Linear algebra (Osnabrück 2024-2025)/Part II/Lecture 57

Linear mappings under change of field

Definition

\varphi \colon V\longrightarrow W

between ${}K$ -vector spaces ${}V$ and ${}W$ , and a field extension ${}K\subseteq L$ , the ${}L$ -linear mapping

\varphi _{L}\colon V_{L}=L\otimes _{K}V\longrightarrow W_{L}=L\otimes _{K}W,b\otimes v\longmapsto b\otimes \varphi (v),

is called the

scalar extended linear mapping.

This mapping is not only ${}K$ -linear but also ${}L$ -linear; this was shown in Proposition 56.13 (3).

Let ${}K$ be a field, and let ${}V$ be an ${}n$ -dimensional ${}K$ -vector space, endowed with a basis ${}{\mathfrak {v}}=v_{1},\ldots ,v_{n}$ , and let ${}W$ be an ${}m$ -dimensional ${}K$ -vector space, endowed with of a basis ${}{\mathfrak {w}}=w_{1},\ldots ,w_{m}$ . Let

\varphi \colon V\longrightarrow W

be a linear mapping with the describing matrix ${}M=M_{\mathfrak {w}}^{\mathfrak {v}}(\varphi )={\left(a_{ij}\right)}$ . Let ${}K\subseteq L$ be a field extension. Then the linear mapping

\varphi _{L}\colon V_{L}\longrightarrow W_{L}

given by extension of scalars

is also described by the matrix

{}M

, considered over

{}L

, with respect to the bases

{}1\otimes v_{1},\ldots ,1\otimes v_{n}

of

{}V_{L}

, and

{}1\otimes w_{1},\ldots ,1\otimes w_{m}

of

{}W_{L}

.

Proof

Because of Lemma 56.14 , the given families are indeed bases. The basis element ${}v_{j}$ of ${}V$ is mapped under ${}\varphi$ to

{}\varphi (v_{j})=\sum _{i=1}^{n}a_{ij}w_{i}\,.

Therefore, the basis element ${}1\otimes v_{j}$ of ${}V_{L}$ is mapped under ${}\varphi _{L}$ to

{}\varphi (1\otimes v_{j})=1\otimes \varphi (v_{j})=1\otimes {\left(\sum _{i=1}^{n}a_{ij}w_{i}\right)}=\sum _{i=1}^{n}a_{ij}{\left(1\otimes w_{i}\right)}\,.

The coefficients ${}a_{ij}$ constitute also the describing matrix of ${}\varphi _{L}$ .

\Box

The wedge product

Among the multilinear mappings, the alternating mappings play a special role, the most important example is the determinant. We construct here the so-called wedge product, which plays for the alternating mappings a similar role as the tensor product for the multilinear mappings.

We recall the definition of an alternating mapping.

Let ${}K$ be a field, let ${}V$ and ${}W$ denote ${}K$ -vector spaces, and let ${}n\in \mathbb {N}$ . A multilinear mapping

\Phi \colon V^{n}=\underbrace {V\times \cdots \times V} _{n{\text{-mal}}}\longrightarrow W

is called alternating if the following holds: Whenever in ${}v=(v_{1},\ldots ,v_{n})$ , two entries are identical, that is ${}v_{i}=v_{j}$ for a pair ${}i\neq j$ , then

{}\Phi (v)=0\,.

Construction

Let ${}K$ a field, let ${}V$ be a ${}K$ -vector space, and ${}n\in \mathbb {N}$ . We construct the so-called ${}n$ -th wedge product of ${}V$ with itself, written as ${}\bigwedge ^{n}V$ . For this, we consider the set ${}S$ of all symbols of the form

{(v_{1},\ldots ,v_{n})}{\text{ with }}v_{i}\in V,

and the corresponding set of the ${}e_{(v_{1},\ldots ,v_{n})}$ . We consider the vector space

{}H=K^{(S)}\,,

this is the set of all (finite) sums

a_{1}e_{s_{1}}+\cdots +a_{k}e_{s_{k}}{\text{ with }}a_{i}\in K{\text{ and }}s_{i}\in S;

the ${}e_{s}$ form a basis of this space. This is, with the natural addition and the natural scalar multiplication, a vector space; it is a linear subspace of the mapping space ${}\operatorname {Map} \,{\left(S,K\right)}$ ( ${}H$ is the set of those vectors where almost all elements ${}s\in S$ have the value ${}0$ ). In ${}H$ , we consider the linear subspace ${}U$ that is generated by the following elements (they are called the standard relations of the wedge product).

e_{(v_{1},\ldots ,v_{i-1},v+w,v_{i+1},\ldots ,v_{n})}-e_{(v_{1},\ldots ,v_{i-1},v,v_{i+1},\ldots ,v_{n})}-e_{(v_{1},\ldots ,v_{i-1},w,v_{i+1},\ldots ,v_{n})}

for arbitrary ${}v_{1},\ldots ,v_{i-1},v_{i+1},\ldots ,v_{n},v,w\in V$ .

e_{(v_{1},\ldots ,v_{i-1},av,v_{i+1},\ldots ,v_{n})}-ae_{(v_{1},\ldots ,v_{i-1},v,v_{i+1},\ldots ,v_{n})}

for arbitrary ${}v_{1},\ldots ,v_{i-1},v_{i+1},\ldots ,v_{n},v\in V$ and ${}a\in K$ .

e_{(v_{1},\ldots ,v_{i-1},v,v_{i+1},\ldots ,v_{j-1},v,v_{j+1},\ldots ,v_{n})}

for ${}i<j$ and arbitrary ${}v_{1},\ldots ,v_{i-1},v_{i+1},\ldots ,,v_{j-1},v_{j+1},\ldots ,v_{n},v\in V$ .

Here, the main idea is to enforce the rules that hold for an alternating multilinear mapping by making these relations to ${}0$ . The first type represents the additivity in every argument, the second represents the compatibility with the scalar multiplication, the third represents the alternating property.

We define now

{}\bigwedge ^{n}V:=H/U\,,

that is, we form the residue class space

of

{}H

modulo the linear subspace

{}U

.

The elements ${}e_{(v_{1},\ldots ,v_{n})}$ form a generating system of ${}H$ . The residue class of ${}e_{(v_{1},\ldots ,v_{n})}$ modulo ${}U$ is denoted by^[1]

v_{1}\wedge \ldots \wedge v_{n}.

The standard relations become calculation rules^[2]

{}{\begin{aligned}v_{1}\wedge \ldots \wedge v_{i-1}\wedge (v+w)\wedge v_{i+1}\wedge \ldots \wedge v_{n}&=v_{1}\wedge \ldots \wedge v_{i-1}\wedge v\wedge v_{i+1}\wedge \ldots \wedge v_{n}\,+\,v_{1}\wedge \ldots \wedge v_{i-1}\wedge w\wedge v_{i+1}\wedge \ldots \wedge v_{n}\\\,\end{aligned}}

v_{1}\wedge \ldots \wedge v_{i-1}\wedge av\wedge v_{i+1}\wedge \ldots \wedge v_{n}=a\cdot v_{1}\wedge \ldots \wedge v_{i-1}\wedge v\wedge v_{i+1}\wedge \ldots \wedge v_{n}\,,

and

{}v_{1}\wedge \ldots \wedge v_{i-1}\wedge v\wedge v_{i+1}\wedge \ldots \wedge v_{j-1}\wedge v\wedge v_{j+1}\wedge \ldots \wedge v_{n}=0\,.

Definition

Let ${}K$ be a field, and let ${}V$ denote a ${}K$ -vector space. The ${}K$ -vector space (constructed in Construction 57.3 ) ${}\bigwedge ^{n}V$ is called the ${}n$ -th wedge product (or the ${}n$ -th exterior power, or exterior product) of ${}V$ . The mapping

V^{n}\longrightarrow \bigwedge ^{n}V,(v_{1},\ldots ,v_{n})\longmapsto v_{1}\wedge \ldots \wedge v_{n},

is called the universal alternating mapping.

Calculating rules for the wedge product

Lemma

Let ${}K$ be a field, and let ${}V$ denote a ${}K$ -vector space. Then the exterior powers

fulfill the following properties.

The elements of the form ${}v_{1}\wedge \ldots \wedge v_{n}$ with ${}v_{i}\in V$ form a generating system of ${}\bigwedge ^{n}V$ .
The mapping
$V^{n}\longrightarrow \bigwedge ^{n}V,(v_{1},\ldots ,v_{n})\longmapsto v_{1}\wedge \ldots \wedge v_{n},$

is multilinear and alternating.
We have
$v_{1}\wedge \ldots \wedge v_{i-1}\wedge v\wedge w\wedge v_{i+2}\wedge \ldots \wedge v_{n}=-v_{1}\wedge \ldots \wedge v_{i-1}\wedge w\wedge v\wedge v_{i+2}\wedge \ldots \wedge v_{n}\,.$
Let ${}u_{1},\ldots ,u_{m}\in V$ be given, and let
${}v_{j}=\sum _{i=1}^{m}a_{ij}u_{i}\,$

for ${}j=1,\ldots ,n$ . Then

${}{\begin{aligned}v_{1}\wedge \ldots \wedge v_{n}&=\sum _{(i_{1},\ldots ,i_{n})\in \{1,\ldots ,m\}^{n}}{\left(\prod _{j=1}^{n}a_{i_{j}j}\right)}u_{i_{1}}\wedge \ldots \wedge u_{i_{n}}\\&=\sum _{1\leq i_{1}<\ldots <i_{n}\leq m}{\left(\sum _{\pi \in S_{n}}\operatorname {sgn} (\pi )\prod _{j=1}^{n}a_{i_{\pi (j)}j}\right)}u_{i_{1}}\wedge \ldots \wedge u_{i_{n}}.\end{aligned}}$

Proof

(1) follows directly from construction.
(2). We consider the composed mapping

V^{n}\longrightarrow H\cong K^{(V^{n})}\longrightarrow H/U,

where ${}(v_{1},\ldots ,v_{n})$ is sent to ${}e_{(v_{1},\ldots ,v_{n})}$ , and this is sent to the residue class ${}v_{1}\wedge \ldots \wedge v_{n}$ . The definition of the linear subspace ${}U$ ensures that the multilinearity and the alternating property are satisfied.
(3) hold, due to Lemma 16.8 , for every alternating mapping.
(4). The first equation holds, due to Lemma 16.6 , for every multilinear mapping. If in the index tuple ${}(i_{1},\ldots ,i_{n})$ , an entry appears twice, then ${}u_{i_{1}}\wedge \ldots \wedge u_{i_{n}}=0$ . Hence, we only have to consider tuples where all entries are different. After reordering, they have the form ${}1\leq i_{1}<\ldots <i_{n}\leq m$ . For a fixed increasing index tuple, the sum over all permuted index tuples equals

\sum _{\pi \in S_{n}}\prod _{j=1}^{n}a_{i_{\pi (j)}j}u_{i_{\pi (1)}}\wedge \ldots \wedge u_{i_{\pi (n)}}=\sum _{\pi \in S_{n}}\operatorname {sgn} (\pi )\prod _{j=1}^{n}a_{i_{\pi (j)}j}u_{i_{1}}\wedge \ldots \wedge u_{i_{n}}\,.

\Box

Corollary

Let ${}K$ be a field, and let ${}V$ denote a ${}K$ -vector space of dimension ${}n$ . Let ${}v_{1},\ldots ,v_{n}$ and ${}w_{1},\ldots ,w_{n}$ be vectors in ${}V$ , fulfilling the relation

{}{\begin{pmatrix}v_{1}\\\vdots \\v_{n}\end{pmatrix}}=M{\begin{pmatrix}w_{1}\\\vdots \\w_{n}\end{pmatrix}}\,,

where ${}M$ is an ${}n\times n$ -matrix. Then we have in in ${}\bigwedge ^{n}V$ the identiy

{}v_{1}\wedge \ldots \wedge v_{n}=(\det M)w_{1}\wedge \ldots \wedge w_{n}\,.

Proof

Setting ${}M={\left(b_{jk}\right)}$ , we have ${}v_{j}=\sum _{k=1}^{n}b_{jk}w_{k}$ , and, with the transposed matrix ${}{M^{\text{tr}}}={\left(a_{ij}\right)}$ , we have ${}v_{j}=\sum _{i=1}^{n}a_{ij}w_{i}$ . Hence, we are in the situation of Lemma 57.5 (4), and we have

{}{\begin{aligned}v_{1}\wedge \ldots \wedge v_{n}&=\sum _{1\leq i_{1}<\ldots <i_{n}\leq n}{\left(\sum _{\pi \in S_{n}}\operatorname {sgn} (\pi )\prod _{j=1}^{n}a_{i_{\pi (j)}j}\right)}w_{i_{1}}\wedge \ldots \wedge w_{i_{n}}\\&=\sum _{\pi \in S_{n}}\operatorname {sgn} (\pi )\prod _{j=1}^{n}a_{{\pi (j)}j}w_{1}\wedge \ldots \wedge w_{n},\end{aligned}}

because ${}i_{j}=j$ must hold. Therefore, the statement follows from the Leibniz formula for the determinant.

\Box

The universal property of the wedge product

The following statement gives the universal property of the wedge product.

Theorem

Let ${}K$ be a field, let ${}V$ be a ${}K$ -vector space, and ${}n\in \mathbb {N}$ . Let

\psi \colon V^{n}\longrightarrow W

be an alternating multilinear mapping in another ${}K$ -vector space ${}W$ . Then there exists a uniquely determined linear mapping

{\tilde {\psi }}\colon \bigwedge ^{n}V\longrightarrow W

such that the diagram

{\begin{matrix}V^{n}&{\stackrel {}{\longrightarrow }}&\bigwedge ^{n}V&\\&\searrow &\downarrow &\\&&W&\!\!\!\!\!\!\!\!\\\end{matrix}}

commutes.

Proof

We use the notation from Construction 57.3 . The assignment

e_{(v_{1},\ldots ,v_{n})}\longmapsto \psi (v_{1},\ldots ,v_{n})

defines, due to Theorem 10.10 , a ${}K$ -linear mapping

{\bar {\psi }}\colon H\longrightarrow W.

As ${}\psi$ is multilinear and alternating, ${}{\bar {\psi }}$ sends the linear subspace ${}U\subseteq H$ to ${}0$ . According to Theorem 47.16 , there exists a ${}K$ -linear mapping

{\tilde {\psi }}\colon H/U\longrightarrow W,

which is compatible with ${}{\bar {\psi }}$ .
The uniqueness follows from the fact that the ${}v_{1}\wedge \ldots \wedge v_{n}$ form a generating system of ${}\bigwedge ^{n}V$ , and these have to be sent to ${}\psi (v_{1},\ldots ,v_{n})$ .

\Box

We denote by ${}\operatorname {Alt} ^{n}(V,K)$ the set of all alternating mappings from ${}V^{n}$ to ${}K$ . This set carries a natural ${}K$ -vector space structure; see Exercise 16.28 .

Corollary

Let ${}K$ be a field, let ${}V$ be a ${}K$ -vector space, and ${}n\in \mathbb {N}$ . Then there exists a natural isomorphism

{\left(\bigwedge ^{n}V\right)}^{*}\longrightarrow \operatorname {Alt} ^{n}(V,K),\psi \longmapsto ((v_{1},\ldots ,v_{n})\mapsto \psi (v_{1}\wedge \ldots \wedge v_{n})).

Proof

The mapping is just the composition ${}\psi \mapsto \psi \circ \delta$ , where ${}\delta \colon V\times \cdots \times V\rightarrow \bigwedge ^{n}V$ denotes the canonical mapping. The linearity of the assignment comes from the linear structures on the dual space, and on the space of alternating forms. The bijectivity of the mapping follows from Theorem 57.7 , applied to ${}W=K$ .

\Box

Theorem

Let ${}K$ be a field, and let ${}V$ denote a ${}K$ -vector space, and ${}k\in \mathbb {N} _{+}$ . Then there exists a canonical surjective linear mapping

V\otimes _{}\cdots \otimes _{}V\longrightarrow V\wedge \ldots \wedge V,v_{1}\otimes _{}\cdots \otimes _{}v_{k}\longmapsto v_{1}\wedge \ldots \wedge v_{k}.

Proof

The multilinear mapping

V^{k}\longrightarrow \bigwedge ^{k}V,(v_{1},\ldots ,v_{k})\longmapsto v_{1}\wedge \ldots \wedge v_{k},

induces, due to Theorem 55.4 (2), the linear mapping. The surjectivity rests on the fact that the generating system ${}v_{1}\wedge \ldots \wedge v_{k}$ belongs to the the image.

\Box

If ${}V$ is finite-dimensional, then the preceding statement and Corollary 55.13 imply that the wedge product has finite dimension. We will determine this dimension in the last lecture.

Footnotes

↑
It is not easy to build up an imagination for the expressions ${}v_{1}\wedge \ldots \wedge v_{n}$ and ${}\wedge$ . More important than the "meaning“ of this symbols is their transformation behavior, and the calculation rules. It is rather the working with these symbols that give them their meaning. However, we can say in a vague sense that ${}v_{1}\wedge \ldots \wedge v_{n}$ represents the "oriented“ parallelotope generated by the vectors ${}v_{1},\ldots ,v_{n}$ . The wedge product ${}\bigwedge ^{n}V$ consists of all linear combinations of such parallelotopes.
↑ We follow the convention that the expression ${}av_{1}\wedge \ldots \wedge v_{n}$ means ${}a(v_{1}\wedge \ldots \wedge v_{n})$ . However, the equations
${}a(v_{1}\wedge \ldots \wedge v_{n})=(av_{1})\wedge \ldots \wedge v_{n}=v_{1}\wedge \ldots \wedge (av_{n})\,$
hold anyway.

<< \| Linear algebra (Osnabrück 2024-2025)/Part II \| >> PDF-version of this lecture Exercise sheet for this lecture (PDF)

[1] 
It is not easy to build up an imagination for the expressions ${}v_{1}\wedge \ldots \wedge v_{n}$ and ${}\wedge$ . More important than the "meaning“ of this symbols is their transformation behavior, and the calculation rules. It is rather the working with these symbols that give them their meaning. However, we can say in a vague sense that ${}v_{1}\wedge \ldots \wedge v_{n}$ represents the "oriented“ parallelotope generated by the vectors ${}v_{1},\ldots ,v_{n}$ . The wedge product ${}\bigwedge ^{n}V$ consists of all linear combinations of such parallelotopes.

[2] We follow the convention that the expression ${}av_{1}\wedge \ldots \wedge v_{n}$ means ${}a(v_{1}\wedge \ldots \wedge v_{n})$ . However, the equations
${}a(v_{1}\wedge \ldots \wedge v_{n})=(av_{1})\wedge \ldots \wedge v_{n}=v_{1}\wedge \ldots \wedge (av_{n})\,$
hold anyway.

[1]

[2]