Vector Spaces: Theorems Part 7

Vector Spaces

Theorems Part 7

Theorem 18

Let V be a vector space and B = { u₁, u₂, ... u_n} be a subset of V. Then B is a basis for V if each v ∈ V can be uniquely expressed as a linear combination of vectors of B. (i.e) It can be expressed in the form,

 v = а₁u₁ + а₂u₂ +...+ a_nu_n, for unique scalars a₁, a₂, ... a_n.

Proof:

Assume that B is a basis.

If v ∈ V then v ∈ L(B)           (`.`L (B) = V)

 v = a₁u₁ + a₂u₂+ ….. + a_nu_n & a_i∈F.

(ie) v is a linear combination of vectors of B.

Suppose that there exists another representation of v,

 v = b₁u₁ + b₂u₂+ ….. + b_nu_n  b_i∈F.

Subtracting the second equation from the first equation of v, we get

  0 = v‒v

 0 = (a₁ − b₁) u₁ + (a₂ − b₂) u₂ + ... + (a_n − b_n) u_n.

Since B is linearly independent, it follows that

 a₁−b₁ = a₂−b₂ = ….. = a_n−b_n = 0

Hence a₁ =b₁, a₂ = b₂, ..., a_n = b_n.

  v is uniquely expressible as a linear combination of vectors of B.

Conversely,

Assume each vector v ∈ V can be uniquely expresses as a linear combination of B

  v = a₁u₁ + а₂u₂ + ... + a_nu_n

To prove: B is a Basis

Let α₁u₁ +α₂u₂ + ... + α_nu_n = 0 also

 0 = 0u₁ + 0u₂ + ... + 0u_n

 α₁u₁ + α₂u₂ + …. + α_nu_n = 0u₁ + 0u₂ + ... + 0u_n

 α₁ = α₂ = ... α_n = 0

{u₁, u₂,...u_n} is linearly independent.

 L(B) = V.

Clearly L(B) is a basis of V.

Theorem 19

If a vector space V is generated by a finite set S, then some subset of S is a basis for V. Hence V has a finite basis.

Proof:

If S= ϕ (or) S={0}, then V={0} & ϕ is a subset of S that is a basis for V.

Otherwise, S contains a non−zero vector u₁.

B={u₁} is a linearly independent set.

Continue, if possible, choosing vectors u₂, u₃,…..u_k in S.

such that { u₁, u₂, u₃,…..u_k } is linearly independent.

Since S is a finite set we must eventually reach a stage at which B = { u₁, u₂, ... u_k} is a linearly independent subset of S,

but adjoining to B any vector in S not in B produces a linearly dependent set.

We claim that B is a basis for V,

because B is linearly independent by construction.

To prove: L(B) = V.

It is enough to prove that S  span (B).

Let v ∈ S. If v ∈ B, then clearly v ∈ span (B) otherwise if v ∉ B then B ∪ {v} is linearly dependent.

So v ∈ span (B)

 S  span (B).

Replacement Theorem

Theorem 20

Let V be a vector space that is generating by a set G containing exactly n vectors, and let L be a linearly independent subset of V containing exactly m vectors. Then m≤n and there exists a subset H of G, containing exactly n−m vectors such that L∪H generates V.

Proof:

The proof is by mathematical induction on m. The induction begins with m = 0; for in this case L= ϕ and so taking H=G gives the desired result.

Suppose that the theorem is true for some integer m ≥0. We prove that the theorem is true for m+ 1. Let L= { v₁, v₂, v₃ ... v_m+1 } be a linearly independent subset of V consisting of of m+1 vectors. {v₁, v₂, v₃... v_m} is linearly independent and apply the induction hypothesis to conclude that m≤n there is a subset {u₁, u₂, u₃... u_n-m} at G such that {v₁, v₂, v₃... v_m} ∪ {u₁, u₂, u₃... u_n-m} generates V. There exists scalars a₁, a₂, a₃ ....a_m, b₁, b₂, b₃ ....b_n-m, such that

a₁v₁ + a₂v₂ + a₃v₃ + ... + a_mv_m + b₁u₁ + b₂u₂ + ... + b_n−mu_{n − m} = V_m+1 note that n−m> 0. Hence n >m, (ie) n ≥m +1.

Let H = { u₂, u₃... u_n-m }. Then u₁ ∈ span (L ∪ H) and because v₁, v₂, v₃... v_m, u₁, u₂, u₃... u_n-m ∈ span (L ∪ H)

 { v₁, v₂, v₃... v_m, u₁, u₂, u₃... u_n-m }  span (L ∪ H)

Because { v₁, v₂, v₃, ... v_m, u₁, u₂, u₃... u_n-m } generates V,

 span (L ∪ H) = V. Since H is a subset of G that contains (n − m) − 1 = n − (m+1) vectors, the theorem is true for m+1. This completes the induction.

Theorem 21

Let V be a vector space having a finite basis, then prove that every basis for V contains the same number of vectors.

Proof:

Suppose that B is a finite basis for V that contains exactly n vectors and let γ be any other basis for V. If γ contains more than n vectors, then we can select a subset S and γ containing exactly n+1 vectors. Since S is linearly independent and B generates V, the replacement theorem implies that n+1≤n a contradiction. Therefore, γ is finite, and the number of m of vectors in γ satisfies m≤n. Reversing at B and γ we obtain n≤m. Hence n = m.

Definition:

Finite Dimensional: A vector space is called finite dimensional if it has a basis consisting of a finite number of vectors. The unique number of vectors in each basis for V is called the dimension of V and is denoted by dim (V).

A vector space that is not finite−dimensional is called infinite−dimensional.

Examples:

1. The vector space {0} has dimension zero.

2. The vector space Fⁿ has dimension n.

3. The vector space M_m×n (F) has dimension mn.

4. The vector space P_n(F) has dimension n + 1.

5. Over the field of complex numbers, the vector space of complex numbers has dimension 1. (A basis is { 1 })

6. Over the field of real numbers, the vector space of complex numbers has dimension 2. {A basis is { 1, i } }

Theorem 22

Let W be a subspace of a finite dimensional vector space V. Then W is finite dimensional and dim (W) ≤ dim (V). Moreover if dim (W) = dim (V) then V=W.

Proof:

Let dim (V)= n.

If W={0}, then dim (W) = 0, then W is finite dimensional and dim(W) = 0 ≤ n ≤ dim(V) otherwise, W contains a non−zero vector x₁. So {x₁} is a linearly independent set.

Continue for choosing vectors x₁, x₂, ... x_k in W such that {x₁, x₂, ... x_k} is linearly independent.

Since no linearly independent subset of V can contain more than n vectors, this process must stop at a stage where k≤n and {x₁, x₂,…x_k } is linearly independent but adjoining any other vector from W produces a linearly independent set.

 {x₁, x₂,…x_k } generates W and hence it is a basis for W.

 dim (W) = k ≤ n = dim (V).

 dim (W) ≤ dim (V).

If dim (W) = n, then { x₁, x₂,..., x_n) is a basis of W which is linearly independent subset of V.

 It is also a basis for V.

  V=W

Maximal linearly independent subsets

Definition

Let S be a subset of a vector space V. A maximal linearly independent subset of S is a subset B of S satisfying both of the following conditions.

(i) B is linearly independent.

(ii) The only linearly independent subset of S that contains B is B itself.

Theorem 23

Let V be a vector space and S a subset that generates V. If B is a maximal linearly independent subset of S then B is a basic for V.

Proof:

Let B be a maximal linearly independent subset of S.

Since B is linearly independent it is enough to prove that B generates V.

Claim: S  span (B)

Clearly span (B)  V.

Suppose there exists a v ∈ S such that v ∉ span (B) then B ∪{v} is linearly independent,

which is contradiction to the maximality of B.

  v ϵ span (B)          ('. Given L(S) = V)

 Hence S  span (B).

L(S)  span (B).

  V  span (B).            (' . ' L(S) = V)

 V=L (B) and B is a basis.

Hence the proof.

Theorem 24

Let S be a linearly independent subset of a vector space V. There exists a maximal linearly independent subset of V that contains S.

Corollary: Every vector space has a basis.