📂Linear Algebra

Tensor Product of Linear Transformations

선형대수

[ 펼치기 · 접기 ] 벡터공간 벡터공간 벡터공간 · 부분공간 · 합$+$ · 직합$\oplus$ · 선형결합 · 생성 · 선형 독립/종속 · 론스키안 기저 기저 · 순서기저 · 차원 · 좌표벡터 · 벡터의 좌표변환 · 선형변환의 좌표변환 잉여류 잉여류와 몫공간 · 몫공간의 기저 · 몫공간으로의 사상 · 몫공간 위의 작용소 선형변환 내용 · 내용 · 내용 고유값 내용 · 내용 · 내용 쌍대공간 내용 · 내용 · 내용 텐서 벡터공간의 텐서곱 · 곱 벡터 · 보편성질 · 선형변환의 텐서곱(행렬표현) · 쌍대공간 · 전치 내적공간 내용 · 내용 · 내용 관련 분야 행렬대수 · 바나흐 공간 · 힐베르트 공간

제목

[ 펼치기 · 접기 ] 대분류 내용 · 내용 · 내용 대분류 중분류 내용 · 내용 · 내용 중분류 내용 · 내용(내용 · 내용 · 내용) · 내용 중분류 내용 · 내용 · 내용 대분류 내용 · 내용 · 내용 대분류 TBD · TBD · TBD 관련 분야 내용 · 내용 · 내용

Buildup¹

Given finite-dimensional vector spaces $V_{1}, V_{2}, W_{1}, W_{2}$ and linear transformations $\phi_{1} : V_{1} \to W_{1}$, $\phi_{2} : V_{2} \to W_{2}$, one can consider the following bilinear transformation.

$$V_{1} \times V_{2} \to W_{1} \otimes W_{2}$$

$$(v_{1}, v_{2}) \mapsto \phi_{1}(v_{1}) \otimes \phi_{2}(v_{2})$$

$\phi_{1}, \phi_{2}$ is a linear transformation, and it is easy to see that this function is bilinear due to the definition of product vectors.

$$\begin{align*} \phi_{1}(\alpha v_{1} + \beta v_{1}^{\prime}) \otimes \phi_{2}(v_{2}) &= \big( \alpha \phi_{1}(v_{1}) + \beta \phi_{1}(v_{1}^{\prime}) \big) \otimes \phi_{2}(v_{2}) \\ &= \alpha \phi_{1}(v_{1}) \otimes \phi_{2}(v_{2}) + \beta \phi_{1}(v_{1}^{\prime}) \otimes \phi_{2}(v_{2}) \\ \end{align*}$$

$$\begin{align*} \phi_{1}(v_{1}) \otimes \phi_{2}(\alpha w_{1} + \beta w_{1}^{\prime}) &= \phi_{1}(v_{1}) \otimes \big( \alpha \phi_{2}(w_{1}) + \beta \phi_{2}(w_{1}^{\prime}) \big) \\ &= \alpha \phi_{1}(v_{1}) \otimes \phi_{2}(w_{1}) + \beta \phi_{1}(v_{1}) \otimes \phi_{2}(w_{1}^{\prime}) \end{align*}$$

Universality of the tensor product

For vector spaces $V_{1}, \dots, V_{r}$ and $W$, suppose we have the following multilinear transformation $\phi$ given. $$\phi : V_{1} \times \cdots \times V_{r} \to W$$ Then, there exists a unique linear transformation $\psi$ satisfying the following. $$\psi : V_{1} \otimes \cdots \otimes V_{r} \to W$$ $$\phi (v_{1}, \dots, v_{r}) = \psi (v_{1} \otimes \cdots \otimes v_{r}),\quad \forall v_{i} \in V_{i}, \forall i$$

Then, according to the theorem above, a unique linear transformation $V_{1} \otimes V_{2} \to W_{1} \otimes W_{2}$ exists.

Definition

The tensor product of two linear transformations $\phi_{1} : V_{1} \to W_{1}$ and $\phi_{2} : V_{2} \to W_{2}$ $\phi_{1} \otimes \phi_{2}$ is defined as follows.

$$\begin{equation} \begin{aligned} \phi_{1} \otimes \phi_{2} : V_{1} \otimes V_{2} &\to W_{1} \otimes W_{2} \\ (v_{1} \otimes v_{2}) &\mapsto \phi_{1}(v_{1}) \otimes \phi_{2}(v_{2}),\quad \forall v_{1} \in V_{1}, \forall v_{2} \in V_{2} \end{aligned} \end{equation}$$

$$(\phi_{1} \otimes \phi_{2})(v_{1} \otimes v_{2}) = \phi_{1}(v_{1}) \otimes \phi_{2}(v_{2})$$

Generalization

Regarding $1 \le i \le n$, the tensor product of linear transformations $\phi_{i} : V_{i} \to W_{i}$ is the unique linear transformation satisfying the following.

$$\phi_{1} \otimes \cdots \otimes \phi_{n} : V_{1} \otimes \cdots \otimes V_{n} \to W_{1} \otimes \cdots \otimes W_{n}$$

$$(\phi_{1} \otimes \cdots \otimes \phi_{n}) (v_{1} \otimes \cdots \otimes v_{n}) = \phi_{1}(v_{1}) \otimes \cdots \otimes \phi_{n}(v_{n}), \forall\quad v_{i} \in V_{i}$$

Properties

For linear operators $\phi_{1}, \phi_{2} : V \to V$ and $\psi_{1}, \psi_{2} : W \to W$, the composition of two tensor products $(\phi_{1} \otimes \psi_{1})$ and $(\phi_{2} \otimes \psi_{2})$ is as follows.

$$(\phi_{1} \otimes \psi_{1}) \circ (\phi_{2} \otimes \psi_{2}) = (\phi_{1} \circ \phi_{2}) \otimes (\psi_{1} \circ \psi_{2})$$

Generalization

$$\begin{align*} &\phi_{1} \otimes \cdots \otimes (\alpha \phi_{k} + \beta \psi_{k}) \otimes \cdots \otimes \phi_{n} \\ &= \alpha (\phi_{1} \otimes \cdots \otimes \phi_{k} \otimes \cdots \otimes \phi_{n}) + \beta (\phi_{1} \otimes \cdots \otimes \psi_{k} \otimes \cdots \otimes \phi_{n})\end{align*}$$

$$(\phi_{1} \otimes \cdots \otimes \phi_{n}) \circ (\psi_{1} \otimes \cdots \otimes \psi_{n}) = (\phi_{1} \circ \psi_{1}) \otimes \cdots \otimes (\phi_{n} \circ \psi_{n})$$

Proof

Linearity

It can easily be shown using the definition $(1)$ and the definition of product vectors. For $\phi : V \to V^{\prime}$, $\psi_{1}, \psi_{2} : W \to W^{\prime}$, and $v \in V, w \in W$,

$$\begin{align*} \left[ \phi \otimes (\alpha \psi_{1} + \beta \psi_{2}) \right] (v \otimes w) &= \phi (v) \otimes \left[(\alpha \psi_{1} + \beta \psi_{2})(w)\right] \\ &= \phi (v) \otimes \left[\alpha \psi_{1}(w) + \beta \psi_{2}(w)\right] \\ &= \alpha [\phi (v) \otimes \psi_{1}(w)] + \beta [\phi (v) \otimes \psi_{2}(w)] \\ &= [\alpha (\phi \otimes \psi_{1})] (v \otimes w) + [\beta (\phi \otimes \psi_{2})] (v \otimes w) \\ &= \left[ \alpha (\phi \otimes \psi_{1}) + \beta (\phi \otimes \psi_{2}) \right] (v \otimes w) \\ \end{align*}$$

$$\implies \phi \otimes (\alpha \psi_{1} + \beta \psi_{2}) = \alpha (\phi \otimes \psi_{1}) + \beta (\phi \otimes \psi_{2})$$

■

Composition

The calculation can easily be done using the definition $(1)$ and the definition of product vectors. For $v \in V$ and $w \in W$,

$$\begin{align*} \left[ (\phi_{2} \otimes \psi_{2}) \circ (\phi_{1} \otimes \psi_{1}) \right] (v \otimes w) &= (\phi_{2} \otimes \psi_{2}) \left[ (\phi_{1} \otimes \psi_{1}) (v \otimes w) \right] \\ &= (\phi_{2} \otimes \psi_{2}) (\phi_{1}(v) \otimes \psi_{1}(w)) \\ &= [\phi_{2} (\phi_{1}(v))] \otimes [\psi_{2}(\psi_{1}(w))] \\ &= [(\phi_{2} \circ \phi_{1})(v)] \otimes [(\psi_{2} \circ \psi_{1})(w)] \\ &= \left[ (\phi_{2} \circ \phi_{1}) \otimes (\psi_{2} \circ \psi_{1})\right] (v \otimes w) \end{align*}$$

$$\implies (\phi_{2} \otimes \psi_{2}) \circ (\phi_{1} \otimes \psi_{1}) = (\phi_{2} \circ \phi_{1}) \otimes (\psi_{2} \circ \psi_{1})$$

■