Linear algebra (Osnabrück 2024-2025)/Part II/Lecture 56

Base change for tensor product

Lemma

Let ${}K$ be a field, and let ${}V_{1},\ldots ,V_{n}$ denote finite-dimensional vector spaces over ${}K$ . Let ${}v_{1j},j\in J_{1},\ldots ,v_{nj},j\in J_{n}$ , and ${}w_{1j},j\in J_{1},\ldots ,w_{nj},j\in J_{n}$ , be bases of ${}V_{1},\ldots ,V_{n}$ , with the base change matrices

{}B_{i}=M_{{\mathfrak {w}}_{i}}^{{\mathfrak {v}}_{i}}={\left(a_{irs}\right)}_{1\leq r,s\leq \dim _{K}{\left(V_{i}\right)}}\,.

Then the base change matrix

(with ${}J=J_{1}\times \cdots \times J_{n}$ ) between the bases

v_{1j_{1}}\otimes _{}\cdots \otimes _{}v_{nj_{n}},\,(j_{1},\ldots ,j_{n})\in J\,\,{\text{  and }}\,\,w_{1j_{1}}\otimes _{}\cdots \otimes _{}w_{nj_{n}},\,(j_{1},\ldots ,j_{n})\in J

of the tensor product is given by the ${}J\times J$ -matrix with the entries

{}c_{(j_{1},\ldots ,j_{n}),(k_{1},\ldots ,k_{n})}=a_{1j_{1}k_{1}}\cdots a_{nj_{n}k_{n}}\,.

Proof

Due to the definition of the base change matrix, we have

{}v_{is}=\sum _{r\in J_{i}}a_{irs}w_{ir}\,.

Therefore, using Lemma 55.9 (3), we obtain

{}{\begin{aligned}v_{1k_{1}}\otimes _{}\cdots \otimes _{}v_{nk_{n}}&={\left(\sum _{j_{1}\in J_{1}}a_{1j_{1}k_{1}}w_{1j_{1}}\right)}\otimes _{}\cdots \otimes _{}{\left(\sum _{j_{n}\in J_{n}}a_{nj_{n}k_{n}}w_{nj_{n}}\right)}\\&=\sum _{(j_{1},\ldots ,j_{n})\in J}a_{1j_{1}k_{1}}\cdots a_{nj_{n}k_{n}}w_{1j_{1}}\otimes _{}\cdots \otimes _{}w_{nj_{n}},\end{aligned}}

and these coefficients constitute the base change matrix.

\Box

Example

We consider ${}\mathbb {R} ^{2}$ with den bases ${}{\mathfrak {v}}={\begin{pmatrix}5\\6\end{pmatrix}},{\begin{pmatrix}-3\\8\end{pmatrix}}$ , and the standard basis ${}{\mathfrak {w}}$ . We consider ${}\mathbb {C}$ as a real vector space with the bases ${}{\mathfrak {x}}=3-2{\mathrm {i} },4+5{\mathrm {i} }$ and ${}{\mathfrak {y}}=1,{\mathrm {i} }$ . Hence, the base change matrices, as they appear in Lemma 56.1 , are

{}B_{1}={\begin{pmatrix}5&-3\\6&8\end{pmatrix}}\,

and

{}B_{2}={\begin{pmatrix}3&4\\-2&5\end{pmatrix}}\,.

We use the ordering ${}(1,1),(1,2),(2,1),(2,2)$ , and obtain the base change matrix

{\begin{pmatrix}15&20&-9&-12\\-10&25&6&-15\\18&24&24&32\\-12&30&-16&40\end{pmatrix}}.

For example, in the second column, we write the coefficients needed to express ${}v_{1}\otimes x_{2}$ as a linear combination of the ${}w_{1}\otimes y_{1},\,w_{1}\otimes y_{2},\,w_{2}\otimes y_{1},\,w_{2}\otimes y_{2}$ .

Tensor product and dual space

Theorem

Let ${}K$ be a field, and let ${}V_{1},\ldots ,V_{n}$ denote finite-dimensional vector spaces over ${}K$ . Then there exists a natural isomorphism

{V_{1}}^{*}\otimes _{}\cdots \otimes _{}{V_{n}}^{*}\longrightarrow {\left(V_{1}\otimes _{}\cdots \otimes _{}V_{n}\right)}^{*},f_{1}\otimes _{}\cdots \otimes _{}f_{n}\longmapsto {\left(v_{1}\otimes _{}\cdots \otimes _{}v_{n}\mapsto f_{1}(v_{1})\cdots f_{n}(v_{n})\right)}.

Proof

For fixed linear forms ${}f_{1}\in {V_{1}}^{*},\ldots ,f_{n}\in {V_{n}}^{*}$ , the mapping

V_{1}\times \cdots \times V_{n}\longrightarrow K,(v_{1},\ldots ,v_{n})\longmapsto f_{1}(v_{1})\cdots f_{n}(v_{n}),

is multilinear due to Exercise 16.39 ; therefore, it defines a linear form on ${}V_{1}\otimes _{}\cdots \otimes _{}V_{n}$ . This yields the mapping

\Psi \colon {V_{1}}^{*}\times \cdots \times {V_{n}}^{*}\longrightarrow {\left(V_{1}\otimes _{}\cdots \otimes _{}V_{n}\right)}^{*},(f_{1},\ldots ,f_{n})\longmapsto {\left((v_{1}\otimes _{}\cdots \otimes _{}v_{n})\mapsto f_{1}(v_{1})\cdots f_{n}(v_{n})\right)}.

This assignment ${}\Psi$ is also multilinear, and gives a linear mapping

{V_{1}}^{*}\otimes _{}\cdots \otimes _{}{V_{n}}^{*}\longrightarrow {\left(V_{1}\otimes _{}\cdots \otimes _{}V_{n}\right)}^{*}.

Due to Corollary 55.13 and Corollary 13.12 , both spaces have the same dimension. Let ${}v_{ij}$ , ${}1\leq j\leq \dim _{K}{\left(V_{i}\right)}$ , be bases of ${}V_{i}$ . Then the ${}v_{1j_{1}}\otimes _{}\cdots \otimes _{}v_{nj_{n}}$ form, according to Theorem 55.12 (3), a basis of ${}V_{1}\otimes _{}\cdots \otimes _{}V_{n}$ , and the dual basis is a basis of the dual space. We claim the equality of the linear mappings

{}\Psi (v_{1j_{1}}^{*}\otimes _{}\cdots \otimes _{}v_{nj_{n}}^{*})={\left(v_{1j_{1}}\otimes _{}\cdots \otimes _{}v_{nj_{n}}\right)}^{*}\,.

This equality follows from the fact that both mappings give, when applied to the basis elements ${}v_{1k_{1}}\otimes _{}\cdots \otimes _{}v_{nk_{n}}$ , in case ${}(k_{1},\ldots ,k_{n})=(j_{1},\ldots ,j_{n})$ the value ${}1$ , and else the value ${}0$ . Therefore, ${}\Psi$ is surjective, and then also injective.

\Box

Lemma

Let ${}K$ be a field, and let ${}U,V,W$ denote ${}K$ -vector spaces. Then the following statements hold.

(in the sense of a canonical isomorphy).

We have
${}U\otimes _{K}V\cong V\otimes _{K}U\,.$
We have
${}U\otimes _{K}{\left(V\otimes _{K}W\right)}\cong {\left(U\otimes _{K}V\right)}\otimes _{K}W\,.$

Proof

See Exercise 56.2 .

\Box

Tensor product of linear mappings

Lemma

Let ${}K$ be a field, and let ${}V_{1},\ldots ,V_{n},W_{1},\ldots ,W_{n}$ denote vector spaces over ${}K$ . Let

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

be ${}K$ -linear mappings. Then there exists a well-defined linear mapping

V_{1}\otimes _{}\cdots \otimes _{}V_{n}\longrightarrow W_{1}\otimes _{}\cdots \otimes _{}W_{n}

with

{}\psi (v_{1}\otimes _{}\cdots \otimes _{}v_{n}):=\varphi _{1}(v_{1})\otimes _{}\cdots \otimes _{}\varphi _{n}(v_{n})\,.

Proof

The composed mapping

V_{1}\times \cdots \times V_{n}{\stackrel {\varphi _{1}\times \cdots \times \varphi _{n}}{\longrightarrow }}W_{1}\times \cdots \times W_{n}{\stackrel {\pi }{\longrightarrow }}W_{1}\otimes _{}\cdots \otimes _{}W_{n}

is multilinear according to Exercise 16.13 . Therefore, due to Theorem 55.4 , this induces a linear mapping

V_{1}\otimes _{}\cdots \otimes _{}V_{n}\longrightarrow W_{1}\otimes _{}\cdots \otimes _{}W_{n},v_{1}\otimes _{}\cdots \otimes _{}v_{n}\longmapsto \varphi (v_{1})\otimes _{}\cdots \otimes _{}\varphi (v_{n}).

\Box

Definition

Let ${}K$ be a field, and let ${}V_{1},\ldots ,V_{n},W_{1},\ldots ,W_{n}$ denote vector spaces over ${}K$ . For ${}K$ -linear mappings

\varphi _{i}\colon V_{i}\longrightarrow W_{i},

the linear mapping

V_{1}\otimes _{}\cdots \otimes _{}V_{n}\longrightarrow W_{1}\otimes _{}\cdots \otimes _{}W_{n},v_{1}\otimes _{}\cdots \otimes _{}v_{n}\longmapsto \varphi _{1}(v_{1})\otimes _{}\cdots \otimes _{}\varphi _{n}(v_{n}),

is called the tensor product

of the

{}\varphi _{i}

. It is denoted by

{}\varphi _{1}\otimes _{}\cdots \otimes _{}\varphi _{n}

.

Proposition

Let ${}K$ be a field, and let ${}U,V,Z$ denote vector spaces

over

{}K

. Then the following statements hold.

For a ${}K$ -linear mapping ${}\varphi \colon U\rightarrow V$ , there exists a natural ${}K$ -linear mapping ${}\varphi \otimes _{K}\operatorname {Id} _{Z}\colon U\otimes _{K}Z\rightarrow V\otimes _{K}Z$ .
If ${}\varphi$ is surjective, then also ${}\varphi \otimes _{K}\operatorname {Id} _{Z}$ is surjective.
If ${}\varphi$ injective is, then also ${}\varphi \otimes _{K}\operatorname {Id} _{Z}$ is injective.

Proof

(1). This is a special case of Lemma 56.5 .

(2). The surjectivity of the mapping

U\otimes _{K}Z\longrightarrow V\otimes _{K}Z

is clear, because the ${}v\otimes z$ form a ${}K$ -generating system of ${}V\otimes _{K}Z$ , and these belong to the image of the mapping.

(3). Because of the injectivity, we may assume that

{}U\subseteq V\,

is a linear subspace. A basis ${}u_{i}$ , ${}i\in I$ , of ${}U$ can be extended to a basis ${}u_{i}$ , ${}i\in J$ , of ${}V$ , with ${}I\subseteq J$ Let ${}z_{\ell }$ , ${}\ell \in L$ , be a basis of ${}Z$ . Then, according to Theorem 55.12 (3), the family ${}v_{j}\otimes z_{\ell }$ , ${}(j,\ell )\in J\times L$ , is a basis of ${}V\otimes Z$ . The subset ${}v_{i}\otimes z_{\ell }$ , ${}(i,\ell )\in I\times L$ , is a basis of ${}U\otimes Z$ . therefore, under

U\otimes Z\longrightarrow V\otimes Z,

a basis is mapped to linearly independent elements, and thus this mapping is injective.

\Box

Therefore, for linear subspaces ${}U_{1}\subseteq V_{1},\ldots ,U_{n}\subseteq V_{n}$ , the tensor product ${}U_{1}\otimes _{}\cdots \otimes _{}U_{n}$ is in a natural way a linear subspace of ${}V_{1}\otimes _{}\cdots \otimes _{}V_{n}$ .

Corollary

Let ${}K$ be a field, and let ${}U,V,W$ denote ${}K$ -vector spaces. Then we have

{}U\otimes _{K}{\left(V\oplus W\right)}\cong {\left(U\otimes _{K}V\right)}\oplus {\left(U\otimes _{K}W\right)}\,.

Proof

See Exercise 56.5 .

\Box

Change of base field

We have several times simplified a real problem by considering the situation over the complex numbers. If the situation can be formulated with a real matrix, then we can consider it directly as a complex matrix, and we can compute its (non-real) complex eigenvalues. Matrices describe mathematical objects, and they depend in general on the choice of bases. The tensor product allows to describe the passage to the complex situation on the level of the objects. We consider here the case that a ${}K$ -vector space ${}V$ and a field extension ${}K\subseteq L$ are given. We fix the terminology that we will use.

Definition

A subset ${}K\subseteq L$ of a field ${}L$ is called a subfield of ${}L$ , if the following properties hold.

We have ${}0,1\in K$ .
For ${}a,b\in K$ also ${}a+b\in K$ holds.
For ${}a,b\in K$ , also ${}a\cdot b\in K$ holds.
For ${}a\in K$ , we also have ${}-a\in K$ .
For ${}a\in K$ , ${}a\neq 0$ , we also have ${}a^{-1}\in K$ .

Definition

Let ${}L$ be a field, and let ${}K\subseteq L$ be a subfield of ${}L$ . Then ${}L$ is called an extension field of ${}K$ , and the inclusion ${}K\subseteq L$ is called a

field extension.

Lemma

Let ${}K\subseteq L$ be a field extension. Then is ${}L$ is in natural way a

{}K

-vector space.

Proof

The scalar multiplication

K\times L\longrightarrow L,(\lambda ,x)\longmapsto \lambda x,

is simply given by the multiplication in ${}L$ . The vector space axioms follow directly from the field axioms.

\Box

Definition

For a ${}K$ -vector space ${}V$ over a field ${}K$ , and a field extension ${}K\subseteq L$ , the tensor product ${}L\otimes _{K}V$ is called the ( ${}L$ -)

vector space given by extension of scalars.

Instead of ${}L\otimes _{K}V$ , we also write ${}V_{L}$ .

Proposition

Let ${}K$ be a field, ${}V$ be a ${}K$ -vector space, and ${}K\subseteq L$ a

field extension. Then the following statements hold.

The tensor product ${}L\otimes _{K}V$ is an ${}L$ -vector space.
There exists a canonical ${}K$ -linear mapping
$V\longrightarrow L\otimes _{K}V,v\longmapsto 1\otimes v.$

For ${}K=L$ , this is an isomorphism.
For a ${}K$ -linear mapping ${}\varphi \colon V\rightarrow W$ , the induced mapping
$\operatorname {Id} _{L}\otimes \varphi \colon L\otimes _{K}V\longrightarrow L\otimes _{K}W$

is ${}L$ -linear.
For ${}V=K^{n}$ , we have
${}L\otimes _{K}K^{n}\cong L^{n}\,.$
For a finite-dimensional ${}K$ -vector space ${}V$ , we have
${}\dim _{K}{\left(V\right)}=\dim _{K}{\left(L\otimes _{K}V\right)}\,.$
For another field extension ${}L\subseteq M$ , we have
${}M\otimes _{K}V\cong M\otimes _{L}{\left(L\otimes _{K}V\right)}\,$

(an isomorphism of $M$ -vector spaces).

Proof

(1). The multiplication

L\times L\longrightarrow L,(r,s)\longmapsto rs,

is ${}L$ -bilinear, and, in particular, ${}K$ -bilinear; therefore, it yields, according to Theorem 55.4 , to a ${}K$ -linear mapping

L\otimes _{K}L\longrightarrow L.

This induces, due to Lemma 56.4 (2) and Proposition 56.7 , a ${}K$ -linear mapping

L\otimes _{K}{\left(L\otimes _{K}V\right)}\cong {\left(L\otimes _{K}L\right)}\otimes _{K}V\longrightarrow L\otimes _{K}V.

This gives a well-defined scalar multiplication

L\times {\left(L\otimes _{K}V\right)}\longrightarrow {\left(L\otimes _{K}V\right)},

which is explicitly given by

{}s\cdot {\left(\sum _{j=1}^{n}r_{j}\otimes m_{j}\right)}=\sum _{j=1}^{n}{\left(sr_{j}\right)}\otimes m_{j}\,.

From this description, we can deduce directly the properties of a scalar multiplication.
(2). The ${}K$ -homomorphism follows directly from the bilinearity of the tensor product. For ${}L=K$ , the mapping is surjective. The scalar multiplication ${}K\times V\rightarrow V$ induces a ${}K$ -linear mapping

K\otimes _{K}V\longrightarrow V.

The composition of the canonical mapping ${}V\rightarrow K\otimes _{K}V$ with this mapping is the identity on ${}V$ , so that the first mapping is also injective.
(3) follows from the explicit description in (1).
(4) follows from Corollary 56.8 .
(5) follows from (4).
(6). Because of part (2), we have, one one hand, a ${}K$ -linear mapping ${}V\rightarrow L\otimes _{K}V$ . This yields a ${}K$ -multilinear mapping

M\times V\longrightarrow M\times {\left(L\otimes _{K}V\right)}\longrightarrow M\otimes _{L}{\left(L\otimes _{K}V\right)},

which induces a ${}K$ -linear mapping

M\otimes _{K}V\longrightarrow M\otimes _{L}{\left(L\otimes _{K}V\right)}.

On the other hand, we have a ${}L$ -linear mapping

L\otimes _{K}V\longrightarrow M\otimes _{K}V.

On the right-hand side, we have an ${}M$ -vector space; therefore, the scalar multiplication may be considered as an ${}L$ -multilinear mapping

M\times {\left(L\otimes _{K}V\right)}\longrightarrow L\otimes _{K}V.

This induces an ${}L$ -linear mapping

M\otimes _{L}{\left(L\otimes _{K}V\right)}\longrightarrow L\otimes _{K}V.

The two mappings are inverse to each other, as this can be checked on the decomposable tensors. This shows also that the ${}M$ -multiplications are the same.

\Box

Lemma

Let ${}K$ be a field, ${}V$ be a ${}K$ -vector space, and ${}K\subseteq L$ be a field extension. Let ${}v_{i}$ , ${}i\in I$ ,

be a family of vectors in

{}V

. Then the following statements hold.

The family ${}v_{i}$ , ${}i\in I$ , is a ${}K$ -generating system of ${}V$ if and only if ${}1\otimes v_{i}$ , ${}i\in I$ , is an ${}L$ -generating system of ${}L\otimes V$ .
The family ${}v_{i}$ , ${}i\in I$ , is ${}K$ -linearly independent in ${}V$ if and only if ${}1\otimes v_{i}$ , ${}i\in I$ , is linearly independent (over ${}L$ ) in ${}L\otimes V$ .
Die family ${}v_{i}$ , ${}i\in I$ , is a ${}K$ -basis of ${}V$ if and only if ${}1\otimes v_{i}$ , ${}i\in I$ , is an ${}L$ -basis of ${}L\otimes V$ .

Proof

See Exercise 56.15 .

\Box

Example

Let ${}V$ be a real vector space. The tensorization with the ${}\mathbb {R}$ -algebra ${}\mathbb {C}$ , that is,

{}V_{\mathbb {C} }:=\mathbb {C} \otimes _{\mathbb {R} }V\,,

is called the complexification of ${}V$ . If ${}V$ has the dimension ${}n$ , then ${}V_{\mathbb {C} }$ has, considered as a complex vector space, also the dimension ${}n$ . If we consider ${}V_{\mathbb {C} }$ as a real vector space, then it has the real dimension ${}2n$ .

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