This is a living document. It will grow throughout the semester.
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
what the set of vectors V is
how to define the sum x+y of two vectors x,y∈V
what we mean by "numbers" (there is a choice)
how to define the product ax of a number a and a vector x .
Fields—what can we multiply a vector with?
In any theory of vectors we would define 2v for any vector v by saying 2v=v+v. But would we have to define 2v, or πv, or (3−i3)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, 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 is one of the following three examples:
the rational numbersQ: a rational number is a number of the form qp where p,q are integers, and where q=0. Two rational numbers p/q and m/n are equal if pn=mq.
the real numbersR: it's a long story. Take math 421 and/or 521 to get a fuller description of the real numbers. We'll assume that they are familiar from calculus and not worry about questions like “do infinitely small number exist?”.
or the complex numbersC: numbers of the form a+bi where a,b∈R and where i2=−1.
Note: the notation R, Q, C has been completely standard since the 1980ies, but our textbook was written in 1979 and writes R, Q, and C instead.
Another note: number and scalar mean the same thing for Friedberg-Insel-Spence.
An example of a vectorspace: Fn
Fn consists of all n-tuples (x1,…,xn) where x1,…,xn∈F. Once we start using matrices it turns out be more convenient to change notation, and write the components of (x1,…,xn) in a column:
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:
(VS1) For all x,y∈V one has x+y=y+x
(VS2) For all x,y,z∈V one has (x+y)+z=x+(y+z)
(VS3) There is a 0V∈V such that for all x∈V one has x+0V=x
(VS4) For each x∈V there is a y∈V such that x+y=0V
(VS5) For each x∈V one has 1x=x
(VS6, 7, 8) For all a,b∈F, and all x,y∈V one has
(ab)x=a(bx),a(x+y)=ax+ay,(a+b)x=ax+bx.
Some first proofs
The following facts are consequences of the axioms (VS1–8):
V=Rn and F=R satisfy the vector space axioms (VS1–8).
there is only one zero vector in V
For each x∈V there is only one additive inverse "−x"
The cancellation theorem (Theorem 1.1 in the book)
0Fx=0V for all x∈V.
a0V=0V for all a∈F
(−1)x=−x for all x∈V
More examples
Mm×n(F): all m×n matrices with entries from F
P(F): all polynomials with coefficients in F
Pn(F): all polynomials with coefficients in F of degree ≤n
T(F): all trigonometric polynomials with coefficients in F
Tn(F): all trigonometric polynomials with coefficients in F of degree ≤n
C as a real vector space
R as a vectorspace over Q
F([a,b]): all real valued functions defined on an interval [a,b]⊂R
C([a,b]): all continuous real valued functions defined on an interval [a,b]⊂R
C∞(a,b): all infinitely often differentiable real valued functions that are defined on the interval (a,b)⊂R.
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 W of a vectorspace V
Definition.If V is a vector space over some field F then a
subset W⊂V is called a subspace if
W is not empty, and
for all x,y∈W one has x+y∈W, and
for all x∈W and a∈F one has ax∈W.
Properties of linear subspaces
If W is a linear subspace of the vector space V, and F is the
number field, then:
0V∈W
for each x∈W one has −x∈W
W with the addition and scalar multiplication from V is a vector space over F
Examples of linear subspaces
a line through the origin—other lines?
the xy plane in R3
symmetric matrices ⊂Mn×n(F)
Pn(F) : polynomials of degree at most n
Solutions to a linear differential equation
Let V=F(R,R) be the set of all functions f:R→R. Then
(V,R) is a vector space.
Let W⊂V be the set of functions f:R→R that satisfy the
differential equation
f′′(x)−4f(x)=0 for all x∈R
Problem: Prove that W a linear subspace of V.
Solution
We have to check three things:
W is not empty. This is true because the zero function, defined by
f(x)=0 for all x∈R satisfies the differential equation, and therefore belongs to W.
W is closed under addition: if f∈W and g∈W then consider the sum
function h=f+g. By definition it satisfies h(x)=f(x)+g(x) for all
x∈R. By the differentiation rules from calculus we have
h′(x)=f′(x)+g′(x) and h′′(x)=f′′(x)+g′′(x).
Substituting this in the differential equation and rearranging terms we get
Theorem.For any subset S⊂V of a vector space V, span(S) is a linear subspace of V.
Theorem.If W⊂V is a linear subspace of V and if S⊂W, then span(S)⊂W.
Definition.If S⊂V and if span(S)=V then "S spans V", or, "S generates V."
Solving linear systems of equations
The book describes the standard method of solving a system of n equations with m unknowns.
Linear independence, Bases, and Dimension
Definition of independence. A set of vectors {u1,…,un}⊂V is linearly independent if for any a1,…,an∈F one has
a1u1+⋯+anun=0⟹a1=a2=⋯=an=0.
More generally a set of vectors β⊂V is linearly independent if
every finite subset {u1,…,un}⊂β is linearly independent.
This second definition allows for the possibility that the set β is
infinite.
Definition of basis. A set of vectors {u1,…,un}⊂V is a basis for V if
{u1,…,un} is linearly independent, and
{u1,…,un} spans V.
Extension Theorem for Independent Sets. If {u1,…,un}⊂V is linearly independent, and v∈V is not one the vectors u1,…,un, then v∈span(u1,…,un) if and only if {u1,…,un,v} is dependent.
Proof
First we prove:v∈span(u1,…,un)⟹{u1,…,un,v} is linearly dependent.
If v∈span(u1,…,un) then there are a1,…,an∈F with v=a1u1+⋯+anun. Therefore
−a1u1−⋯−anun+v=0,
so we have a nontrivial linear combination of {u1,…,un,v} that adds
up to zero. Hence {u1,…,un,v} is linearly dependent.
Conversely, we now show:{u1,…,un,v} is linearly dependent ⟹v∈span(u1,…,un).
Suppose {u1,…,un,v} is linearly dependent.
Then there exist a1,…,an,b∈F such that
a1u1+⋯+anun+bv=0,
and such that at least one of the coefficients a1,…,an,b is nonzero.
If b=0 then we have a1u1+⋯+anun=0. Since {u1,…,un} is
independent, this implies a1=⋯=an=0, which is impossible because at
least one of a1,…,an,b does not vanish.
Therefore b=0. This implies that
v=−ba1u1−⋯−banun∈span(u1,…,un).
Dimension Theorem. If {v1,…,vm}⊂span(u1,…,un) and if {v1,…,vm} is linearly independent then m≤n.
Proof of the dimension theorem
We will show that if m>n and {v1,…,vm}⊂span({u1,…,un}) then {v1,…,vm} is linearly dependent.
We begin with the case n=1.
There is only one vector
u1, and each v1,v2,…,vm is a linear combination of
u1, i.e. a multiple of u1.
Thus for certain numbers a1,…,am∈F we have
v1=a1u1,v2=a2u1,…,vm=amu1.
If a1=0 then v1=0 and {v1,…,vm} is dependent.
If a1=0 then we have
−a2v1+a1v2+0⋅v3+⋯+0⋅vm=0.
Since a1=0 this is a nontrivial linear combination of v1,…,vm that adds up to zero. Hence {v1,…,vm} is
dependent.
Next we consider the general case n>1, and we assume that
the case n−1 has already been proven.
In this case each vi is a linear combination of the
vectors u1,…,un.
So we have
If all the coefficients a1n,a2n,…,amn=0 then the
vi are linear combinations of u1,…,un−1. Since m>n we have m>n−1 and therefore the induction hypothesis tells us
that v1,…,vm are linearly dependent.
We are left with the case in which one of the coefficients a1n,…,amn does not vanish.
Assume that a1n=0. Then we consider the vectors
w2=v2−a1na2nv1,…wm=vm−a1namnv1.
The vectors w2,…,wm are linear combinations of u1,…,un−1. Since m−1>n−1 the induction hypothesis applies. We
therefore know that w2,…,wm are linearly dependent, i.e.
there exist c2,…,cm∈F such that
c2w2+⋯+cmwm=0,
and such that at least one of c2,…,cm does not vanish. By
substituting the definition of the wi in this linear combination, we
find after some simplification that
c1v1+c2v2+⋯+cmvm=0,
where
c1=−a1na2nc2−⋯−a1namncm.
Thus v1,…,vm is linearly dependent.
Corollaries to the Dimension Theorem.
If {u1,…,un} and {v1,…,vm}
both are bases of a vector space V, then m=n.
If L⊂V is a linear subspace and V is finite dimensional then dimL≤dimV. If dimL=dimV then L=V.
If L⊂V is a linear subspace and V is finite dimensional, and if {v1,…,vk}⊂L is a basis for L, then there exist vectors vk+1,…,vn∈V such that {v1,…,vk,vk+1,…,vn} is a basis for V.
Proof
This is true because {v1,…,vm}⊂span(u1,…,un)
and {v1,…,vm} is linearly independent, so the Dimension Theorem
implies m≤n. On the other hand, {u1,…,un}⊂span(v1,…,vn) and {u1,…,un} is linearly
independent, so the Dimension Theorem implies n≤m
Definition.
If a vector space V has a basis
{u1,…,un} with n elements, then n is the
dimension of V.
If a vector space V has a basis with finitely many vectors then V is called finite dimensional.
The Dimension Theorem and its corollary imply that the dimension as defined above does not depend on which basis of V you consider.
Components of a vector with respect to a basis
If {u1,…,un} is a basis for a vector space V and x∈V is any
other given vector in V, then there exist numbers x1,…,xn∈F such that
x=x1u1+⋯+xnun.
The numbers x1,…,xn are called the components, or coefficients, of the vector x with respect to the basis {u1,…,un}.
The coefficients x1,…,xn are completely determined by the vector x and the basis {u1,…,un}. Namely, if
x1′u1+⋯+xn′un=x1u1+⋯+xnun
then we could subtract x1u1, …, xnun from both sides and we would end up with
(x1′−x1)u1+⋯+(xn′−xn)un=0.
Since u1,…,un are independent, this implies x1=x1′, … ,
xn=xn′.
Examples—bases for R2
{e1,e2} is a basis for R2
(12),(21) is a basis for R2
Examples—dimension depends on the number field F
Consider C as vector space over the complex numbers.
It has a basis with one “vector,” namely {1}.
Now consider C as vector space over the real numbers. Then the
set {1,i} is linearly independent, and spans C, i.e. {1,i} is a basis for C over the real numbers.
Consider V=R as vector space over the rational numbers Q. Then {1,2} is linearly independent.
Examples—bases for spaces of polynomials
{1,x,x2,x3,…} is a basis for P(R). P(R) is infinite dimensional.