Lecture 5: Span, linear independence, bases, and dimension

Lecture 5: Span, linear independence, bases, and dimension

Travis Schedler

Thurs, Sep 23, 2010 (version: 9/21 9:55 PM)

1 Motivation

Motivation

• To understand what it means that R has dimension one, R²dimension 2, etc.;

– More generally, what does it mean that Fⁿ and F{0,_n¹, . . . ,ⁿ⁻¹_n } have dimension n?

– What does it mean to be infinite-dimensional?

• To understand the relationship between vector spaces and special spanning lists.

Main goals: To understand span, linear independence, bases, and dimension.

Span

Fix a field F throughout. Recall from last time and the homework:

Definition 1. Let v1, . . . , vn ∈ V . Span(v1, . . . , vn) = {λ1v1+ · · · + λnvn | λ1, . . . , λn∈ F}.

Definition 2. A vector space V is finite-dimensional if V = Span(v₁, . . . , v_n) for some v1, . . . , vn ∈ V and some n ≥ 1.

Definition 3. The dimension dim(V ) of a finite-dimensional vector space is the minimum n such that V = Span(v₁, . . . , v_n).

Finite-dimensional examples

• Let V = R³. Then V = Span((1, 0, 0), (0, 1, 0), (0, 0, 1)). So dim(V ) ≤ 3.

• If V = Span(v1, . . . , vn), then also V = Span(v1, . . . , vn, 0). Also, for all v ∈ V , V = Span(v1, . . . , vn, v).

• We have Fⁿ= Span(δ1, . . . , δn) where δi = (0, 0, . . . , 0, 1, 0, 0, . . . , 0), with 1 in the i-th coordinate and 0 elsewhere. So, dim(Fⁿ) ≤ n.

• Similarly, F{0,_n¹, . . . ,ⁿ⁻¹_n } = Span(δ0, δ¹

n, . . . , δn−1

n ). So, dim(F{0,¹_n, . . . ,ⁿ⁻¹_n }) ≤ n.

Infinite-dimensional examples

• The vector space of polynomials P(F).

• Continuous functions [0, 1] → R.

• Infinite lists F^∞.

• F(X) where X is infinite.

• We can prove the first directly (x^d+1 does not appear in any finite list of polynomials of maximum degree d).

• Second one: homework problem (!)

• Last two: There are linearly independent lists of arbitrary length. We will show that, if V is finite-dimensional, then linearly independent lists are at most dim V in size.

Linear (in)dependence Recall from last time:

Definition 4. A linearly independent list of vectors (v1, . . . , vn) is one such that, if λ1v1+ · · · + λnvn = 0, then λ1= · · · = λn = 0.

• A list which is not linearly independent is linearly dependent. Then, λ1v1+

· · · + λnvn= 0 for some λi which are not all zero.

Examples:

• ((v1, . . . , vn, 0)) is always linearly dependent.

• For every vector space V and nonzero vector v ∈ V , (v) is linearly inde- pendent.

• In R², ((1, 0), (1, 1)) is linearly independent.

• In Fⁿ, the list ((1, 0, 0, . . . , 0), (0, 1, 0, . . . , 0), . . . , (0, 0, . . . , 0, 1)) is linearly independent. Similarly (δ0, δ1

n, . . . , δn−1

n ) in F{0,¹_n, . . . ,ⁿ⁻¹_n }.

Reducing spanning lists to linearly independent ones

Lemma 5 (Lemma 2.4 + more). Let V = Span(v1, . . . , vn). Then, (v1, . . . , vn) is linearly independent if and only if, for all i,

Span(v₁, . . . , v_i−1, v_i+1, . . . , v_n) ( Span(v1, . . . , v_n). (1.1) In other words, a spanning list is linearly independent if and only if it is minimal. Examples:

• (1, 0), (0, 1) is a minimal spanning list, hence linearly independent. Same is true for (1, 0), (1, 1), or any spanning list of two vectors in R².

• If the zero vector is in a spanning list, it is not minimal (hence linearly dependent).

Definition 6. A linearly independent spanning set is called a basis.

By the lemma, every spanning set can be reduced to a basis (Theorem 2.10).

Just discard vectors until (1.1) is true!

Maximal linearly independent lists

Lemma 7 (Part of Theorem 2.12). A linearly independent list (v₁, . . . , v_n) of vectors of V spans V if and only if it is maximal: (v₁, . . . , v_n, v) is not linearly independent for all v ∈ V .

Examples:

• (1, 0, 0), (1, 1, 0) is linearly independent, but not maximal.

• (1, 1), (−2, 1) is a maximal linearly independent list, hence it spans.

Consequence: Any linearly independent list which does not span can be ex- tended.

The fundamental inequality and bases

Theorem 8 (Theorem 2.6). The length of every linearly independent list is less than or equal to the length of every spanning list.

Corollary 9 (Theorem 2.14). Every basis has the same length.

Proof. Let (v₁, . . . , v_m) and (w₁, . . . , w_n) be bases.

• Since (v1, . . . , vm) is linearly independent and (w1, . . . , wn) spans, m ≤ n.

• Since (v1, . . . , vm) spans and (w1, . . . , wn) is linearly independent, m ≥ n.

Bases, continued

Corollary 10 (Corollary 2.11 + more). Every finite-dimensional vector space V has a basis, and every basis has length dim V .

Proof. Every spanning list of length dim V must be minimal, hence linearly independent, i.e., a basis. By the previous corollary, all bases have the same length.

Corollary 11 (Theorem 2.12). Every linearly independent list in a finite- dimensional vector space can be extended to a basis.

Proof. • Linearly independent lists in V of length < dim V don’t span, so can be extended.

• Linearly independent lists of length ≥ dim V must have length exactly dim V . Thus, they are maximal and span.

Examples

• For R², every pair of nonzero vectors, neither of which is a multiple of the other, is a basis.

• In Fⁿ, the delta vectors form a basis. Also any lin. ind. list of length n (or any spanning list of length n).

• In F{0,_n¹, . . . ,ⁿ⁻¹_n }, again, the delta functions form a basis.

• For F = C, another basis important for Fourier transform: {1, e^2πit, e^4πit, . . . , e^2(n−1)πit, where e^iy = cos(y) + i sin(y).

• Lots of bases important for MP3s, auto-tune, DVDs, etc!

The basis theorem

Theorem 12 (Propositions 2.8, 2.16, and 2.17). The following conditions for a list (v1, . . . , vn) in V are equivalent:

(a) It is a basis.

(b) It is linearly independent of length dim V . (c) It is spanning of length dim V .

(d) For all v ∈ V , there exist unique λ₁, . . . , λ_n such that v = λ₁v₁+· · ·+λ_nv_n. Proof. • By the corollary, (a) implies (b) and (c).

• (b) implies (a) since linearly independent sets of length dim V are maximal.

• Similarly, (c) implies (a) since spanning sets of length dim V are minimal.

• (a) implies (d): There exist λ1, . . . , λn since v1, . . . , vn span. They are

Dimension

Theorem 13 (Propositions 2.7 and 2.15). If V is finite-dimensional and U ⊆ V is a subspace, then U is finite-dimensional, and dim U ≤ dim V .

Proof. • Any linearly independent list in U is also linearly independent in V , so has length at most dim V .

• So, there is a maximal such list, which must be a basis.

• Since it has length at most dim V , we obtain that dim U ≤ dim V .