# Mathematics for Applied Sciences (Osnabrück 2023-2024)/Part I/Lecture 23/refcontrol

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 "linear combinations“ of finitely many (simple) solutions. In this lecture, we develop the concepts to make this precise.
    Generating systems

## Definition

Let $K$ be a field,MDLD/field and let $V$ be a $K$-vector space.MDLD/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,MDLD/field and let $V$ be a $K$-vector space.MDLD/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,MDLD/field and let $V$ be a $K$-vector space.MDLD/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.

## LemmaCreate referencenumber

Let $K$ be a field,MDLD/field and let $V$ be a
$K$
* For a family $v_{i}$, $i\in I$, of elements in $V$, the linear spanMDLD/linear span is a linear subspaceMDLD/linear subspace of $V$.
* The family $v_{i}$, $i\in I$, is a spanning system of $V$, if and only if
    $\langle v_{i},\,i\in I\rangle =V\,.$

### Proof

$\Box$
    Linear independence

## Definition

Let $K$ be a field,MDLD/field and let $V$ be a $K$-vector space.MDLD/vector space A family of vectors $v_{i}$, $i\in I$, (where $I$ denotes a finite index set) is called linearly independent, if an equation of the form
    $\sum _{i\in I}s_{i}v_{i}=0{\text{ with }}s_{i}\in K$
is only possible when $s_{i}=0$
$i$
If a family is not linear independent, then it is called linearly dependent. A linear combination $\sum _{i\in I}s_{i}v_{i}=0$ is called a representation of the null vector. It is called the trivial representation, if all coefficients $s_{i}$ equal $0$, and, if at least one coefficient is not $0$, a 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.
== Example Example 23.6
The standard vectorsMDLD/standard vectors in $K^{n}$ are linearly independent.MDLD/linearly independent A representation
    $\sum _{i=1}^{n}s_{i}e_{i}=0\,$
just means
    $s_{1}{\begin{pmatrix}1\\0\\\vdots \\0\end{pmatrix}}+s_{2}{\begin{pmatrix}0\\1\\\vdots \\0\end{pmatrix}}+\cdots +s_{n}{\begin{pmatrix}0\\0\\\vdots \\1\end{pmatrix}}={\begin{pmatrix}0\\0\\\vdots \\0\end{pmatrix}}\,.$
The $i$-th row yields directly $s_{i}=0$.

## Example Create referencenumber

The three vectors
    ${\begin{pmatrix}3\\3\\3\end{pmatrix}},\,\,{\begin{pmatrix}0\\4\\5\end{pmatrix}}\,{\text{ and }}\,{\begin{pmatrix}4\\8\\9\end{pmatrix}}$
are linearly dependent.MDLD/linearly dependent The equation
    $4{\begin{pmatrix}3\\3\\3\end{pmatrix}}+3{\begin{pmatrix}0\\4\\5\end{pmatrix}}-3{\begin{pmatrix}4\\8\\9\end{pmatrix}}={\begin{pmatrix}0\\0\\0\end{pmatrix}}\,$
is a nontrivial representation of the null vector.

## LemmaCreate referencenumber

Let $K$ be a field,MDLD/field let $V$ be a $K$-vector space,MDLD/vector space and let $v_{i}$, $i\in I$,
$V$
* If the family is linearly independent, then for each subset $J\subseteq I$, also the family $v_{i}$ , $i\in J$, 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 $v\neq 0$.
* Two vectors $v$ and $u$ are linearly independent if and only if $u$ is not a scalar multiple of $v$ and vice versa.

### Proof

$\Box$

## Remark

The vectors $v_{1}={\begin{pmatrix}a_{11}\\\vdots \\a_{m1}\end{pmatrix}},\ldots ,v_{n}={\begin{pmatrix}a_{1n}\\\vdots \\a_{mn}\end{pmatrix}}\in K^{m}$ are linearly dependent,MDLD/linearly dependent if and only if the homogeneous linear systemMDLD/homogeneous linear system
    ${\begin{matrix}a_{11}x_{1}+a_{12}x_{2}+\cdots +a_{1n}x_{n}&=&0\\a_{21}x_{1}+a_{22}x_{2}+\cdots +a_{2n}x_{n}&=&0\\\vdots &\vdots &\vdots \\a_{m1}x_{1}+a_{m2}x_{2}+\cdots +a_{mn}x_{n}&=&0\end{matrix}}$
has a nontrivial solution.
    Basis

## Definition

Let $K$ be a field,MDLD/field and let $V$ be a $K$-vector space.MDLD/vector space Then a linearly independentMDLD/linearly independent generating systemMDLD/generating system (vs) $v_{i}\in V$, $i\in I$,
$V$
$V$

## Example Create referencenumber

The standard vectorsMDLD/standard vectors in $K^{n}$ form a basis.MDLD/basis (vs) The linear independenceMDLD/linear independence was shown in Example 23.6 . To show that they also form a generating system,MDLD/generating system (vs) let
    $v={\begin{pmatrix}b_{1}\\b_{2}\\\vdots \\b_{n}\end{pmatrix}}\in K^{n}\,$
be an arbitrary vector. Then we have immediately
    $v=\sum _{i=1}^{n}b_{i}e_{i}\,.$
Hence, we have a basis, which is called the standard basis of $K^{n}$.

## TheoremTheorem 23.12 change

Let $K$ be a field,MDLD/field and let $V$ be a $K$-vector space.MDLD/vector space Let $v_{1},\ldots ,v_{n}\in V$
* The family is a basisMDLD/basis (vs) of $V$.
* The family is a minimal generating system,MDLD/generating system (vs) that is, as soon as we remove one vector $v_{i}$, the remaining family is not a generating system any more.
* For every vector $u\in V$, there is exactly one representation
    $u=s_{1}v_{1}+\cdots +s_{n}v_{n}\,.$
* The family is maximal linearly independent,MDLD/linearly independent that is, as soon as some vector is added, the family is not linearly independent any more.

### Proof

$\Box$

## Remark

Let a basisMDLD/basis (vs) $v_{1},\ldots ,v_{n}$ of a $K$-vector spaceMDLD/vector space $V$ be given. Due to Theorem 23.12 , this means that for every vector $u\in V$, there exists a uniquely determined representation
    $u=s_{1}v_{1}+s_{2}v_{2}+\cdots +s_{n}v_{n}\,.$
The elements $s_{i}\in K$ (scalars) are called the coordinates of $u$ with respect to the given basis. Thus, for a fixed basis, we have a (bijective) correspondence between the vectors from $V$, and the coordinate tuples $(s_{1},s_{2},\ldots ,s_{n})\in K^{n}$. We express this by saying that a basis determines a linear coordinate system.

## TheoremCreate referencenumber

Let $K$ be a field,MDLD/field and let $V$ be a $K$-vector spaceMDLD/vector space with a finite generating system.MDLD/generating system (vs) Then $V$ has a finite

### Proof

Let $v_{i}$, $i\in I$, be a finite generating system of $V$, with a finiteMDLD/finite (set) 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 $k\in I$, such that the remaining family where $v_{k}$ is removed, that is $v_{i}$, $i\in I\setminus \{k\}$, 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 $J\subseteq I$ such that $v_{i}$, $i\in J$, is a minimal generating set, hence a basis.
$\Box$
    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.

## TheoremTheorem 23.15 change

Let $K$ be a field,MDLD/field and let $V$ be a $K$-vector spaceMDLD/vector space with a finite generating system.MDLD/generating system (vs) Then any two basesMDLD/bases (vs)
$V$

### Proof

$\Box$
This theorem enables the following definition.

## Definition

Let $K$ be a field,MDLD/field and let $V$ be a $K$-vector spaceMDLD/vector space with a finite generating system.MDLD/generating system (vs) Then the number of vectors in any basisMDLD/basis (vs) of $V$, is called the dimension of $V$, written
    $\dim _{\left(V\right)}.$
Due to the preceding theorem, the dimension is well-defined. If a vector space is not finitely generated, then one puts $\dim _{\left(V\right)}=\infty$. The null space $0$ has dimension $0$. A one-dimensional vector space is called a line, a two-dimensional vector space a plane, a three-dimensional vector space a space (in the strict sense) but every vector space is called a space.

## CorollaryCreate referencenumber

Let $K$ be a field and $n\in \mathbb {N}$. Then the standard spaceMDLD/standard space $K^{n}$ has the
$n$

### Proof

The standard basisMDLD/standard basis (vs) $e_{i}$, $i=1,\ldots ,n$, consists of $n$ vectors, hence the dimension is $n$.
$\Box$

## Example Create referencenumber

The complex numbersMDLD/complex numbers form a two-dimensional real vector space,MDLD/vector space a basis is $1$ and ${\mathrm {i} }$.

## Example Create referencenumber

The polynomial ringMDLD/polynomial ring (K1) $R=K[X]$ over a fieldMDLD/field $K$ is not a finite-dimensional vector space.MDLD/finite-dimensional vector space To see this, we have to show that there is no finite generating systemMDLD/generating system (vs) for the polynomial ring. Consider $n$ polynomials $P_{1},\ldots ,P_{n}$. Let $d$ be the maximum of the degreesMDLD/degrees (polynomial) of these polynomials. Then every $K$-linear combinationMDLD/linear combination $\sum _{i=1}^{n}a_{i}P_{i}$ has at most degree $d$. In particular, polynomials of larger degree can not be presented by $P_{1},\ldots ,P_{n}$, so these do not form a generating system for all polynomials.

## CorollaryCorollary 23.20 change

Let $V$ denote a finite-dimensional vector space over a field $K$. Let $U\subseteq V$ denote a linear subspace.MDLD/linear subspace Then $U$ is also finite-dimensional, and the estimate
    $\dim _{\left(U\right)}\leq \dim _{\left(V\right)}\,$

### Proof

$\Box$

## CorollaryCorollary 23.21 change

Let $K$ be a field,MDLD/field and let $V$ be a $K$-vector spaceMDLD/vector space with finite dimensionMDLD/dimension (fgvs) $n=\dim _{\left(V\right)}$. Let $n$ vectors $v_{1},\ldots ,v_{n}$ in $V$ be given. Then the following properties are equivalent.
* $v_{1},\ldots ,v_{n}$ form a basisMDLD/basis (vs) of $V$.
* $v_{1},\ldots ,v_{n}$ form a generating systemMDLD/generating system (vs) of $V$.
* $v_{1},\ldots ,v_{n}$ are linearly independent.MDLD/linearly independent

### Proof

$\Box$

## Example Create referencenumber

Let $K$ be a field.MDLD/field It is easy to get an overview over the linear subspacesMDLD/linear subspaces of $K^{n}$, as the dimensionMDLD/dimension (fgvs) of a linear subspace equals $k$ with $0\leq k\leq n$, due to Corollary 23.20 . For $n=0$, there is only the null space itself, for $n=1$ there is the null space and $K$ itself. For $n=2$, there is the null space, the whole plane $K^{2}$, and the one-dimensional lines through the origin. Every line $G$ has the form
    $G=Kv={\left\{sv\mid s\in K\right\}}\,,$
with a vector $v\neq 0$. Two vectors different from $0$ define the same line if and only if they are linearly dependent.MDLD/linearly dependent For $n=3$, 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.

## TheoremTheorem 23.23 change

Let $V$ denote a finite-dimensional vector space over a field $K$. Let
    $u_{1},\ldots ,u_{k}$
denote linearly independentMDLD/linearly independent vectors in $V$. Then there exist vectors
    $u_{k+1},\ldots ,u_{n}$
such that
    $u_{1},\ldots ,u_{k},u_{k+1},\ldots ,u_{n}$
form a basisMDLD/basis (vs)
$V$

### Proof

$\Box$
    Footnotes
* ↑ This follows from the definition, if we use the convention that the empty sum equals $0$.
* ↑ 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 (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.