# Linear algebra (Osnabrück 2024-2025)/Part I/Lecture 14

    Linear forms

## Definition

Let $K$ be a field and let $V$ be a $K$-vector space. A linear mapping
    $V\longrightarrow K$
$V$

## Example

A linear form on $K^{n}$ is of the form
    $K^{n}\longrightarrow K,\left(x_{1},\,\ldots ,\,x_{n}\right)\longmapsto \sum _{i=1}^{n}a_{i}x_{i},$
for a tuple $\left(a_{1},\,\ldots ,\,a_{n}\right)$. The projections
    $p_{j}\colon K^{n}\longrightarrow K,\left(x_{1},\,\ldots ,\,x_{n}\right)\longmapsto x_{j},$
are the easiest linear forms.
The zero mapping to $K$ is also a linear form, called the zero form.
We have encountered many linear forma already, for example, the price function for a purchase of several products, or the content of vitamin of fruit salads of various fruits. With respect to a basis $v_{1},\ldots ,v_{n}$ of $V$ and a basis $w$ of $K$ (where $w$ is just an element from $K$ different from $0$), the describing matrix of a linear form is simply a row with $n$ entries.

## Example

Many important examples of linear forms on some vector spaces of infinite dimension arise in analysis. For a real interval $[a,b]$, the set of functions $\operatorname {Map} \,{\left([a,b],\mathbb {R} \right)}$, or the set of continuous functions $C([a,b],\mathbb {R} )$, or the set of continuously differentiable functions $C^{1}([a,b],\mathbb {R} )$ form real vector spaces. For a point $P\in [a,b]$, the evaluation $f\mapsto f(P)$ is a linear form (because addition and scalar multiplication is defined pointwisely on these spaces). Also, the evaluation of the derivative at $P$,
    $C^{1}([a,b],\mathbb {R} )\longrightarrow \mathbb {R} ,f\longmapsto f'(P),$
is a linear form. For $C([a,b],\mathbb {R} )$, the integral, that is, the mapping
    $C([a,b],\mathbb {R} )\longrightarrow \mathbb {R} ,f\longmapsto \int _{a}^{b}f(t)dt,$
is a linear form. This rests on the linearity of the integral.

## Remark

Let $K$ be a field, and let $V$ and $W$ denote vector spaces over $K$. For a linear form
    $f\colon V\longrightarrow K$
and a vector $w\in W$, the mapping
    $fw\colon V\longrightarrow W,v\longmapsto f(v)w,$
is linear. It is just the composition
    $V{\stackrel {f}{\longrightarrow }}K{\stackrel {\iota _{w}}{\longrightarrow }}W,$
where $\iota _{w}$ denotes the mapping $s\rightarrow sw$.
The kernel of the zero form is the total space; for any other linear form $f\in \operatorname {Hom} _{K}{\left(V,K\right)}$ with $f\neq 0$, the dimension is $\dim _{K}{\left(V\right)}-1$. This follows from the dimension formula. With the exception of the zero form, a linear form is always surjective.

## Lemma

Let $V$ be an $n$-dimensional $K$-vector space, and let $U\subseteq V$ denote an $n-1$-dimensional linear subspace. Then there exists a linear form $f\colon V\rightarrow K$ such that
$U=\operatorname {kern} f$

### Proof

$\Box$

## Lemma

Let $V$ denote a $K$-vector space and let $v\in V$ be a vector different from $0$. Then there exists a linear form $f\colon V\rightarrow K$ such that
$f(v)\neq 0$

### Proof

The one-dimensional $K$-linear subspace $Kv\subseteq V$ contains a direct complement, that is,
    $V=Kv\oplus U\,$
with some linear subspace $U\subseteq V$. The projection onto $Kv$ for this decomposition sends $v$ to $1$.
$\Box$

## Lemma

Let $K$ be a field, and let $V$ be a $K$-vector space. Let $v_{1},\ldots ,v_{n}\in V$ be vectors. Suppose that for every $k$, there exists a linear form
    $\varphi _{k}\colon V\longrightarrow K$
such that
    $\varphi _{k}(v_{k})\neq 0{\text{ and }}\varphi _{k}(v_{i})=0{\text{ for }}i\neq k.$
$v_{1},\ldots ,v_{n}$

### Proof

$\Box$
    The dual space

## Definition

Let $K$ be a field and let $V$ denote an $K$-vector space. Then the space of homomorphisms
    ${V}^{*}:=\operatorname {Hom} _{K}{\left(V,K\right)}\,$
$V$
Addition and scalar multiplication are defined as in the general case of a space of homomorphisms, thus $(f+g)(v):=f(v)+g(v)$ and $(sf)(v):=s\cdot f(v)$. For a finite-dimensional $V$, we obtain, due to Corollary 13.12 , that the dimension of the dual space ${V}^{*}$ equals the dimension of $V$.

## Definition

Let $V$ denote a finite-dimensional $K$-vector space, endowed with a basis $v_{1},\ldots ,v_{n}$. Then the linear forms
    $v_{1}^{*},\ldots ,v_{n}^{*}\in {V}^{*}\,,$
defined by
    $v_{i}^{*}(v_{j})=\delta _{ij}={\begin{cases}1,{\text{ falls }}i=j\,,\\0,{\text{ falls }}i\neq j\,,\end{cases}}\,$
are called the dual basis
Because of Theorem 10.10 , this rule defines indeed a linear form. The linear form $v_{i}^{*}$ assigns to an arbitrary vector $v\in V$ the $i$-th coordinate of $v$ with respect to the given basis. Note that for $v=\sum _{j=1}^{n}s_{j}v_{j}$, we have
    $v_{i}^{*}(v)=v_{i}^{*}{\left(\sum _{j=1}^{n}s_{j}v_{j}\right)}=\sum _{j=1}^{n}s_{j}v_{i}^{*}(v_{j})=s_{i}\,.$
It is important to stress that $v_{i}^{*}$ does not only depend on the vector $v_{i}$, but on the basis. There doe not exist something like a "dual vector“ for a vector. This looks different in the situation where an inner product is given on $V$.

## Example

For the standard basis $e_{1},\ldots ,e_{n}$ of $K^{n}$, the dual basis consists in the projections onto some component, that is, we have $e_{i}^{*}=p_{i}$, where
    $p_{i}\colon K^{n}\longrightarrow K,\left(x_{1},\,\ldots ,\,x_{n}\right)\longmapsto x_{i}.$
This basis is called the standard dual basis.

## Lemma

Let $V$ be a finite-dimensional $K$-vector space, endowed with a basis $v_{1},\ldots ,v_{n}$. Then the dual basis
    $v_{1}^{*},\ldots ,v_{n}^{*}\in {V}^{*}\,$
is a basis of the

### Proof

Suppose that
    $\sum _{j=1}^{n}a_{j}v_{j}^{*}=0\,,$
where $a_{j}\in K$. If we apply this linear form to $v_{i}$, we get directly
    $a_{i}=0\,.$
Therefore, the $v_{1}^{*},\ldots ,v_{n}^{*}$ are linearly independent. Due to Corollary 13.12 , the dual space has dimension $n$, thus we have a basis already.
$\Box$

## Lemma

Let $V$ be a finite-dimensional $K$-vector space, endowed with a basis $v_{1},\ldots ,v_{n}$, and the corresponding dual basis
    $v_{1}^{*},\ldots ,v_{n}^{*}\in {V}^{*}\,.$
$v\in V$, the equality
    $v=\sum _{i=1}^{n}v_{i}^{*}(v)v_{i}\,.$
holds. The linear forms $v_{i}^{*}$ yield the scalars (coordinates)

### Proof

The vector $v$ has a unique representation
    $v=\sum _{j=1}^{n}s_{j}v_{j}\,$
with $s_{j}\in K$. The right hand side of the claimed equality is therefore
    $\sum _{i=1}^{n}v_{i}^{*}(v)v_{i}=\sum _{i=1}^{n}v_{i}^{*}{\left(\sum _{j=1}^{n}s_{j}v_{j}\right)}v_{i}=\sum _{i=1}^{n}s_{i}v_{i}=v\,.$
$\Box$

## Lemma

Let $V$ be a finite-dimensional $K$-vector space. Let $v_{1},\ldots ,v_{n}$ be a basis of $V$ with the dual basis $v_{1}^{*},\ldots ,v_{n}^{*}$, and let $w_{1},\ldots ,w_{n}$ be another basis with the dual basis $w_{1}^{*},\ldots ,w_{n}^{*}$, and with
    $w_{r}=\sum _{k=1}^{n}a_{kr}v_{k}\,.$
    $w_{j}^{*}=\sum _{i=1}^{n}b_{ij}v_{i}^{*}\,,$
where ${\left(b_{ij}\right)}_{ij}={{\left(A^{-1}\right)}^{\text{tr}}}$ is the transposed matrix of the
inverse matrix of
$A={\left(a_{kr}\right)}_{kr}$

### Proof

We have
    ${\begin{aligned}{\left(\sum _{i=1}^{n}b_{ij}v_{i}^{*}\right)}(w_{r})&={\left(\sum _{i=1}^{n}b_{ij}v_{i}^{*}\right)}{\left(\sum _{k=1}^{n}a_{kr}v_{k}\right)}\\&=\sum _{1\leq i,k\leq n}b_{ij}a_{kr}v_{i}^{*}(v_{k})\\&=\sum _{i=1}^{n}b_{ij}a_{ir}.\end{aligned}}$
Here, we have the "product“ of the $j$-th column of $B$ and the $r$-th column of $A$, which is also the product of the $j$-th row of ${B^{\text{tr}}}=A^{-1}$ and the $r$-te column of $A$. For $r=j$, this is $1$, and for $r\neq j$, this is $0$. Therefore, the given linear form coincides with $w_{j}^{*}$.
$\Box$
With the help of transformation matrices, this can be expressed as
    $M_{\mathfrak {v^{*}}}^{\mathfrak {w^{*}}}={({\left(M_{\mathfrak {v}}^{\mathfrak {w}}\right)}^{-1})^{\text{tr}}}={(M_{\mathfrak {w}}^{\mathfrak {v}})^{\text{tr}}}\,.$

## Example

We consider $\mathbb {R} ^{2}$ with the standard basis $e_{1},e_{2}$, its dual basis $e_{1}^{*},e_{2}^{*}$, and the basis consisting in $u_{1}={\begin{pmatrix}2\\1\end{pmatrix}}$ and $u_{2}={\begin{pmatrix}-1\\3\end{pmatrix}}$. We want to express the dual basis $u_{1}^{*}$ and $u_{2}^{*}$ as a linear combination of the standard dual basis, that is, we want to determine the coefficients $a$ and $b$ (and $c$ and $d$) in
    $u_{1}^{*}=ae_{1}^{*}+be_{2}^{*}\,$
(and in $u_{2}^{*}=ce_{1}^{*}+de_{2}^{*}$). Here, $a=u_{1}^{*}(e_{1})$ and $b=u_{1}^{*}(e_{2})$. In order to compute this, we have to express $e_{1}$ and $e_{2}$ as a linear combination of $u_{1}$ and $u_{2}$. This is
    $e_{1}={\frac {3}{7}}{\begin{pmatrix}2\\1\end{pmatrix}}-{\frac {1}{7}}{\begin{pmatrix}-1\\3\end{pmatrix}}\,$
and
    $e_{2}={\frac {1}{7}}{\begin{pmatrix}2\\1\end{pmatrix}}+{\frac {2}{7}}{\begin{pmatrix}-1\\3\end{pmatrix}}\,.$
Therefore, we have
    $a=u_{1}^{*}(e_{1})=u_{1}^{*}{\left({\frac {3}{7}}u_{1}-{\frac {1}{7}}u_{2}\right)}={\frac {3}{7}}\,$
and
    $b=u_{1}^{*}(e_{2})=u_{1}^{*}{\left({\frac {1}{7}}u_{1}+{\frac {2}{7}}u_{2}\right)}={\frac {1}{7}}\,.$
Hence,
    $u_{1}^{*}={\frac {3}{7}}e_{1}^{*}+{\frac {1}{7}}e_{2}^{*}\,.$
With similar computations we get
    $u_{2}^{*}=-{\frac {1}{7}}e_{1}^{*}+{\frac {2}{7}}e_{2}^{*}\,.$
The transformation matrix from $u^{*}$ to $e^{*}$ is thus
    $M_{\mathfrak {e^{*}}}^{\mathfrak {u^{*}}}={\begin{pmatrix}{\frac {3}{7}}&-{\frac {1}{7}}\\{\frac {1}{7}}&{\frac {2}{7}}\end{pmatrix}}\,.$
The transposed matrix of this is
    ${{\left(M_{\mathfrak {e^{*}}}^{\mathfrak {u^{*}}}\right)}^{\text{tr}}}={\begin{pmatrix}{\frac {3}{7}}&{\frac {1}{7}}\\-{\frac {1}{7}}&{\frac {2}{7}}\end{pmatrix}}={\begin{pmatrix}2&-1\\1&3\end{pmatrix}}^{-1}={\left(M_{\mathfrak {e}}^{\mathfrak {u}}\right)}^{-1}\,.$
The inverse task to express the standard dual basis with $u_{1}^{*}$ and $u_{2}^{*}$, is easier to solve, because we can read of directly the representations of the $u_{i}$ with respect to the standard basis. We have
    $e_{1}^{*}=2u_{1}^{*}+u_{2}^{*}\,$
and
    $e_{2}^{*}=-u_{1}^{*}+3u_{2}^{*}\,,$
as becomes clear by evaluation $u_{1},u_{2}$ on both sides.
    The trace

## Definition

Let $K$ be a field and let $M={\left(a_{ij}\right)}_{ij}$ be an $n\times n$-matrix over $K$. Then
    $\operatorname {trace} {\left(M\right)}:=\sum _{i=1}^{n}a_{ii}\,$
$M$

## Definition

Let $K$ be a field, and let $V$ denote a finite-dimensional $K$-vector space. Let $\varphi \colon V\rightarrow V$ be a linear mapping, which is described by the matrix $M$ with respect to a basis. Then $\operatorname {trace} {\left(M\right)}$ is called the trace
$\varphi$
$\operatorname {trace} {\left(\varphi \right)}$
Because of Exercise 14.15 , this is independent of the basis chosen. The trace is a linear form on the vector space of all square matrices, and on the vector space of all endomorphisms.
    Footnotes
* ↑ This symbol is called Kronecker-Delta.