# Linear Span / Generator of Vector Space

Linear Algebra / Saturday, October 27th, 2018
(Last Updated On: November 8, 2018)

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 α1, α2, …, αn ∈ V and c1, c2, …, cn ∈ F, then

${{c}_{1}}{{\alpha }_{1}}+{{c}_{2}}{{\alpha }_{2}}+…+{{c}_{n}}{{\alpha }_{n}}$

is called a linear combination of the vectors α1, α2, …, αn

## 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 = c1 α1 + c2 α2 + … + cn αn  αi ∈ S, ci ∈ F  i=1, 2, …, n

and β = d1 β1 + d2 β 2 + … + dm β m  β i ∈ S, di ∈ 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 = c1 α1 + c2 α2 + … + cn αn + d1 β1 + d2 β 2 + … + dm β m

Clearly, α + β is a finite linear combination of the vectors α1, α2, …, αn, β 1, β 2, …, β m ∈ S, and so α + β ∈ L(S)

Again, cα = cc1 α1 + cc2 α2 + … + ccn αn  = c’1 α1 + c’2 α2 + … + c’n αn, c’i = cci

Since c’i = cci ∈ 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 = x1 α1 + x2 α2 + … + xn αn for some xi (i=1, 2, …, n) ∈ F.

Since W is a subspace of V containing αi and xi, xi αi ∈ W for i=1, 2, …, n

Since W is a subspace and xi αi ∈ W, follows that

x1 α1 + x2 α2 + … + xn αn ∈ 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 c1, c2, c3 such that

(2, -1, -8) = c1 (1, 2, 1) + c2 (1, 1, -1) + c3 (4, 5, -2)

i.e., if (2, -1, -8) = (c1 + c2 + 4c3, 2c1 + c2 +5c3, c1 – c2 – 2c3)

This will be true if

c1 + c2 + 4c3 = 2

2c1 + c2 +5c3 = -1

c1 – c2 – 2c3 = -8

Solving we get, c1 = -14/3, c2 = 0, c3 = 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 R3 is generated by (0, 1, 1) and (0, 2, -1)

To show this we have to find real numbers c1 and c2 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 c1 and c2 such that

(0, y, z) = c1 (0, 1, 1) + c2 (0, 2, -1)

This is true if

y = c1 + 2c2

and z = c1 – c2

⇒ c1 = (y + 2z)/3 and c2 = (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 V3(R) is a vector space over R and W1 = {{a, b,, 0) | a, b ∈ R} and W2 = {(0, 0, c) | c ∈ R} are two subspaces of V3(R), then V3(R) = W1 ⊕ W2.

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