Linear algebra (Osnabrück 2024-2025)/Part II/Lecture 58

We show first that we have a generating system. Because the elements of the form ${\displaystyle {}w_{1}\wedge \ldots \wedge w_{n}}$ form a generating system of ${\displaystyle {}\bigwedge ^{n}V}$ due to Lemma 57.5   (1), it is enough to show that these elements can be represented. For every ${\displaystyle {}w_{j}}$, there exists a representation ${\displaystyle {}w_{j}=\sum _{i=1}^{m}a_{ij}v_{i}}$; therefore, according to Lemma 57.5   (4). we can write the ${\displaystyle {}w_{1}\wedge \ldots \wedge w_{n}}$ as linear combinations of wedge products of the basis elements; however, every ordering may occur. Hence, let ${\displaystyle {}v_{k_{1}}\wedge \ldots \wedge v_{k_{n}}}$ be given, with ${\displaystyle {}k_{j}\in \{1,\ldots ,m\}}$. By swapping neighboring vectors, using Lemma 57.5   (3), we may achieve (maybe with another sign) that the indices are (not necessarily strict) increasing. If an index appears twice, the wedge product is ${\displaystyle {}0}$, due to Lemma 57.5   (2). Hence, no index occurs twice, and this wedge product is in the form asked for.

To show that the family is linearly independent, we show, using Lemma 14.8 , that for every subset ${\displaystyle {}I=\{i_{1},\ldots ,i_{n}\}\subseteq \{1,\ldots ,m\}}$ with ${\displaystyle {}n}$ elements (where ${\displaystyle {}i_{1}<\ldots <i_{n}}$), there exists a ${\displaystyle {}K}$-linear mapping

${\displaystyle \bigwedge ^{n}V\longrightarrow K}$

such that ${\displaystyle {}v_{i_{1}}\wedge \ldots \wedge v_{i_{n}}}$ is not mapped to ${\displaystyle {}0}$, but all other wedge products in the family are mapped to ${\displaystyle {}0}$. To show this, it is enough, by Theorem 57.7 , to give an alternating multilinear mapping

${\displaystyle \triangle \colon V^{n}\longrightarrow K}$

satisfying ${\displaystyle {}\triangle {\left(v_{i_{1}},\ldots ,v_{i_{n}}\right)}\neq 0}$ but ${\displaystyle {}\triangle {\left(v_{j_{1}},\ldots ,v_{j_{n}}\right)}=0}$ for every other strictly increasing index tuple. Let ${\displaystyle {}U}$ be the linear subspace generated by ${\displaystyle {}v_{i}}$, ${\displaystyle {}i\neq i_{k}}$, of ${\displaystyle {}V}$, and let ${\displaystyle {}W=V/U}$ denote the residue class space. Then the images of the ${\displaystyle {}v_{i_{k}}}$, ${\displaystyle {}k=1,\ldots ,n}$, form a basis of ${\displaystyle {}W}$, and the images of all other subsets with ${\displaystyle {}n}$ elements of the given basis do ot form a basis of ${\displaystyle {}W}$, because at least one element is mapped to ${\displaystyle {}0}$. We consider now the composed mapping

${\displaystyle \triangle :V^{n}\longrightarrow W^{n}\cong (K^{n})^{n}{\stackrel {\det }{\longrightarrow }}K.}$

This mapping is multilinear and alternating, due to Theorem 16.9 and Theorem 16.10 . Due to Theorem 16.11 , we have ${\displaystyle {}\triangle (z_{1},\ldots ,z_{n})=0}$ if and only if the images of ${\displaystyle {}z_{i}}$ in ${\displaystyle {}W}$ do not form a basis.

This follows directly from Theorem 58.1 and Fact *****.

We consider the mapping (with ${\displaystyle {}k}$ factors)

${\displaystyle V^{*}\times \cdots \times V^{*}\longrightarrow \operatorname {Map} \,(V\times \cdots \times V,K)}$

mit

${\displaystyle (f_{1},\ldots ,f_{k})\longmapsto {\left((v_{1},\ldots ,v_{k})\longmapsto \det {\left(f_{i}(v_{j})\right)}_{ij}\right)}.}$

For fixed ${\displaystyle {}f_{1},\ldots ,f_{k}}$, the mapping on the right is multilinear and alternating, as a direct verification using the determinant rules shows. Therefore, according to Corollary 57.8 , we obtain an element in ${\displaystyle {}{\left(\bigwedge ^{k}V\right)}^{*}}$. Hence, we get altogether a mapping

${\displaystyle V^{*}\times \cdots \times V^{*}\longrightarrow {\left(\bigwedge ^{k}V\right)}^{*}.}$

A direct inspection shows that this assignment is also multilinear and alternating. Due to the universal property, there exists a linear mapping

${\displaystyle \psi \colon \bigwedge ^{k}V^{*}\longrightarrow {\left(\bigwedge ^{k}V\right)}^{*}.}$

We have to show that this mapping is an isomorphism. To show this, let ${\displaystyle {}v_{1},\ldots ,v_{n}}$ be a basis of ${\displaystyle {}V}$, with the corresponding dual basis ${\displaystyle {}v_{1}^{*},\ldots ,v_{n}^{*}}$. Because of Theorem 58.1 , the family

${\displaystyle v_{i_{1}}^{*}\wedge \ldots \wedge v_{i_{k}}^{*},1\leq i_{1}<\ldots <i_{k}\leq n,}$

is a basis of ${\displaystyle {}\bigwedge ^{k}V^{*}}$. Moreover, the family

${\displaystyle v_{i_{1}}\wedge \ldots \wedge v_{i_{k}},1\leq i_{1}<\ldots <i_{k}\leq n,}$

is a basis of ${\displaystyle {}\bigwedge ^{k}V}$, with corresponding dual basis ${\displaystyle {}{\left(v_{i_{1}}\wedge \ldots \wedge v_{i_{k}}\right)}^{*}}$. We show that ${\displaystyle {}v_{i_{1}}^{*}\wedge \ldots \wedge v_{i_{k}}^{*}}$ is mapped under ${\displaystyle {}\psi }$ to ${\displaystyle {}(v_{i_{1}}\wedge \ldots \wedge v_{i_{k}})^{*}}$. For ${\displaystyle {}1\leq j_{1}<\ldots <j_{k}\leq n}$, we have

${\displaystyle {\left(\psi {\left(v_{i_{1}}^{*}\wedge \ldots \wedge v_{i_{k}}^{*}\right)}\right)}{\left(v_{j_{1}}\wedge \ldots \wedge v_{j_{k}}\right)}=\det {\left(v_{i_{r}}^{*}{\left(v_{j_{s}}\right)}_{1\leq r,s\leq k}\right)}\,.}$

If ${\displaystyle {}\{i_{1},\ldots ,i_{k}\}\neq \{j_{1},\ldots ,j_{k}\}}$, then there exists an ${\displaystyle {}i_{r}}$ that is different from all ${\displaystyle {}j_{s}}$. Therefore, the ${\displaystyle {}r}$-th row of the matrix is ${\displaystyle {}0}$; hence, its determinant is ${\displaystyle {}0}$. If the index sets coincide, then we obtain the identity matrix with determinant ${\displaystyle {}1}$. This effect coincides with the effect of ${\displaystyle {}{\left(v_{i_{1}}\wedge \ldots \wedge v_{i_{k}}\right)}^{*}}$.

The mapping

${\displaystyle V^{n}{\stackrel {\varphi \times \cdots \times \varphi }{\longrightarrow }}W^{n}{\stackrel {\delta }{\longrightarrow }}\bigwedge ^{n}W}$

is, due to Exercise 16.29 , multilinear and alternating. Due to Theorem 57.7 , there exists a uniquely determined linear mapping

${\displaystyle \bigwedge ^{n}V\longrightarrow \bigwedge ^{n}W,}$

with ${\displaystyle {}v_{1}\wedge \ldots \wedge v_{n}\mapsto \varphi (v_{1})\wedge \ldots \wedge \varphi (v_{n})}$.

(1). Let ${\displaystyle {}w_{1},\ldots ,w_{n}\in W}$ be given, and let ${\displaystyle {}v_{1},\ldots ,v_{n}\in V}$ be preimages, that is, ${\displaystyle {}\varphi (v_{i})=w_{i}}$. We have

${\displaystyle {}{\left(\bigwedge ^{n}\varphi \right)}(v_{1}\wedge \ldots \wedge v_{n})=w_{1}\wedge \ldots \wedge w_{n}\,.}$

Surjectivity follows from Lemma 57.5   (1).
(2). We may assume, due to the construction of the wedge product, that ${\displaystyle {}V}$ and ${\displaystyle {}W}$ have finite dimension. The statement follows from the explicit description of the bases in Theorem 58.1 .
(3). It is enough to show the equality for the generating system ${\displaystyle {}u_{1}\wedge \ldots \wedge u_{n}}$ with ${\displaystyle {}u_{i}\in U}$; but this is clear due to the explicit description.

Let ${\displaystyle {}v_{1},\ldots ,v_{n}}$ and ${\displaystyle {}w_{1},\ldots ,w_{n}}$ be two bases of ${\displaystyle {}V}$, fulfilling the relation

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

Due to Corollary 57.6 , we have

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

This shows the well-definedness of the mapping, and the statement follows.