Mathematics for Applied Sciences (Osnabrück 2023-2024)/Part I/Lecture 23/latex
\setcounter{section}{23}
The solution set of a homogeneous system of linear equations in $n$ variables over a field $K$ is a linear subspace of $K^n$. Quite often, this solution space is described as the set of all \anfuehrung{linear combinations}{} of finitely many (simple) solutions. In this lecture, we develop the concepts to make this precise.
\zwischenueberschrift{Generating systems}
\mavergleichskette
{\vergleichskette
{ u }
{ = }{ xv_1+yv_2 }
{ }{ }
{ }{ }
{ }{ }
} {}{}{.}
\inputdefinition
{ }
{
Let $K$ be a
field,
and let $V$ be a
$K$-vector space.
Let \mathl{v_1 , \ldots , v_n}{} denote a family of vectors in $V$. Then the vector
\mathdisp {s_1v_1+s_2v_2 + \cdots + s_nv_n \text{ with } s_i \in K} { }
is called a \definitionswort {linear combination}{} of this vectors
}
Two different coefficient tuples can define the same vector.
\inputdefinition
{ }
{
Let $K$ be a
field,
and let $V$ be a
$K$-vector space.
A family
\mathbed {v_i \in V} {}
{i \in I} {}
{} {} {} {,}
is called a \definitionswort {generating system}{}
\zusatzklammer {or \definitionswort {spanning system}{}} {} {}
of $V$, if every vector
\mavergleichskette
{\vergleichskette
{v
}
{ \in }{V
}
{ }{
}
{ }{
}
{ }{
}
}
{}{}{}
can be written as
\mavergleichskettedisp
{\vergleichskette
{v
}
{ =} {\sum_{j \in J} s_j v_j
}
{ } {
}
{ } {
}
{ } {
}
}
{}{}{,}
with a finite subfamily
\mavergleichskette
{\vergleichskette
{J
}
{ \subseteq }{I
}
{ }{
}
{ }{
}
{ }{
}
}
{}{}{,}
and with
\mavergleichskette
{\vergleichskette
{ s_j
}
{ \in }{ K
}
{ }{
}
{ }{
}
{ }{
}
}
}
In $K^n$, the standard vectors
\mathbed {e_i} {}
{1 \leq i \leq n} {}
{} {} {} {,}
form a generating system. In the polynomial ring \mathl{K[X]}{,} the powers
\mathbed {X^n} {}
{n \in \N} {}
{} {} {} {,}
form an
\zusatzklammer {infinite} {} {}
generating system.
\inputdefinition
{ }
{
Let $K$ be a
field,
and let $V$ be a
$K$-vector space.
For a family
\mathbed {v_i} {}
{i \in I} {}
{} {} {} {,}
we set
\mavergleichskettedisp
{\vergleichskette
{ \langle v_i ,\, i \in I \rangle
}
{ =} { { \left\{ \sum_{i \in J} s_i v_i \mid s_i \in K , \, J \subseteq I \text{ finite subset} \right\} }
}
{ } {
}
{ } {
}
{ } {
}
}
{}{}{,}
}
The empty set generates the null space\zusatzfussnote {This follows from the definition, if we use the convention that the empty sum equals $0$} {.} {.}
The null space is also generated by the element $0$. A single vector $v$ spans the space
\mavergleichskette
{\vergleichskette
{ Kv
}
{ = }{ { \left\{ s v \mid s \in K \right\} }
}
{ }{
}
{ }{
}
{ }{
}
}
{}{}{.}
For
\mavergleichskette
{\vergleichskette
{v
}
{ \neq }{ 0
}
{ }{
}
{ }{
}
{ }{
}
}
{}{}{,}
this is a \stichwort {line} {,} a term we will make more precise in the framework of dimension theory. For two vectors
\mathkor {} {v} {and} {w} {,}
the \anfuehrung{form}{} of the spanned space depends on how the two vectors are related to each other. If they both lie on a line, say
\mavergleichskette
{\vergleichskette
{w
}
{ = }{ s v
}
{ }{
}
{ }{
}
{ }{
}
}
{}{}{,}
then $w$ is superfluous, and the linear subspace generated by the two vectors equals the linear subspace generated by $v$. If this is not the case
\zusatzklammer {and
\mathkor {} {v} {and} {w} {}
are not $0$} {} {,}
then the two vectors span a \anfuehrung{plane}{.}
We list some simple properties for generating systems and linear subspaces.
\inputfaktbeweis
{Vector space/Generating system and spanned subspace/Fact}
{Lemma}
{}
{
\faktsituation {Let $K$ be a
field,
and let $V$ be a
$K$-vector space.}
\faktuebergang {Then the following statements hold.}
\faktfolgerung {\aufzaehlungzwei {For a family
\mathbed {v_i} {}
{i \in I} {}
{} {} {} {,}
of elements in $V$, the
linear span
is a
linear subspace
of $V$.
} {The family
\mathbed {v_i} {}
{i \in I} {}
{} {} {} {,}
is a spanning system of $V$, if and only if
\mavergleichskettedisp
{\vergleichskette
{ \langle v_i ,\, i\in I \rangle
}
{ =} { V
}
{ } {
}
{ } {
}
{ } {
}
}
{}{}{.}
}}
\faktzusatz {}
{See Exercise 23.3 . }
\zwischenueberschrift{Linear independence}
\inputdefinition
{ }
{
Let $K$ be a
field,
and let $V$ be a
$K$-vector space.
A family of vectors
\mathbed {v_i} {}
{i \in I} {}
{} {} {} {,}
\zusatzklammer {where $I$ denotes a finite index set} {} {}
is called \definitionswort {linearly independent}{,} if an equation of the form
\mathdisp {\sum_{i \in I} s_i v_i =0 \text{ with } s_i \in K} { }
is only possible when
\mavergleichskette
{\vergleichskette
{ s_i
}
{ = }{ 0
}
{ }{
}
{ }{
}
{ }{
}
}
{}{}{}
}
If a family is not linear independent, then it is called \stichwort {linearly dependent} {.} A linear combination
\mavergleichskette
{\vergleichskette
{ \sum_{i \in I} s_i v_i
}
{ = }{ 0
}
{ }{
}
{ }{
}
{ }{
}
}
{}{}{}
is called a \stichwort { representation of the null vector} {.} It is called the \stichwort {trivial representation} {,} if all coefficients $s_i$ equal $0$, and, if at least one coefficient is not $0$, a \stichwort {nontrivial representation of the null vector} {.} A family of vectors is linearly independent, if and only if one can represent with it the null vector only in the trivial way. This is equivalent with the property that no vector of the family can be expressed as a linear combination by the others.
\inputbeispiel{}
{
The
standard vectors
in $K^n$ are
linearly independent.
A representation
\mavergleichskettedisp
{\vergleichskette
{\sum_{i = 1}^n s_i e_i
}
{ =} { 0
}
{ } {
}
{ } {
}
{ } {
}
}
{}{}{}
just means
\mavergleichskettedisp
{\vergleichskette
{ s_1 \begin{pmatrix} 1 \\0\\ \vdots\\0 \end{pmatrix} + s_2 \begin{pmatrix} 0 \\1\\ \vdots\\0 \end{pmatrix} + \cdots + s_n \begin{pmatrix} 0 \\0\\ \vdots\\1 \end{pmatrix}
}
{ =} { \begin{pmatrix} 0 \\0\\ \vdots\\0 \end{pmatrix}
}
{ } {
}
{ } {
}
{ } {
}
}
{}{}{.}
The $i$-th row yields directly
\mavergleichskette
{\vergleichskette
{ s_i
}
{ = }{ 0
}
{ }{
}
{ }{
}
{ }{
}
}
{}{}{.}
}
\inputbeispiel{}
{
The three vectors
\mathlistdisp {\begin{pmatrix} 3 \\3\\ 3 \end{pmatrix}} {} {\begin{pmatrix} 0 \\4\\ 5 \end{pmatrix}} {and} {\begin{pmatrix} 4 \\8\\ 9 \end{pmatrix}} {}
are
linearly dependent.
The equation
\mavergleichskettedisp
{\vergleichskette
{ 4 \begin{pmatrix} 3 \\3\\ 3 \end{pmatrix} + 3 \begin{pmatrix} 0 \\4\\ 5 \end{pmatrix} -3 \begin{pmatrix} 4 \\8\\ 9 \end{pmatrix}
}
{ =} { \begin{pmatrix} 0 \\0\\ 0 \end{pmatrix}
}
{ } {
}
{ } {
}
{ } {
}
}
{}{}{}
is a nontrivial representation of the null vector.
}
\inputfaktbeweis
{Linearly independent/Simple properties/Fact}
{Lemma}
{}
{
\faktsituation {Let $K$ be a
field,
let $V$ be a
$K$-vector space,
and let
\mathbed {v_{ i }} {}
{i \in I} {}
{} {} {} {,}
be a family of vectors in $V$.}
\faktuebergang {Then the following statements hold.}
\faktfolgerung {\aufzaehlungsechs {If the family is linearly independent, then for each subset
\mavergleichskette
{\vergleichskette
{ J
}
{ \subseteq }{ I
}
{ }{
}
{ }{
}
{ }{
}
}
{}{}{,}
also the family
\mathbed {v_i} {,}
{i \in J} {}
{} {} {} {,}
is linearly independent.
} {The empty family is linearly independent.
} {If the family contains the null vector, then it is not linearly independent.
} {If a vector appears several times in the family, then the family is not linearly independent.
} {A single vector $v$ is linearly independent if and only if
\mavergleichskette
{\vergleichskette
{ v
}
{ \neq }{ 0
}
{ }{
}
{ }{
}
{ }{
}
}
{}{}{.}
} {Two vectors
\mathkor {} {v} {and} {u} {}
are linearly independent if and only if $u$ is not a scalar multiple of $v$ and vice versa.
}}
\faktzusatz {}
{See Exercise 23.10 . }
\inputremark {}
{
The vectors \mathl{v_1 = \begin{pmatrix} a_{11} \\\vdots\\ a_{m1} \end{pmatrix} , \ldots , v_n = \begin{pmatrix} a_{1n} \\\vdots\\ a_{mn} \end{pmatrix} \in K^m}{} are
linearly dependent,
if and only if the
homogeneous linear system
\mathdisp {\begin{matrix} a _{ 1 1 } x _1 + a _{ 1 2 } x _2 + \cdots + a _{ 1 n } x _{ n } & = & 0 \\ a _{ 2 1 } x _1 + a _{ 2 2 } x _2 + \cdots + a _{ 2 n } x _{ n } & = & 0 \\ \vdots & \vdots & \vdots \\ a _{ m 1 } x _1 + a _{ m 2 } x _2 + \cdots + a _{ m n } x _{ n } & = & 0 \end{matrix}} { }
has a nontrivial solution.
}
\zwischenueberschrift{Basis}
\inputdefinition
{ }
{
Let $K$ be a
field,
and let $V$ be a
$K$-vector space.
Then a
linearly independent
generating system
\mathbed {v_i \in V} {}
{i \in I} {}
{} {} {} {,}
}
\inputbeispiel{}
{
The
standard vectors
in $K^n$ form a
basis.
The
linear independence
was shown in
Example 23.6
.
To show that they also form a
generating system,
let
\mavergleichskettedisp
{\vergleichskette
{v
}
{ =} { \begin{pmatrix} b_1 \\b_2\\ \vdots\\b_n \end{pmatrix}
}
{ \in} { K^n
}
{ } {
}
{ } {
}
}
{}{}{}
be an arbitrary vector. Then we have immediately
\mavergleichskettedisp
{\vergleichskette
{v
}
{ =} { \sum_{i = 1}^n b_i e_i
}
{ } {
}
{ } {
}
{ } {
}
}
{}{}{.}
Hence, we have a basis, which is called the \stichwort {standard basis} {} of $K^n$.
}
\inputfaktbeweisnichtvorgefuehrt
{Vector space/Characterizations of basis/Maximal/Minimal/Fact}
{Theorem}
{}
{
\faktsituation {Let $K$ be a
field,
and let $V$ be a
$K$-vector space.
Let
\mavergleichskette
{\vergleichskette
{ v_1 , \ldots , v_n
}
{ \in }{ V
}
{ }{
}
{ }{
}
{ }{
}
}
{}{}{}
be a family of vectors.}
\faktuebergang {Then the following statements are equivalent.}
\faktfolgerung {\aufzaehlungvier {The family is a
basis
of $V$.
} {The family is a minimal
generating system,
that is, as soon as we remove one vector $v_i$, the remaining family is not a generating system any more.
} {For every vector
\mavergleichskette
{\vergleichskette
{u
}
{ \in }{V
}
{ }{
}
{ }{
}
{ }{
}
}
{}{}{,}
there is exactly one representation
\mavergleichskettedisp
{\vergleichskette
{u
}
{ =} { s_1 v_1 + \cdots + s_n v_n
}
{ } {
}
{ } {
}
{ } {
}
}
{}{}{.}
} {The family is maximal
linearly independent,
that is, as soon as some vector is added, the family is not linearly independent any more.
}}
\faktzusatz {}
}
{Vector space/Characterizations of basis/Maximal/Minimal/Fact/Proof
\inputremark {{{{2}}}}
{
Let a
basis
\mathl{v_1 , \ldots , v_n}{} of a
$K$-vector space
$V$ be given. Due to
Theorem 23.12
,
this means that for every vector
\mavergleichskette
{\vergleichskette
{u
}
{ \in }{V
}
{ }{
}
{ }{
}
{ }{
}
}
{}{}{,}
there exists a uniquely determined representation
\mavergleichskettedisp
{\vergleichskette
{u
}
{ =} { s_1 v_1 + s_2 v_2 + \cdots + s_n v_n
}
{ } {
}
{ } {
}
{ } {
}
}
{}{}{.}
The elements
\mavergleichskette
{\vergleichskette
{s_i
}
{ \in }{K
}
{ }{
}
{ }{
}
{ }{
}
}
{}{}{}
\zusatzklammer {scalars} {} {}
are called the \stichwort {coordinates} {} of $u$ with respect to the given basis. Thus, for a fixed basis, we have a
\zusatzklammer {bijective} {} {}
correspondence between the vectors from $V$, and the coordinate tuples
\mavergleichskette
{\vergleichskette
{ (s_1,s_2 , \ldots , s_n)
}
{ \in }{ K^n
}
{ }{
}
{ }{
}
{ }{
}
}
{}{}{.}
We express this by saying that a basis determines a \stichwort {linear coordinate system} {\zusatzfussnote {Linear coordinates give a bijective relation between points and number tuples. Due to linearity, such a bijection respects the addition and the scalar multiplication. In many different contexts, also nonlinear
\zusatzklammer {curvilinear} {} {}
coordinates are important. These put points of a space and number tuples into a bijective relation. Examples are polar coordinates, cylindrical coordinates and spherical coordinates. By choosing suitable coordinates, mathematical problems, like the computation of volumes, can be simplified.} {} {.}}
}
\inputfaktbeweis
{Vector space/Finitely generated/Basis/Fact}
{Theorem}
{}
{
\faktsituation {Let $K$ be a
field,
and let $V$ be a
$K$-vector space
with a finite
generating system.}
\faktfolgerung {Then $V$ has a finite
basis.}
\faktzusatz {}
}
{
Let
\mathbed {v_i} {}
{i \in I} {}
{} {} {} {,}
be a finite generating system of $V$, with a
finite
index set $I$. We argue with the characterization from
Theorem 23.12
.
If the family is minimal, then we have a basis. If not, then there exists some
\mavergleichskette
{\vergleichskette
{k
}
{ \in }{I
}
{ }{
}
{ }{
}
{ }{
}
}
{}{}{,}
such that the remaining family where $v_k$ is removed, that is
\mathbed {v_i} {}
{i \in I \setminus \{k\}} {}
{} {} {} {,}
is also a generating system. In this case, we can go on with this smaller index set. With this method, we arrive at a subset
\mavergleichskette
{\vergleichskette
{J
}
{ \subseteq }{I
}
{ }{
}
{ }{
}
{ }{
}
}
{}{}{}
such that
\mathbed {v_i} {}
{i \in J} {}
{} {} {} {,}
is a minimal generating set, hence a basis.
\zwischenueberschrift{Dimension theory}
A finitely generated vector space has many quite different bases. However, the number of elements in a basis is constant and depends only on the vector space. We will formulate this important property now and take it as the departure for the definition of dimension of a vector space.
\inputfaktbeweisnichtvorgefuehrt
{Vector space/Finitely generated/Length of every basis/Fact}
{Theorem}
{}
{
\faktsituation {Let $K$ be a
field,
and let $V$ be a
$K$-vector space
with a finite
generating system.}
\faktfolgerung {Then any two
bases
of $V$ have the same number of vectors.}
\faktzusatz {}
}
{Vector space/Finitely generated/Length of every basis/Fact/Proof
This theorem enables the following definition.
\inputdefinition
{ }
{
Let $K$ be a field, and let $V$ be a $K$-vector space with a finite generating system. Then the number of vectors in any basis of $V$, is called the \definitionswort {dimension}{} of $V$, written
\mathdisp {\dim_{ } { \left( V \right) }} { . }
}
Due to the preceding theorem, the dimension is \stichwort {well-defined} {.} If a vector space is not finitely generated, then one puts
\mavergleichskette
{\vergleichskette
{ \dim_{ } { \left( V \right) }
}
{ = }{ \infty
}
{ }{
}
{ }{
}
{ }{
}
}
{}{}{.}
The null space $0$ has dimension $0$. A one-dimensional vector space is called a \stichwort {line} {,} a two-dimensional vector space a \stichwort {plane} {,} a three-dimensional vector space a \stichwort {space} {}
\zusatzklammer {in the strict sense} {} {}
but every vector space is called a space.
\inputfaktbeweis
{Standard space/K^n/Dimension n/Fact}
{Corollary}
{}
{
\faktsituation {Let $K$ be a field and
\mavergleichskette
{\vergleichskette
{n
}
{ \in }{\N
}
{ }{
}
{ }{
}
{ }{
}
}
{}{}{.}}
\faktfolgerung {Then the
standard space $K^n$ has the
dimension $n$.}
\faktzusatz {}
}
{
The
standard basis
\mathbed {e_i} {}
{i = 1 , \ldots , n} {}
{} {} {} {,}
consists of $n$ vectors, hence the dimension is $n$.
\inputbeispiel{}
{
The complex numbers form a two-dimensional real vector space, a basis is \mathkor {} {1} {and} {{ \mathrm i}} {.}
}
\inputbeispiel{}
{
The
polynomial ring
\mavergleichskette
{\vergleichskette
{R
}
{ = }{K[X]
}
{ }{
}
{ }{
}
{ }{
}
}
{}{}{}
over a
field
$K$ is not a
finite-dimensional vector space.
To see this, we have to show that there is no finite
generating system
for the polynomial ring. Consider $n$ polynomials \mathl{P_1 , \ldots , P_n}{.} Let $d$ be the maximum of the
degrees
of these polynomials. Then every
$K$-linear combination
\mathl{\sum_{i=1}^n a_i P_i}{} has at most degree $d$. In particular, polynomials of larger degree can not be presented by \mathl{P_1 , \ldots , P_n}{,} so these do not form a generating system for all polynomials.
}
\inputfaktbeweisnichtvorgefuehrt
{Vector space/Linear subspace/Dimensions/Compare/Fact}
{Corollary}
{}
{
\faktsituation {Let $V$ denote a finite-dimensional vector space over a field $K$. Let
\mavergleichskette
{\vergleichskette
{U
}
{ \subseteq }{V
}
{ }{
}
{ }{
}
{ }{
}
}
{}{}{}
denote a
linear subspace.}
\faktfolgerung {Then $U$ is also finite-dimensional, and the estimate
\mavergleichskettedisp
{\vergleichskette
{ \dim_{ } { \left( U \right) }
}
{ \leq} { \dim_{ } { \left( V \right) }
}
{ } {
}
{ } {
}
{ } {
}
}
{}{}{}
holds.}
\faktzusatz {}
}
{Vector space/Linear subspace/Dimensions/Compare/Fact/Proof
\inputfaktbeweis
{Vector space/Dimension n and n vectors/Equivalences/Fact}
{Corollary}
{}
{
\faktsituation {Let $K$ be a
field,
and let $V$ be a
$K$-vector space
with finite
dimension
\mavergleichskette
{\vergleichskette
{n
}
{ = }{ \dim_{ } { \left( V \right) }
}
{ }{
}
{ }{
}
{ }{
}
}
{}{}{.}}
\faktvoraussetzung {Let $n$ vectors \mathl{v_1 , \ldots , v_n}{} in $V$ be given.}
\faktfolgerung {Then the following properties are equivalent.
\aufzaehlungdrei {\mathl{v_1 , \ldots , v_n}{} form a
basis
of $V$.
} {\mathl{v_1 , \ldots , v_n}{} form a
generating system
of $V$.
} {\mathl{v_1 , \ldots , v_n}{} are
linearly independent.
}}
\faktzusatz {}
{See Exercise 23.18 . }
\inputbeispiel{}
{
Let $K$ be a
field.
It is easy to get an overview over the
linear subspaces
of $K^n$, as the
dimension
of a linear subspace equals
\mathbed {k} {with}
{0 \leq k \leq n} {}
{} {} {} {,}
due to
Corollary 23.20
.
For
\mavergleichskette
{\vergleichskette
{n
}
{ = }{ 0
}
{ }{
}
{ }{
}
{ }{
}
}
{}{}{,}
there is only the null space itself, for
\mavergleichskette
{\vergleichskette
{n
}
{ = }{1
}
{ }{
}
{ }{
}
{ }{
}
}
{}{}{}
there is the null space and $K$ itself. For
\mavergleichskette
{\vergleichskette
{n
}
{ = }{2
}
{ }{
}
{ }{
}
{ }{
}
}
{}{}{,}
there is the null space, the whole plane $K^2$, and the one-dimensional lines through the origin. Every line $G$ has the form
\mavergleichskettedisp
{\vergleichskette
{G
}
{ =} { Kv
}
{ =} { { \left\{ s v \mid s \in K \right\} }
}
{ } {
}
{ } {
}
}
{}{}{,}
with a vector
\mavergleichskette
{\vergleichskette
{ v
}
{ \neq }{ 0
}
{ }{
}
{ }{
}
{ }{
}
}
{}{}{.}
Two vectors different from $0$ define the same line if and only if they are
linearly dependent.
For
\mavergleichskette
{\vergleichskette
{n
}
{ = }{3
}
{ }{
}
{ }{
}
{ }{
}
}
{}{}{,}
there is the null space, the whole space $K^3$, the one-dimensional lines through the origin, and the two-dimensional planes through the origin.
}
\inputfaktbeweisnichtvorgefuehrt
{Vector space/Basis complement/Fact}
{Theorem}
{}
{
\faktsituation {Let $V$ denote a finite-dimensional vector space over a field $K$.}
\faktvoraussetzung {Let
\mathdisp {u_1 , \ldots , u_k} { }
denote
linearly independent
vectors in $V$.}
\faktfolgerung {Then there exist vectors
\mathdisp {u_{k+1} , \ldots , u_n} { }
such that
\mathdisp {u_1 , \ldots , u_k, u_{k+1} , \ldots , u_n} { }
form a
basis
of $V$.}
\faktzusatz {}