{m} β_{ m}

Clearly, α + β is a finite linear combination of the vectors α_{1}, α_{2}, …, α_{n}, β_{ 1}, β_{ 2}, …, β_{ m} ∈ S, and so α + β ∈ L(S)

Again, cα = cc_{1} α_{1} + cc_{2} α_{2} + … + cc_{n} α_{n } = c’_{1} α_{1} + c’_{2} α_{2} + … + c’_{n} α_{n}, c’_{i} = cc_{i}

Since c’_{i} = cc_{i} ∈ F for i=1, 2, …, n, it follows that cα ∈ L(S).

Hence L(S) is a subspace of V.

To show that L(S) is the smallest subspace containing S, we consider any other subspace W of V containing S and we shall show that L(S) ⊆ W

Let ξ ∈ L(S). Then ξ = x_{1} α_{1} + x_{2} α_{2} + … + x_{n} α_{n} for some x_{i} (i=1, 2, …, n) ∈ F.

Since W is a subspace of V containing α_{i} and x_{i}, x_{i} α_{i} ∈ W for i=1, 2, …, n

Since W is a subspace and x_{i} α_{i} ∈ W, follows that

x_{1} α_{1} + x_{2} α_{2} + … + x_{n} α_{n} ∈ W

or, ξ ∈ W

Thus ξ ∈ L(S) ⇒ ξ ∈ W

Hence L(S) ⊆ W.

In other words, L(S) is the smallest subspace of V containing S.

The subspace L(S) is called the subspace spanned or generated by S.

Example 01 |

**Let S = {(1, 2, 1), (1, 1, -1), (4, 5, -2)}. Examine if the vector (2, -1, -8) is in L(S).**

The vector (2, -1, -8) ∈ L(S), if there exists scalars c_{1}, c_{2}, c_{3} such that

(2, -1, -8) = c_{1} (1, 2, 1) + c_{2} (1, 1, -1) + c_{3} (4, 5, -2)

i.e., if (2, -1, -8) = (c_{1} + c_{2} + 4c_{3}, 2c_{1 }+ c_{2} +5c_{3}, c_{1} – c_{2} – 2c_{3})

This will be true if

c_{1} + c_{2} + 4c_{3} = 2

2c_{1 }+ c_{2} +5c_{3} = -1

c_{1} – c_{2} – 2c_{3} = -8

Solving we get, c_{1} = -14/3, c_{2} = 0, c_{3} = 5/3

This shows that the vector (2, -1, -8) is a linear combination of the vectors (1, 2, 1), (1, 1, -1), (4, 5, -2) such that

(2, -1, -8) = -14/3 (1, 2, 1) + 0 (1, 1, -1) + 5/3 (4, 5, -2)

Hence the vector (2, -1, -8) ∈ L(S).

Example 02 |

**Show that the yz plane W = {(0, y, z)} in R ^{3} is generated by (0, 1, 1) and (0, 2, -1)**

To show this we have to find real numbers c_{1 }and c_{2} such that, any vector α = (0, y, z) ∈ W can be expressed as a linear combination of (0, 1, 1) and (0, 2, -1). That is to find c_{1 }and c_{2} such that

(0, y, z) = c_{1} (0, 1, 1) + c_{2} (0, 2, -1)

This is true if

y = c_{1} + 2c_{2}

and z = c_{1} – c_{2}

⇒ c_{1} = (y + 2z)/3 and c_{2} = (y – z)/3

Thus for any given vector α = (0, y, z) can be expressed as a linear combination of vectors (0, 1, 1) and (0, 2, -1). Hence W is generated by (0, 1, 1) and (0, 2, -1).

### Definition

Let U and W be subspaces of a vector space V. The sum of U and W, written as U + W is defined as follows

\[U+W=\left\{ u+w|u\in U,w\in W \right\}\]

Theorem |

**The sum U + W of the subspaces U and W of V, is a subspace of V.**

**Proof:**

Since 0 ∈U and 0 ∈ W, 0 = 0 + 0 ∈ U + W

Further, suppose that u + w ∈ U + W and u’ + w’ ∈ U + W with u, u’ ∈ U and w, w’ ∈ W.

Then (u + w) + (u’ + w’ ) = (u + u’) + (w + w’) ∈ U + W and for any scalar c, c(u + w) = cu + cw

Since cu ∈ U and cw ∈ W, we have c(u + w) ∈ U + W.

Hence U + W is a subspace of V.

### Definition

The vector space V is said to be the direct sum of its subspaces U and W, denoted by

\[V=U\oplus W\]

if every v ∈ V can be written in one and only one way say v = u + w where u ∈ U and w ∈ W.

Example |

**If V ^{3}(R) is a vector space over R and W_{1} = {{a, b,, 0) | a, b ∈ R} and W_{2} = {(0, 0, c) | c ∈ R} are two subspaces of V^{3}(R), then V^{3}(R) = W_{1} ⊕ W_{2}.**