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

Linear forms

Example

In a shop, there are ${}n$ different products. A purchase is described by an ${}n$-tuple (purchase tuple), where the ${}i$-th entry says how much of the ${}i$-th product is bought (with respect to a certain unit). The set of all purchases (including refunds) forms an ${}n$-dimensional vector space. A price list for the products is also described by an ${}n$-tuple (price tuple), where the ${}i$-th entry tells the price of the ${}i$-th product (with respect to the same unit). The set of all price tuples forms also an ${}n$-dimensional vector space (think about comparing prices, inflation, tax, etc.). It is obviously nonsense to consider a purchase tuple and a price tuple in the same vector space and to add them. To the contrary, the right way to process a purchase tuple ${}\left(x_{1},\,x_{2},\,\ldots ,\,x_{n}\right)$ and a price tuple ${}\left(p_{1},\,p_{2},\,\ldots ,\,p_{n}\right)$ is to compute the total price ${}\sum _{i=1}^{n}p_{i}x_{i}$, which belongs to ${}\mathbb {R}$. Purchase tuple and price tuple are dual to each other, the vector space of price tuples is dual to the vector space of purchase tuples.

Definition

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

$V\longrightarrow K$

is called a Linear form on ${}V$.

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

See Exercise 14.5 .

$\Box$

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

$\varphi _{k}(v_{k})\neq 0{\text{ and }}\varphi _{k}(v_{i})=0{\text{ for }}i\neq k.$

Then the ${}v_{1},\ldots ,v_{n}$ are linearly independent.

Proof

See Exercise 14.7 .

$\Box$

The dual space

Definition

Let ${}K$ be a field and let ${}V$ denote an ${}K$-vector space. Then the space of homomorphisms

${}{V}^{*}:=\operatorname {Hom} _{K}{\left(V,K\right)}\,$

is called the dual space of ${}V$.

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

of the given 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

dual space.

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

${}v_{1}^{*},\ldots ,v_{n}^{*}\in {V}^{*}\,.$

Then, for every vector

${}v\in V$, the equality

${}v=\sum _{i=1}^{n}v_{i}^{*}(v)v_{i}\,.$

holds. The linear forms ${}v_{i}^{*}$ yield the scalars (coordinates)

of a vector with respect to a basis.

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

${}w_{r}=\sum _{k=1}^{n}a_{kr}v_{k}\,.$

Then

${}w_{j}^{*}=\sum _{i=1}^{n}b_{ij}v_{i}^{*}\,,$

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

${}\operatorname {trace} {\left(M\right)}:=\sum _{i=1}^{n}a_{ii}\,$

is called the trace of ${}M$.

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

of ${}\varphi$, written as ${}\operatorname {trace} {\left(\varphi \right)}$.

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

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