# Linear algebra (Osnabrück 2024-2025)/Part I/Lecture 6

    Vector spaces
The central concept of linear algebra is a vector space.

## Definition

Let $K$ denote a field, and $V$ a set with a distinguished element $0\in V$, and with two mappings
    $+\colon V\times V\longrightarrow V,(u,v)\longmapsto u+v,$
and
    $\cdot \colon K\times V\longrightarrow V,(s,v)\longmapsto sv=s\cdot v.$
Then $V$ is called a $K$-vector space (or a vector space over $K$), if the following axioms hold (where $r,s\in K$ and $u,v,w\in V$ are arbitrary).
* $u+v=v+u$,
* $(u+v)+w=u+(v+w)$,
* $v+0=v$,
* For every $v$, there exists a $z$ such that $v+z=0$,
* $1\cdot u=u$,
* $r(su)=(rs)u$,
* $r(u+v)=ru+rv$,
* $(r+s)u=ru+su$.
The binary operation in $V$ is called (vector-)addition, and the operation $K\times V\rightarrow V$ is called scalar multiplication. The elements in a vector space are called vectors, and the elements $r\in K$ are called scalars. The null element $0\in V$ is called null vector, and for $v\in V$, the inverse element is called the negative of $v$, denoted by $-v$.
The field which occurs in the definition of a vector space is called the base field. All the concepts of linear algebra refer to such a base field. In case $K=\mathbb {R}$, we talk about a real vector space, and in case $K=\mathbb {C}$, we talk about a 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.

## Example

Let $K$ denote a field, and let $n\in \mathbb {N} _{+}$. Then the product set
    $K^{n}=\underbrace {K\times \cdots \times K} _{n{\text{-times}}}={\left\{(x_{1},\ldots ,x_{n})\mid x_{i}\in K\right\}}\,,$
with componentwise addition and with scalar multiplication given by
    $s(x_{1},\ldots ,x_{n})=(sx_{1},\ldots ,sx_{n})\,,$
is a vector space. This space is called the $n$-dimensional standard space. In particular, $K^{1}=K$ is a vector space.
The null space $0$, consisting of just one element $0$, is a vector space. It might be considered as $K^{0}=0$.
The vectors in the standard space $K^{n}$ can be written as row vectors
    $\left(a_{1},\,a_{2},\,\ldots ,\,a_{n}\right)$
or as column vectors
    ${\begin{pmatrix}a_{1}\\a_{2}\\\vdots \\a_{n}\end{pmatrix}}.$
The vector
    $e_{i}:={\begin{pmatrix}0\\\vdots \\0\\1\\0\\\vdots \\0\end{pmatrix}}\,,$
where the $1$ is at the $i$-th position, is called $i$-th standard vector.

## Example

Let $E$ be a "plane“ with a fixed "origin“ $Q\in E$. We identify a point $P\in E$ with the connecting vector ${\overrightarrow {QP}}$ (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 ${\overrightarrow {QP}}$ and ${\overrightarrow {QR}}$ 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 ${\overrightarrow {QP}}$ through $R$ and a line parallel to ${\overrightarrow {QR}}$ through $P$. The intersection point is the point sought after. An accompanying idea is that we move the vector ${\overrightarrow {QP}}$ in a parallel way so that the starting point of ${\overrightarrow {QP}}$ becomes the ending point of ${\overrightarrow {QR}}$.
In order to describe the multiplication of a vector ${\overrightarrow {QP}}$ with a scalar $s$, this number has to be given on a line $G$, which is also marked with a zero point $0\in G$ and a one point $1\in G$. This line lies somehow in the plane. We move this line
(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 ${\overrightarrow {QP}}$ (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 $L'$ and $H$ is $s{\overrightarrow {QP}}$.
These considerations can also be done in higher dimensions, but everything takes place already in the plane spanned by these vectors.

## Example

The complex numbers $\mathbb {C}$ form a field, and therefore they form also a vector space over the field itself. However, the set of complex numbers equals $\mathbb {R} ^{2}$ as an additive group. The multiplication of a complex number $a+b{\mathrm {i} }$ with a real number $s=(s,0)$ is componentwise, so this multiplication coincides with the scalar multiplication on $\mathbb {R} ^{2}$. Hence, the complex numbers are also a real vector space.

## Example

For a field $K$, and given natural numbers $m,n$, the set
    $\operatorname {Mat} _{m\times n}(K)$
of all $m\times n$-matrices with componentwise addition and componentwise scalar multiplication, is a $K$-vector space. The null element in this vector space is the null matrix
    $0={\begin{pmatrix}0&\ldots &0\\\vdots &\ddots &\vdots \\0&\ldots &0\end{pmatrix}}\,.$
We will introduce polynomials later, they are probably known from school.

## Example

Let $R=K[X]$ be the polynomial ring in one variable over the field $K$, consisting of all polynomials, that is expressions of the form
    $a_{n}X^{n}+a_{n-1}X^{n-1}+\cdots +a_{2}X^{2}+a_{1}X+a_{0},$
with $a_{i}\in K$. Using componentwise addition and componentwise multiplication with a scalar $s\in K$ (this is also multiplication with the constant polynomial $s$), the polynomial ring is a $K$-vector space.

## Example

We consider the inclusion $\mathbb {Q} \subseteq \mathbb {R}$ of the rational numbers inside the real numbers. Using the real addition and the multiplication of rational numbers with real numbers, we see that $\mathbb {R}$ is a vector space, as follows directly from the field axioms. This is a quite crazy vector space.

## Lemma

Let $K$ be a field, and let $V$ be a $K$-vector space. Then the following properties hold (for
$v\in V$
$s\in K$
* We have $0v=0$.
* We have $s0=0$.
* We have $(-1)v=-v$.
* If $s\neq 0$ and $v\neq 0$, then $sv\neq 0$.

### Proof

$\Box$
    Linear subspaces

## Definition

Let $K$ be a field, and $V$ be a $K$-vector space. A subset $U\subseteq V$ is called a linear subspace, if the following properties hold.
* $0\in U$.
* If $u,v\in U$, then also $u+v\in U$.
* If $u\in U$ and $s\in K$, then also $su\in U$ holds.
Addition and scalar multiplication can be restricted to such a linear subspace. Hence, the linear subspace is itself a vector space, see exercise. The simplest linear subspaces in a vector space $V$ are the null space $0$ and the whole vector space $V$.

## Lemma

Let $K$ be a field, and let
    ${\begin{matrix}a_{11}x_{1}+a_{12}x_{2}+\cdots +a_{1n}x_{n}&=&0\\a_{21}x_{1}+a_{22}x_{2}+\cdots +a_{2n}x_{n}&=&0\\\vdots &\vdots &\vdots \\a_{m1}x_{1}+a_{m2}x_{2}+\cdots +a_{mn}x_{n}&=&0\end{matrix}}$
be a homogeneous system of linear equations over $K$. Then the set of all solutions to the system is a linear subspace
$K^{n}$

### Proof

$\Box$
Therefore, we talk about the 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 to the inhomogeneous system again.

## Example

We have a look at the homogeneous version of example, so we consider the homogeneous linear system
    ${\begin{matrix}2x&+5y&+2z&&-v&=&0\\\,3x&-4y&&+u&+2v&=&0\\\,4x&&-2z&+2u&&=&0\,.\end{matrix}}$
over $\mathbb {R}$. Due to fact, the solution set $L$ is a linear subspace of $\mathbb {R} ^{5}$. We have described it explicitly in example as
    ${\left\{u{\left(-{\frac {1}{3}},0,{\frac {1}{3}},1,0\right)}+v{\left(-{\frac {2}{13}},{\frac {5}{13}},-{\frac {4}{13}},0,1\right)}\mid u,v\in \mathbb {R} \right\}},$
which also shows that the solution set is a vector space. With this description, it is clear that $L$ is in bijection with $\mathbb {R} ^{2}$, and this bijection respects the addition and also the scalar multiplication (the solution set $L'$ of the inhomogeneous system is also in bijection with $\mathbb {R} ^{2}$, but there is no reasonable addition nor scalar multiplication on $L'$). However, this bijection depends heavily on the chosen "basic solutions“ ${\left(-{\frac {1}{3}},0,{\frac {1}{3}},1,0\right)}$ 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 "in natural way“, that means independent on any choice, a linear subspace of $K^{n}$ (where $n$ is the number of variables). For this solution space, there always exists a "linear bijection“ (an "isomorphism“) to some $K^{d}$ ($d\leq n$), but for is no natural choice for such a bijection. This is one of the main reasons to work with abstract vector spaces, instead of just $K^{n}$.
    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 "linear combinations“ of finitely many
(simple) solutions. In this and the next lecture, we will develop the concepts to make this precise.

## Definition

Let $K$ be a field, and let $V$ be a $K$-vector space. Let $v_{1},\ldots ,v_{n}$ denote a family of vectors in $V$. Then the vector
    $s_{1}v_{1}+s_{2}v_{2}+\cdots +s_{n}v_{n}{\text{ with }}s_{i}\in K$
is called a linear combination of this vectors
$(s_{1},\ldots ,s_{n})$
Two different coefficient tuples can define the same vector.

## Definition

Let $K$ be a field, and let $V$ be a $K$-vector space. A family $v_{i}\in V$, $i\in I$, is called a generating system (or spanning system) of $V$, if every vector $v\in V$ can be written as
    $v=\sum _{j\in J}s_{j}v_{j}\,,$
with a finite subfamily $J\subseteq I$, and with
$s_{j}\in K$
In $K^{n}$, the standard vectors $e_{i}$, $1\leq i\leq n$, form a generating system. In the polynomial ring $K[X]$, the powers $X^{n}$, $n\in \mathbb {N}$, form an (infinite) generating system.

## Definition

Let $K$ be a field, and let $V$ be a $K$-vector space. For a family $v_{i}$, $i\in I$, we set
    $\langle v_{i},\,i\in I\rangle ={\left\{\sum _{i\in J}s_{i}v_{i}\mid s_{i}\in K,\,J\subseteq I{\text{ finite subset}}\right\}}\,,$
The empty set generates the null space. The null space is also generated by the element $0$. A single vector $v$ spans the space $Kv={\left\{sv\mid s\in K\right\}}$. For $v\neq 0$, this is a line, a term we will make more precise in the framework of dimension theory. For two vectors $v$ and $w$, the "form“ of the spanned space depends on how the two vectors are related to each other. If they both lie on a line, say $w=sv$, 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 (and $v$ and $w$ are not $0$), then the two vectors span a "plane“.
We list some simple properties for generating systems and linear subspaces.

## Lemma

Let $K$ be a field and let $V$ be a
$K$
* Let $U_{j}$, $j\in J$, be a family of subspaces of $V$. Then the intersection is a linear subspace.
    $U=\bigcap _{j\in J}U_{j}\,$
* Let $v_{i}$, $i\in I$, 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 $v_{i}$, $i\in I$, is a system of generators of $V$ if and only if
    $\langle v_{i},\,i\in I\rangle =V\,.$

### Proof

$\Box$
    Footnotes
* ↑ The first four axioms, which are independent of $K$, mean that $(V,0,+)$ is a commutative group.
* ↑ Also for vector spaces, there is the convention that multiplication binds stronger than addition.
* ↑ This follows from the definition, if we use the convention that the empty sum equals $0$.