Linear algebra (Osnabrück 2024-2025)/Part I/Lecture 6/latex
\setcounter{section}{6}
\subtitle {Vector spaces}
\image{ \begin{center}
\includegraphics[width=5.5cm]{\imageinclude {Vector Addition.svg} }
\end{center}
\imagetext {The addition of two arrows $a$ and $b$, a typical example for vectors.} }
\imagelicense { Vector Addition.svg } {} {Booyabazooka} {Commons} {PD} {}
The central concept of linear algebra is a vector space.
\inputdefinition
{ }
{
Let $K$ denote a
field,
and $V$ a set with a distinguished element
\mathrelationchain
{\relationchain
{0
}
{ \in }{V
}
{ }{
}
{ }{
}
{ }{
}
}
{}{}{,}
and with two mappings
\mathdisp {+ \colon V \times V \longrightarrow V
, (u,v) \longmapsto u+v} { , }
and
\mathdisp {\cdot \colon K \times V \longrightarrow V
, (s,v) \longmapsto s v = s \cdot v} { . }
Then $V$ is called a
\definitionwordpremath {K}{ vector space }{}
\extrabracket {or a vector space over $K$} {} {,}
if the following axioms hold
\extrabracket {where
\mathcor {} {r,s \in K} {and} {u,v,w \in V} {}
are arbitrary} {} {.}
\enumerationeight {
\mathrelationchain
{\relationchain
{ u+v
}
{ = }{v+u
}
{ }{
}
{ }{
}
{ }{
}
}
{}{}{,}
} {
\mathrelationchain
{\relationchain
{(u+v)+w
}
{ = }{ u +(v+w)
}
{ }{
}
{ }{
}
{ }{
}
}
{}{}{,}
} {
\mathrelationchain
{\relationchain
{ v+0
}
{ = }{v
}
{ }{
}
{ }{
}
{ }{
}
}
{}{}{,}
} {For every $v$, there exists a $z$ such that
\mathrelationchain
{\relationchain
{v+z
}
{ = }{ 0
}
{ }{
}
{ }{
}
{ }{
}
}
{}{}{,}
} {
\mathrelationchain
{\relationchain
{1 \cdot u
}
{ = }{ u
}
{ }{
}
{ }{
}
{ }{
}
}
{}{}{,}
} {
\mathrelationchain
{\relationchain
{ r(su)
}
{ = }{ (rs) u
}
{ }{
}
{ }{
}
{ }{
}
}
{}{}{,}
} {
\mathrelationchain
{\relationchain
{ r(u+v)
}
{ = }{ru + rv
}
{ }{
}
{ }{
}
{ }{
}
}
{}{}{,}
} {
\mathrelationchain
{\relationchain
{ (r+s) u
}
{ = }{ru + su
}
{ }{
}
{ }{
}
{ }{
}
}
{}{}{.}
}
The binary operation in $V$ is called (vector-)addition, and the operation
$K \times V \rightarrow V$
is called \keyword {scalar multiplication} {.} The elements in a vector space are called \keyword {vectors} {,} and the elements
\mathrelationchain
{\relationchain
{r
}
{ \in }{K
}
{ }{
}
{ }{
}
{ }{
}
}
{}{}{}
are called \keyword {scalars} {.} The null element
\mathrelationchain
{\relationchain
{ 0
}
{ \in }{V
}
{ }{
}
{ }{
}
{ }{
}
}
{}{}{}
is called \keyword {null vector} {,} and for
\mathrelationchain
{\relationchain
{v
}
{ \in }{V
}
{ }{
}
{ }{
}
{ }{
}
}
{}{}{,}
the inverse element, with respect to the addition, is called the \keyword {negative} {} of $v$, denoted by $-v$.
The field that occurs in the definition of a vector space is called the \keyword {base field} {.} All the concepts of linear algebra refer to such a base field. In case
\mathrelationchain
{\relationchain
{K
}
{ = }{\R
}
{ }{
}
{ }{
}
{ }{
}
}
{}{}{,}
we talk about a \keyword {real vector space} {,} and in case
\mathrelationchain
{\relationchain
{K
}
{ = }{ \Complex
}
{ }{
}
{ }{
}
{ }{
}
}
{}{}{,}
we talk about a \keyword {complex vector space} {.} For real and complex vector spaces, there exist further structures like length, angle, inner product. But first we develop the algebraic theory of vector spaces over an arbitrary field.
\image{ \begin{center}
\includegraphics[width=5.5cm]{\imageinclude {Vector_space_illust.svg} }
\end{center}
\imagetext {} }
\imagelicense { Vector space illust.svg } {} {Oleg Alexandrov} {Commons} {PD} {}
\inputexample{}
{
Let $K$ denote a
field,
and let
\mathrelationchain
{\relationchain
{ n
}
{ \in }{ \N_+
}
{ }{
}
{ }{
}
{ }{
}
}
{}{}{.}
Then the
product set
\mathrelationchaindisplay
{\relationchain
{ K^n
}
{ =} { \underbrace{K \times \cdots \times K }_{n\text{-times} }
}
{ =} { { \left\{ (x_1 , \ldots , x_{ n }) \mid x_i \in K \right\} }
}
{ } {}
{ } {}
}
{}{}{,}
with componentwise addition and with scalar multiplication given by
\mathrelationchaindisplay
{\relationchain
{ s (x_1 , \ldots , x_{ n })
}
{ =} { (s x_1 , \ldots , s x_{ n })
}
{ } {
}
{ } {
}
{ } {
}
}
{}{}{,}
is a
vector space.
This space is called the $n$-dimensional \keyword {standard space} {.} In particular,
\mathrelationchain
{\relationchain
{K^1
}
{ = }{K
}
{ }{
}
{ }{
}
{ }{
}
}
{}{}{}
is a vector space.
}
The null space $0$, consisting of just one element $0$, is a vector space. It might be considered as
\mathrelationchain
{\relationchain
{K^0
}
{ = }{0
}
{ }{
}
{ }{
}
{ }{
}
}
{}{}{.}
The vectors in the standard space $K^n$ can be written as row vectors
\mathdisp {\left( a_1 , \, a_2 , \, \ldots , \, a_n \right)} { }
or as column vectors
\mathdisp {\begin{pmatrix} a_1 \\a_2\\ \vdots\\a_n \end{pmatrix}} { . }
The vector
\mathrelationchaindisplay
{\relationchain
{ e_i
}
{ \defeq} { \begin{pmatrix} 0 \\ \vdots\\ 0\\1\\ 0\\ \vdots\\ 0 \end{pmatrix}
}
{ } {
}
{ } {
}
{ } {
}
}
{}{}{,}
where the $1$ is at the $i$-th position, is called $i$-th \keyword {standard vector} {.}
\inputexample{}
{
Let $E$ be a \quotationshort{plane}{} with a fixed \quotationshort{origin}{}
\mathrelationchain
{\relationchain
{ Q
}
{ \in }{ E
}
{ }{
}
{ }{
}
{ }{
}
}
{}{}{.}
We identify a point
\mathrelationchain
{\relationchain
{ P
}
{ \in }{E
}
{ }{
}
{ }{
}
{ }{
}
}
{}{}{}
with the connecting vector \mathl{\overrightarrow{ Q P }}{}
\extrabracket {the arrow from $Q$ to $P$} {} {.}
In this situation, we can introduce an intuitive coordinate-free vector addition and a coordinate-free scalar multiplication. Two vectors
\mathcor {} {\overrightarrow{ Q P }} {and} {\overrightarrow{ Q R }} {}
are added together by constructing the parallelogram of these vectors. The result of the addition is the corner of the parallelogram which lies in opposition to $Q$. In this construction, we have to draw a line parallel to \mathl{\overrightarrow{ Q P }}{} through $R$ and a line parallel to \mathl{\overrightarrow{ Q R }}{} through $P$. The intersection point is the point sought after. An accompanying idea is that we move the vector \mathl{\overrightarrow{ Q P }}{} in a parallel way so that the new starting point of \mathl{\overrightarrow{ Q P }}{} becomes the ending point of \mathl{\overrightarrow{ Q R }}{.}
In order to describe the multiplication of a vector \mathl{\overrightarrow{ Q P }}{} with a scalar $s$, this number has to be given on a line $G$ that is also marked with a zero point
\mathrelationchain
{\relationchain
{ 0
}
{ \in }{ G
}
{ }{
}
{ }{
}
{ }{
}
}
{}{}{}
and a unit point
\mathrelationchain
{\relationchain
{ 1
}
{ \in }{ G
}
{ }{
}
{ }{
}
{ }{
}
}
{}{}{.}
This line lies somewhere in the plane. We move this line
\extrabracket {by translating and rotating} {} {}
in such a way that $0$ becomes $Q$, and we avoid that the line is identical with the line given by \mathl{\overrightarrow{ Q P }}{}
\extrabracket {which we call $H$} {} {.}
Now we connect $1$ and $P$ with a line $L$, and we draw the line $L'$ parallel to $L$ through $s$. The intersecting point of \mathcor {} {L'} {and} {H} {} is \mathl{s \overrightarrow{ Q P }}{.}
These considerations can also be done in higher dimensions, but everything takes place already in the plane spanned by these vectors.
}
\inputexample{
}
{
The
complex numbers
$$ form a
field,
and therefore they form also a
vector space
over the field $\Complex$ itself. However, the set of complex numbers equals $\R^2$ as an additive group. The multiplication of a complex number \mathl{a+b { \mathrm i}}{} with a real number
\mathrelationchain
{\relationchain
{ s
}
{ = }{ (s,0)
}
{ }{
}
{ }{
}
{ }{}
}
{}{}{}
is componentwise, so this multiplication coincides with the scalar multiplication on $\R^2$. Hence, the set of complex numbers is also a real vector space.
}
\inputexample{}
{
For a
field
$K$, and given natural numbers \mathl{m,n}{,} the set
\mathdisp {\operatorname{Mat}_{ m \times n } (K)} { }
of all \mathl{m \times n}{-}matrices, endowed with componentwise addition and componentwise scalar multiplication, is a
$K$-vector space.
The null element in this vector space is the \keyword {null matrix} {}
\mathrelationchaindisplay
{\relationchain
{0
}
{ =} { \begin{pmatrix} 0 & \ldots & 0 \\ \vdots & \ddots & \vdots \\0 & \ldots & 0 \end{pmatrix}
}
{ } {
}
{ } {
}
{ } {
}
}
{}{}{.}
}
We will introduce polynomials later, they are probably known from school.
\inputexample{}
{
Let
\mathrelationchain
{\relationchain
{ R
}
{ = }{ K[X]
}
{ }{
}
{ }{
}
{ }{
}
}
{}{}{}
be the
polynomial ring
in one variable over the
field
$K$, consisting of all polynomials, that is, expressions of the form
\mathdisp {a_nX^n+a_{n-1}X^{n-1} + \cdots + a_2X^2+a_1X+a_0} { , }
with
\mathrelationchain
{\relationchain
{ a_i
}
{ \in }{ K
}
{ }{
}
{ }{
}
{ }{
}
}
{}{}{.}
Using componentwise addition and componentwise multiplication with a scalar
\mathrelationchain
{\relationchain
{s
}
{ \in }{ K
}
{ }{
}
{ }{
}
{ }{
}
}
{}{}{}
\extrabracket {this is also multiplication with the constant polynomial $s$} {} {,}
the polynomial ring is a
$K$-vector space.
}
\inputexample{}
{
We consider the inclusion
\mathrelationchain
{\relationchain
{ \Q
}
{ \subseteq }{ \R
}
{ }{
}
{ }{
}
{ }{
}
}
{}{}{}
of the
rational numbers
inside the
real numbers.
Using the real addition and the multiplication of rational numbers with real numbers, we see that $\R$ is a
$\Q$-vector space,
as follows directly from the
field axioms.
This is a quite crazy vector space.
}
\inputfactproof
{Vector space/Simple properties/Fact}
{Lemma}
{}
{
\factsituation {Let $K$ be a
field,
and let $V$ be a
$K$-vector space.}
\factsegue {Then the following properties hold
\extrabracket {for
\mathcor {} {v \in V} {and} {s \in K} {}} {} {.}}
\factconclusion {\enumerationfour {We have
\mathrelationchain
{\relationchain
{ 0v
}
{ = }{ 0
}
{ }{
}
{ }{
}
{ }{
}
}
{}{}{.}
} {We have
\mathrelationchain
{\relationchain
{ s 0
}
{ = }{ 0
}
{ }{
}
{ }{
}
{ }{
}
}
{}{}{.}
} {We have
\mathrelationchain
{\relationchain
{ (-1) v
}
{ = }{ -v
}
{ }{
}
{ }{
}
{ }{
}
}
{}{}{.}
} {If
\mathrelationchain
{\relationchain
{ s
}
{ \neq }{ 0
}
{ }{
}
{ }{
}
{ }{
}
}
{}{}{}
and
\mathrelationchain
{\relationchain
{ v
}
{ \neq }{ 0
}
{ }{
}
{ }{
}
{ }{
}
}
{}{}{,}
then
\mathrelationchain
{\relationchain
{ s v
}
{ \neq }{ 0
}
{ }{
}
{ }{
}
{ }{
}
}
{}{}{.}
}}
\factextra {}
{See Exercise 6.27 .}
\subtitle {Linear subspaces}
\inputdefinition
{ }
{
Let $K$ be a
field,
and let $V$ be a
$K$-vector space.
A subset
\mathrelationchain
{\relationchain
{ U
}
{ \subseteq }{ V
}
{ }{
}
{ }{
}
{ }{
}
}
{}{}{}
is called a \definitionword {linear subspace}{} if the following properties hold.
\enumerationthree {
\mathrelationchain
{\relationchain
{ 0
}
{ \in }{ U
}
{ }{
}
{ }{
}
{ }{
}
}
{}{}{.}
} {If
\mathrelationchain
{\relationchain
{ u,v
}
{ \in }{U
}
{ }{
}
{ }{
}
{ }{
}
}
{}{}{,}
then also
\mathrelationchain
{\relationchain
{ u+v
}
{ \in }{U
}
{ }{
}
{ }{
}
{ }{
}
}
{}{}{.}
} {If
\mathrelationchain
{\relationchain
{ u
}
{ \in }{ U
}
{ }{
}
{ }{
}
{ }{
}
}
{}{}{}
and
\mathrelationchain
{\relationchain
{ s
}
{ \in }{ K
}
{ }{
}
{ }{
}
{ }{
}
}
{}{}{,}
then also
\mathrelationchain
{\relationchain
{ s u
}
{ \in }{ U
}
{ }{
}
{ }{
}
{ }{
}
}
{}{}{}
holds.
}
Addition and scalar multiplication can be restricted to such a linear subspace. Hence, the linear subspace is itself a vector space, see Exercise 6.10 . The simplest linear subspaces in a vector space $V$ are the null space $0$ and the whole vector space $V$.
\inputfactproof
{System of linear equations/Set of variables/Solution space is vector space/Fact}
{Lemma}
{}
{
\factsituation {Let $K$ be a
field,
and let
\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}} { }
be a
homogeneous system of linear equations
over $K$.}
\factconclusion {Then the set of all solutions to the system is a
linear subspace
of the standard space $K^n$.}
\factextra {}
{See Exercise 6.3 .}
Therefore, we talk about the \keyword {solution space} {} of the linear system. In particular, the sum of two solutions of a system of linear equations is again a solution. The solution set of an inhomogeneous linear system is not a vector space. However, one can add, to a solution of an inhomogeneous system, a solution of the corresponding homogeneous system, and get a solution of the inhomogeneous system again.
\inputexample{}
{
We take a look at the homogeneous version of
Example 5.1
,
so we consider the homogeneous linear system
\mathdisp {\begin{matrix}
2x & +5y & +2z & & -v & = & 0 \\ \,
3x & -4y & & +u & +2v & = & 0 \\ \,
4x & & -2z & +2u & & = & 0 \,
\end{matrix}} { }
over $\R$. Due to
Lemma 6.11
,
the solution set $L$ is a linear subspace of $\R^5$. We have described it explicitly in
Example 5.1
as
\mathdisp {{ \left\{ u { \left(- { \frac{ 1 }{ 3 } }, 0 , { \frac{ 1 }{ 3 } } ,1,0\right) } + v { \left(- { \frac{ 2 }{ 13 } }, { \frac{ 5 }{ 13 } }, -{ \frac{ 4 }{ 13 } },0,1\right) } \mid u,v \in \R \right\} }} { . }
This description also shows that the solution set is a vector space. Moreover, with this description, it is clear that $L$ is in bijection with $\R^2$, and this bijection respects the addition and also the scalar multiplication
\extrabracket {the solution set $L'$ of the inhomogeneous system is also in bijection with $\R^2$, but there is no reasonable addition nor scalar multiplication on $L'$} {} {.}
However, this bijection depends heavily on the chosen \quotationshort{basic solutions}{}
\mathcor {} {{ \left(- { \frac{ 1 }{ 3 } }, 0 , { \frac{ 1 }{ 3 } } ,1,0\right) }} {and} {{ \left(- { \frac{ 2 }{ 13 } }, { \frac{ 5 }{ 13 } }, -{ \frac{ 4 }{ 13 } },0,1\right) }} {,}
which depends on the order of elimination. There are several equally good basic solutions for $L$.
}
This example shows also the following: the solution space of a linear system over $K$ is \quotationshort{in a natural way}{,} that means, independent on any choice, a linear subspace of $K^n$
\extrabracket {where $n$ is the number of variables} {} {.}
For this solution space, there always exists a \quotationshort{linear bijection}{} (an \quotationshort{isomorphism}{})
to some \mathl{K^{d}}{}
\extrabracket {\mathrelationchainb
{\relationchainb
{d
}
{ \leq }{n
}
{ }{
}
{ }{
}
{ }{
}
}
{}{}{}} {} {,}
but there is no natural choice for such a bijection. This is one of the main reasons to work with abstract vector spaces, instead of just $K^n$.
\subtitle {Generating systems}
The solution set of a homogeneous linear system in $n$ variables over a field $K$ is a linear subspace of $K^n$. This solution space is often described as the set of all \quotationshort{linear combinations}{} of finitely many \extrabracket {simple} {} {} solutions. In this and the next lecture, we will develop the concepts to make this precise.
\image{ \begin{center}
\includegraphics[width=5.5cm]{\imageinclude {VectorGenerado.gif} }
\end{center}
\imagetext {The plane generated by two vectors $v_1$ and $v_2$ consists of all linear combinations
\mathrelationchain
{\relationchain
{ u
}
{ = }{ s v_1+ t v_2
}
{ }{
}
{ }{
}
{ }{
}
}
{}{}{.}} }
\imagelicense { VectorGenerado.gif } {} {Marianov} {Commons} {PD} {}
\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 \definitionword {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
\mathcond {v_i \in V} {}
{i \in I} {}
{} {} {} {,}
is called a \definitionword {generating system}{}
\extrabracket {or \definitionword {spanning system}{}} {} {}
of $V$, if every vector
\mathrelationchain
{\relationchain
{v
}
{ \in }{V
}
{ }{
}
{ }{
}
{ }{
}
}
{}{}{}
can be written as
\mathrelationchaindisplay
{\relationchain
{v
}
{ =} {\sum_{j \in J} s_j v_j
}
{ } {
}
{ } {
}
{ } {
}
}
{}{}{,}
with a finite subfamily
\mathrelationchain
{\relationchain
{J
}
{ \subseteq }{I
}
{ }{
}
{ }{
}
{ }{
}
}
{}{}{,}
and with
\mathrelationchain
{\relationchain
{ s_j
}
{ \in }{ K
}
{ }{
}
{ }{
}
{ }{
}
}
}
In $K^n$, the standard vectors
\mathcond {e_i} {}
{1 \leq i \leq n} {}
{} {} {} {,}
form a generating system. In the polynomial ring \mathl{K[X]}{,} the powers
\mathcond {X^n} {}
{n \in \N} {}
{} {} {} {,}
form an
\extrabracket {infinite} {} {}
generating system.
\inputdefinition
{ }
{
Let $K$ be a
field,
and let $V$ be a
$K$-vector space.
For a family
\mathcond {v_i} {}
{i \in I} {}
{} {} {} {,}
we set
\mathrelationchaindisplay
{\relationchain
{ \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\extrafootnote {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
\mathrelationchain
{\relationchain
{ Kv
}
{ = }{ { \left\{ s v \mid s \in K \right\} }
}
{ }{
}
{ }{
}
{ }{
}
}
{}{}{.}
For
\mathrelationchain
{\relationchain
{v
}
{ \neq }{ 0
}
{ }{
}
{ }{
}
{ }{
}
}
{}{}{,}
this is a \keyword {line} {,} a term we will make more precise in the framework of dimension theory. For two vectors
\mathcor {} {v} {and} {w} {,}
the \quotationshort{form}{} of the spanned space depends on how the two vectors are related to each other. If they both lie on a line, say
\mathrelationchain
{\relationchain
{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
\extrabracket {and
\mathcor {} {v} {and} {w} {}
are not $0$} {} {,}
then the two vectors span a \quotationshort{plane}{.}
We list some simple properties for generating systems and linear subspaces.
\inputfactproof
{Vector space/Generating system and spanning subspace/Fact}
{Lemma}
{}
{
\factsituation {Let $K$ be a
field,
and let $V$ be a
$K$-vector space.}
\factsegue {Then the following hold.}
\factconclusion {\enumerationthree {Let
\mathcond {U_j} {}
{j \in J} {}
{} {} {} {,}
be a family of
linear subspaces
of $V$. Then the intersection
\mathrelationchaindisplay
{\relationchain
{ U
}
{ =} { \bigcap_{j \in J} U_j
}
{ } {
}
{ } {
}
{ } {
}
}
{}{}{}
is a linear subspace.
} {Let
\mathcond {v_i} {}
{i \in I} {}
{} {} {} {,}
be a family of elements of $V$, and consider the subset $W$ of $V$ which is given by all
linear combinations
of these elements. Then $W$ is a linear subspace of $V$.
} {The family
\mathcond {v_i} {}
{i \in I} {}
{} {} {} {,}
is a
system of generators
of $V$ if and only if
\mathrelationchaindisplay
{\relationchain
{ \langle v_i ,\, i\in I \rangle
}
{ =} { V
}
{ } {
}
{ } {
}
{ } {
}
}
{}{}{.}
}}
\factextra {}
{See Exercise 6.26 .}