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Vector spaces

A vector space is a set of mathematical objects that we would like to call "vectors." These "vectors" can be added to each other, and they can be multiplied with "numbers."

To specify a vector space we therefore have to say

Fields—what can we multiply a vector with?

In any theory of vectors we would define 2v2v for any vector vv by saying 2v=v+v2v = v+v. But would we have to define 2 v\sqrt{2}\, v, or πv\pi v, or (3−i3)v(3-i\sqrt3)v?

What kind of numbers do we allow in our theory of vectors?

The full answer to this question is that we get to choose what we call a number. The only restriction is that the set of numbers we use should form a "field" F\mathbb{F}, a mathematical concept which is itself defined in terms of axioms that you can find in Appendix C of the textbook.

In this course we will ignore the field axioms and avoid this level of generality. Instead we will always assume the number field F\mathbb F is one of the following three examples:

Note: the notation R\R, Q\mathbb{Q}, C\mathbb{C} has been completely standard since the 1980ies, but our textbook was written in 1979 and writes RR, Q{Q}, and CC instead.

Another note: number and scalar mean the same thing for Friedberg-Insel-Spence.

An example of a vectorspace: Fn\mathbb F^n

Fn\mathbb{F}^n consists of all nn-tuples (x1,…,xn)(x_1, \dots, x_n) where x1,…,xn∈Fx_1, \dots, x_n\in\mathbb F. Once we start using matrices it turns out be more convenient to change notation, and write the components of (x1,…,xn)(x_1, \dots, x_n) in a column:

(x1,…,xn)=(x1⋮xn).(x_1, \dots, x_n) = \begin{pmatrix} x_1 \\ \vdots \\ x_n \end{pmatrix}.

Addition and scalar multiplication are defined by

(x1⋮xn)+(y1⋮yn)=(x1+y1⋮xn+yn),a(x1⋮xn)=(a x1⋮a xn).\begin{pmatrix} x_1 \\ \vdots \\ x_n \end{pmatrix} +\begin{pmatrix} y_1 \\ \vdots \\ y_n \end{pmatrix} =\begin{pmatrix} x_1+y_1 \\ \vdots \\ x_n+y_n \end{pmatrix},\qquad a\begin{pmatrix} x_1 \\ \vdots \\ x_n \end{pmatrix} =\begin{pmatrix} a\,x_1 \\ \vdots \\ a\,x_n \end{pmatrix}.

The standard basis vectors are

e1=(10⋮0),e2=(01⋮0),…en=(00⋮1).\mathbf{e}_1=\begin{pmatrix} 1 \\0 \\ \vdots\\0 \end{pmatrix},\quad \mathbf{e}_2=\begin{pmatrix} 0 \\1 \\ \vdots\\0 \end{pmatrix},\quad\dots\quad \mathbf{e}_n=\begin{pmatrix} 0\\0 \\ \vdots\\1 \end{pmatrix}.

Every vector x=(x1,…,xn)x=(x_1, \dots, x_n) can be written as a linear combination of the standard basis vectors:

(x1,…,xn)=(x1⋮xn)=x1e1+x2e2+⋯+xnen(x_1, \dots, x_n) = \begin{pmatrix} x_1 \\ \vdots \\ x_n \end{pmatrix} =x_1\mathbf{e}_1 + x_2\mathbf{e}_2 + \cdots + x_n\mathbf{e}_n

For example, if

x=(32227),y=(32π),z=(1i),x=\begin{pmatrix} \frac32 \\[6pt] \frac{22}{7} \end{pmatrix},\quad y=\begin{pmatrix} \frac32 \\[6pt] \pi \end{pmatrix},\quad z=\begin{pmatrix} 1 \\ i \end{pmatrix},\quad

then

x∈Q2,x∈R2,x∈C2,y∈C2,y∈R2,y∉Q2,z∈C2,z∉R2,z∉Q2.x\in \mathbb{Q}^2, x\in\R^2, x\in\mathbb{C}^2,\qquad y\in\mathbb{C}^2, y\in\R^2, y\not\in\mathbb{Q}^2,\qquad z\in\mathbb{C}^2, z\not\in\R^2, z\not\in\mathbb{Q}^2.

Vector Space Axioms

The axioms themselves are fairly standard, but the numbering differs from book to book. In our text book the authors wrote the axioms like this:

Some first proofs

The following facts are consequences of the axioms (VS1–8):

More examples

Linear subspaces

Terminology: a linear subspace is the same thing as a subspace

Instead of using the definition in the book we will take the following equivalent definition of a (linear) subspace WW of a vectorspace VV

Definition. If VV is a vector space over some field F\mathbb{F} then a subset W⊂VW\subset V is called a subspace if

Properties of linear subspaces

If WW is a linear subspace of the vector space VV, and F\mathbb F is the number field, then:

  1. 0V∈W0_V\in W
  2. for each x∈Wx\in W one has −x∈W-x\in W
  3. WW with the addition and scalar multiplication from VV is a vector space over F\mathbb{F}

Examples of linear subspaces

Solutions to a linear differential equation

Let V=F(R,R)V=\mathcal{F}(\R,\R) be the set of all functions f:R→Rf:\R\to\R. Then (V,R)(V,\R) is a vector space.

Let W⊂VW\subset V be the set of functions f:R→Rf:\R\to\R that satisfy the differential equation

f′′(x)−4f(x)=0 for all x∈Rf''(x)−4f(x)=0 \text{ for all $x\in\R$}

Problem: Prove that WW a linear subspace of VV.

Solution

We have to check three things:

  1. WW is not empty. This is true because the zero function, defined by f(x)=0f(x)=0 for all x∈Rx\in\R satisfies the differential equation, and therefore belongs to WW.

  2. WW is closed under addition: if f∈Wf\in W and g∈Wg\in W then consider the sum function h=f+gh=f+g. By definition it satisfies h(x)=f(x)+g(x)h(x)=f(x)+g(x) for all x∈Rx\in\R. By the differentiation rules from calculus we have

    h′(x)=f′(x)+g′(x) and h′′(x)=f′′(x)+g′′(x).h'(x)=f'(x)+g'(x) \text{ and } h''(x) = f''(x)+g''(x).

    Substituting this in the differential equation and rearranging terms we get

    h′′(x)−4h(x)=f′′(x)+g′′(x)−4(f(x)+g(x))=f′′(x)−4f(x)+g′′(x)−4g(x)\begin{array}{rcl} h''(x)-4h(x) &=& f''(x)+g''(x) - 4(f(x)+g(x)) \\[4pt] &=& f''(x)-4f(x) + g''(x)-4g(x) \end{array}

    Since both ff and gg satisfy the differential equation we have

    f′′(x)−4f(x)=0 and g′′(x)−4g(x)=0 for all x∈R.f''(x)-4f(x)=0 \text{ and } g''(x)-4g(x)=0 \text{ for all }x\in\R.

    Therefore we get h′′−4h(x)=0h''-4h(x)=0 for all x∈Rx\in\R.
    Hence h∈Wh\in W and we have shown that WW is closed under addition.

  3. WW is closed under scalar multiplication. We have to show that for any f∈Wf\in W and any a∈Ra\in\R the function h(x)=af(x)h(x) = a f(x) also belongs to WW.

Let U⊂VU\subset V be the set of functions f:R→Rf:\R\to\R that satisfy the differential equation

f′′(x)−4f(x)=16 for all x∈Rf''(x)−4f(x)=16 \text{ for all $x\in\R$}

Question: is UU a linear subspace of VV?

Theorem (1.4). If W1,…,Wn⊂VW_1, \dots, W_n\subset V are linear subspaces, then W=W1∩W2∩⋯∩WnW = W_1\cap W_2\cap \cdots \cap W_n is also a linear subspace.

Linear combinations

Definition. A linear combination of vectors v1,v2,…,vn∈Vv_1, v_2, \dots, v_n\in V is any vector of the form a1v1+⋯+anvna_1v_1+\cdots+a_nv_n, for any choice of numbers a1,…,an∈Fa_1, \dots, a_n\in\mathbb F.

Definition. The span of a set of vectors S⊂VS\subset V is the set of all linear combinations of vectors in SS.

span(S)={a1v1+⋯+anvn∣a1,…,an∈F,v1,…,vn∈V,n≥0}\mathrm{span}(S) = \left\{a_1v_1+\cdots+a_nv_n \mid a_1, \dots, a_n\in\mathbb F, v_1, \dots, v_n\in V, n\geq 0\right\}

Theorem. For any subset S⊂VS\subset V of a vector space VV, span(S)\mathrm{span}(S) is a linear subspace of VV.

Theorem. If W⊂VW\subset V is a linear subspace of VV and if S⊂WS\subset W, then span(S)⊂W\mathrm{span}(S)\subset W.

Definition. If S⊂VS\subset V and if span(S)=V\mathrm{span}(S)=V then "SS spans VV", or, "SS generates VV."

Solving linear systems of equations

The book describes the standard method of solving a system of nn equations with mm unknowns.

Linear independence, Bases, and Dimension

Definition of independence. A set of vectors {u1,…,un}⊂V\{u_1, \dots, u_n\}\subset V is linearly independent if for any a1,…,an∈Fa_1,\dots, a_n\in\mathbb{F} one has

a1u1+⋯+anun=0  ⟹  a1=a2=⋯=an=0.a_1u_1+\cdots+a_nu_n = 0 \implies a_1=a_2=\cdots=a_n=0.

More generally a set of vectors β⊂V\beta\subset V is linearly independent if every finite subset {u1,…,un}⊂β\{u_1, \dots, u_n\}\subset \beta is linearly independent. This second definition allows for the possibility that the set β\beta is infinite.

Definition of basis. A set of vectors {u1,…,un}⊂V\{u_1, \dots, u_n\}\subset V is a basis for VV if

Extension Theorem for Independent Sets. If {u1,…,un}⊂V\{u_1, \dots, u_n\}\subset V is linearly independent, and v∈Vv\in V is not one the vectors u1,…,unu_1, \dots, u_n, then v∈span(u1,…,un)v\in\mathrm{span}(u_1, \dots, u_n) if and only if {u1,…,un,v}\{u_1, \dots, u_n, v\} is dependent.

Proof

First we prove: v∈span(u1,…,un)  ⟹  {u1,…,un,v}v\in \mathrm{span}(u_1, \dots, u_n) \implies \{u_1, \dots, u_n, v\} is linearly dependent.

If v∈span(u1,…,un)v\in \mathrm{span}(u_1, \dots, u_n) then there are a1,…,an∈Fa_1, \dots, a_n\in\mathbb{F} with v=a1u1+⋯+anunv=a_1u_1+\cdots+a_nu_n. Therefore

−a1u1−⋯−anun+v=0,-a_1u_1-\cdots - a_nu_n + v = 0,

so we have a nontrivial linear combination of {u1,…,un,v}\{u_1, \dots, u_n, v\} that adds up to zero. Hence {u1,…,un,v}\{u_1, \dots, u_n, v\} is linearly dependent.

Conversely, we now show: {u1,…,un,v}\{u_1, \dots, u_n, v\} is linearly dependent   ⟹  v∈span(u1,…,un)\implies v\in \mathrm{span}(u_1, \dots, u_n).

Suppose {u1,…,un,v}\{u_1, \dots, u_n, v\} is linearly dependent. Then there exist a1,…,an,b∈Fa_1, \dots, a_n, b\in\mathbb{F} such that

a1u1+⋯+anun+bv=0,a_1u_1+\cdots + a_nu_n+bv=0,

and such that at least one of the coefficients a1,…,an,ba_1, \dots, a_n, b is nonzero.

If b=0b=0 then we have a1u1+⋯+anun=0a_1u_1+\cdots + a_nu_n=0. Since {u1,…,un}\{u_1, \dots, u_n\} is independent, this implies a1=⋯=an=0a_1=\cdots=a_n=0, which is impossible because at least one of a1,…,an,ba_1, \dots, a_n, b does not vanish.

Therefore b≠0b\neq0. This implies that

v=−a1bu1−⋯−anbun∈span(u1,…,un).v = -\frac{a_1}{b}u_1-\cdots-\frac{a_n}{b}u_n\in\mathrm{span}(u_1, \dots, u_n).

Dimension Theorem. If {v1,…,vm}⊂span(u1,…,un)\{v_1, \dots, v_m\}\subset \mathrm{span}( u_1, \dots, u_n) and if {v1,…,vm}\{v_1, \dots, v_m\} is linearly independent then m≤nm\leq n.

Proof of the dimension theorem

We will show that if m>nm\gt n and {v1,…,vm}⊂span({u1,…,un})\{v_1, \dots, v_m\}\subset\mathrm{span}\left(\{u_1, \dots, u_n\}\right) then {v1,…,vm}\{v_1, \dots, v_m\} is linearly dependent.

To do this we use mathematical induction on nn.

We begin with the case n=1n=1. There is only one vector u1u_1, and each v1,v2,…,vmv_1, v_2, \dots, v_m is a linear combination of u1u_1, i.e. a multiple of u1u_1. Thus for certain numbers a1,…,am∈Fa_1, \dots, a_m\in\mathbb F we have

v1=a1u1,v2=a2u1,…,vm=amu1.v_1 = a_1u_1, \quad v_2 = a_2u_1, \quad \dots, \quad v_m = a_mu_1.

If a1=0a_1=0 then v1=0v_1=0 and {v1,…,vm}\{v_1, \dots, v_m\} is dependent.

If a1≠0a_1\neq 0 then we have

−a2v1+a1v2+0⋅v3+⋯+0⋅vm=0.-a_2v_1+a_1v_2 + 0\cdot v_3 + \cdots + 0\cdot v_m = 0.

Since a1≠0a_1\neq0 this is a nontrivial linear combination of v1,…,vmv_1, \dots, v_m that adds up to zero. Hence {v1,…,vm}\{v_1, \dots, v_m\} is dependent.

Next we consider the general case n>1n\gt 1, and we assume that the case n−1n-1 has already been proven.

In this case each viv_i is a linear combination of the vectors u1,…,unu_1, \dots, u_n. So we have

v1=a11u1+⋯+a1nunv2=a21u1+⋯+a2nun    ⋮vm=am1u1+⋯+amnun\begin{aligned} v_1 &= a_{11}u_1 + \cdots + a_{1n}u_n \\ v_2 &= a_{21}u_1 + \cdots + a_{2n}u_n \\ &\;\;\vdots \\ v_m &= a_{m1}u_1 + \cdots + a_{mn}u_n \\ \end{aligned}

for certain numbers a11,…,amn∈Fa_{11}, \dots, a_{mn}\in\mathbb F.

If all the coefficients a1n,a2n,…,amn=0a_{1n}, a_{2n}, \dots, a_{mn}=0 then the viv_i are linear combinations of u1,…,un−1u_1, \dots, u_{n-1}. Since m>nm\gt n we have m>n−1m\gt n-1 and therefore the induction hypothesis tells us that v1,…,vmv_1, \dots, v_m are linearly dependent.

We are left with the case in which one of the coefficients a1n,…,amna_{1n}, \dots, a_{mn} does not vanish. Assume that a1n≠0a_{1n}\neq 0. Then we consider the vectors

w2=v2−a2na1nv1,…wm=vm−amna1nv1.w_2 = v_2-\frac{a_{2n}}{a_{1n}}v_1,\quad\dots\quad w_m = v_m-\frac{a_{mn}}{a_{1n}}v_1.

The vectors w2,…,wmw_2, \dots, w_m are linear combinations of u1,…,un−1u_1, \dots, u_{n-1}. Since m−1>n−1m-1>n-1 the induction hypothesis applies. We therefore know that w2,…,wmw_2, \dots, w_m are linearly dependent, i.e. there exist c2,…,cm∈Fc_2, \dots, c_m\in\mathbb F such that

c2w2+⋯+cmwm=0,c_2w_2+\cdots+c_mw_m = 0,

and such that at least one of c2,…,cmc_2, \dots, c_m does not vanish. By substituting the definition of the wiw_i in this linear combination, we find after some simplification that

c1v1+c2v2+⋯+cmvm=0,c_1v_1+c_2v_2+\cdots+c_mv_m = 0,

where

c1=−a2na1nc2−⋯−amna1ncm.c_1 = - \frac{a_{2n}}{a_{1n}}c_2 - \cdots - \frac{a_{mn}}{a_{1n}}c_m.

Thus v1,…,vmv_1, \dots, v_m is linearly dependent.

Corollaries to the Dimension Theorem.

  1. If {u1,…,un}\{u_1, \dots, u_n\} and {v1,…,vm}\{v_1, \dots, v_m\} both are bases of a vector space VV, then m=nm=n.

  2. If L⊂VL\subset V is a linear subspace and VV is finite dimensional then dim⁡L≤dim⁡V\dim L\leq \dim V. If dim⁡L=dim⁡V\dim L=\dim V then L=VL=V.

  3. If L⊂VL\subset V is a linear subspace and VV is finite dimensional, and if {v1,…,vk}⊂L\{v_1, \dots, v_k\}\subset L is a basis for LL, then there exist vectors vk+1,…,vn∈Vv_{k+1}, \dots, v_n\in V such that {v1,…,vk,vk+1,…,vn}\{v_1, \dots, v_k, v_{k+1}, \dots, v_n\} is a basis for VV.

Proof

This is true because {v1,…,vm}⊂span(u1,…,un)\{v_1, \dots, v_m\}\subset \mathrm{span}(u_1, \dots, u_n) and {v1,…,vm}\{v_1, \dots, v_m\} is linearly independent, so the Dimension Theorem implies m≤nm\leq n. On the other hand, {u1,…,un}⊂span(v1,…,vn)\{u_1, \dots, u_n\}\subset \mathrm{span}(v_1, \dots, v_n) and {u1,…,un}\{u_1, \dots, u_n\} is linearly independent, so the Dimension Theorem implies n≤mn\leq m

Definition.

The Dimension Theorem and its corollary imply that the dimension as defined above does not depend on which basis of VV you consider.

Components of a vector with respect to a basis

If {u1,…,un}\{u_1, \dots, u_n\} is a basis for a vector space VV and x∈Vx\in V is any other given vector in VV, then there exist numbers x1,…,xn∈Fx_1, \dots, x_n\in \mathbb{F} such that

x=x1u1+⋯+xnun.x = x_1u_1+\cdots+x_nu_n.

The numbers x1,…,xnx_1, \dots, x_n are called the components, or coefficients, of the vector xx with respect to the basis {u1,…,un}\{u_1, \dots, u_n\}.

The coefficients x1,…,xnx_1, \dots, x_n are completely determined by the vector xx and the basis {u1,…,un}\{u_1, \dots, u_n\}. Namely, if

x1′u1+⋯+xn′un=x1u1+⋯+xnunx_1'u_1+\cdots+x_n'u_n = x_1u_1+\cdots+x_nu_n

then we could subtract x1u1x_1u_1, …, xnunx_nu_n from both sides and we would end up with

(x1′−x1)u1+⋯+(xn′−xn)un=0.(x_1'-x_1)u_1+\cdots+(x_n'-x_n)u_n = 0.

Since u1,…,unu_1, \dots, u_n are independent, this implies x1=x1′x_1=x_1', … , xn=xn′x_n=x_n'.

Examples—bases for R2\R^2

Examples—dimension depends on the number field F\mathbb{F}

Consider C\mathbb{C} as vector space over the complex numbers. It has a basis with one “vector,” namely {1}\{1\}.

Now consider C\mathbb{C} as vector space over the real numbers. Then the set {1,i}\{1, i\} is linearly independent, and spans C\mathbb{C}, i.e. {1,i}\{1,i\} is a basis for C\mathbb{C} over the real numbers.

Consider V=RV=\R as vector space over the rational numbers Q\mathbb{Q}. Then {1,2}\{1, \sqrt2\} is linearly independent.

Examples—bases for spaces of polynomials

{1,x,x2,x3,… }\{1, x, x^2, x^3, \dots\} is a basis for P(R)\mathcal{P}(\R). P(R)\mathcal{P}(\R) is infinite dimensional.

{1,x,x2}\{1, x, x^2\} is a basis for P2(R)\mathcal{P}_2(\R)

If we define

p0(x)=(x−1)(x−2)(−1)(−2),p1(x)=x(x−2)1⋅(−1),p2(x)=x(x−1)2⋅1p_0(x) = \frac{(x-1)(x-2)}{(-1)(-2)},\qquad p_1(x) =\frac{x(x-2)}{1\cdot(-1)},\qquad p_2(x) =\frac{x(x-1)}{2\cdot1}

then {p0(x),p1(x),p2(x)}\{p_0(x), p_1(x), p_2(x)\} is a basis for P2(R)\mathcal{P}_2(\R)

Given a quadratic polynomial f(x)∈P2(R)f(x)\in \mathcal{P}_2(\R) what are the coefficients in

f(x)=c0p0(x)+c1p1(x)+c2p2(x)??f(x) = c_0p_0(x)+c_1p_1(x)+c_2p_2(x) \quad ??