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## 数学代写|微分几何代写Differential Geometry代考|Hom Space and Dual

Definition 4.1.1. Let $V$ and $W$ be two vector spaces over $K$. Denote the set of linear transformations from $V$ to $W$ by $\operatorname{Hom}_{K}(V, W)$, or simply $\operatorname{Hom}(V, W)$ if the field $K$ is understood by context.

We can define addition and multiplication by a $K$-scalar on $\operatorname{Hom}(V, W)$ in the following way. If $T_{1}, T_{2} \in \operatorname{Hom}(V, W)$, then $T_{1}+T_{2}$ is the linear transformation given by
$$\left(T_{1}+T_{2}\right)(v) \stackrel{\text { def }}{=} T_{1}(v)+T_{1}(v) \quad \text { for all } v \in V \text {. }$$
Also, if $\lambda \in K$ and $T \in \operatorname{Hom}(V, W)$, define the linear transformation $\lambda T$ by
$$(\lambda T)(v) \stackrel{\text { def }}{=} \lambda(T(v)) \quad \text { for all } v \in V$$
These definitions lead us to the following foundational proposition.
Proposition 4.1.2. Let $V$ and $W$ be vector spaces over $K$ of dimension $m$ and $n$, respectively. Then $\operatorname{Hom}(V, W)$ is a vector space over $K$, with $\operatorname{dim} \operatorname{Hom}(V, W)=$ $m n$.

Proof. We leave it to the reader to check that $\operatorname{Hom}(V, W)$ satisfies all the axioms of a vector space over $K$.

To prove that $\operatorname{dim} \operatorname{Hom}(V, W)=m n$, first choose an ordered basis $\mathcal{B}=\left(e_{1}, e_{2}, \ldots, e_{m}\right)$ of $V$ and an ordered basis $\mathcal{B}^{\prime}=\left(f_{1}, f_{2}, \ldots, f_{n}\right)$ of $W$. Define $T_{i j} \in \operatorname{Hom}(V, W)$ as the linear transformations defined by
$$T_{i j}\left(e_{k}\right)= \begin{cases}f_{i}, & \text { if } j=k, \ 0, & \text { if } j \neq k,\end{cases}$$
and extended by linearity over all $V$. We show that the set $\left{T_{i j}\right}$ for $1 \leq i \leq m$ and $1 \leq j \leq n$ forms a basis of $\operatorname{Hom}(V, W)$.

Because of linearity, any linear transformation $L \in \operatorname{Hom}(V, W)$ is completely defined given the knowledge of $L\left(e_{j}\right)$ for all $1 \leq j \leq m$. Suppose that for each $j$, there exist $m n$ constants $a_{j}^{i}$ in $K$, indexed by $i=1,2, \ldots, m$ and $j=1,2, \ldots, n$, such that
$$L\left(e_{j}\right)=\sum_{i=1}^{n} a_{j}^{i} f_{i} .$$
Then
$$L=\sum_{i=1}^{n} \sum_{j=1}^{m} a_{j}^{i} T_{i j},$$and hence, $\left{T_{i j}\right}$ spans $\operatorname{Hom}(V, W)$. Furthermore, suppose that for some constants $c_{i j}$
$$\sum_{i=1}^{n} \sum_{j=1}^{m} c_{i j} T_{i j}=\mathbf{0}$$
the trivial linear transformation. Then for all $1 \leq k \leq m$,
$$\sum_{i=1}^{n} \sum_{j=1}^{m} c_{i j} T_{i j}\left(e_{k}\right)=\mathbf{0} \Longleftrightarrow \sum_{i=1}^{n} c_{i k} f_{i}=\mathbf{0} .$$

## 数学代写|微分几何代写Differential Geometry代考|Bilinear Forms and Inner Products

Definition 4.2.1. Let $V$ and $W$ be vector spaces over a field $K$. A bilinear form $\langle\cdot,$,$rangle on V \times W$ is a function $V \times W \rightarrow K$ such that for all $v \in V, w \in W$, and $\lambda \in K$,
\begin{aligned} \left\langle v_{1}+v_{2}, w\right\rangle=\left\langle v_{1}, w\right\rangle+\left\langle v_{2}, w\right\rangle, &\langle\lambda v, w\rangle=\lambda\langle v, w\rangle \ \left\langle v, w_{1}+w_{2}\right\rangle=\left\langle v, w_{1}\right\rangle+\left\langle v, w_{2}\right\rangle \end{aligned} \quad\langle v, \lambda w\rangle=\lambda\langle v, w\rangle .
If $V=W$, then we say $\langle,,\rangle$, is a bilinear form on $V$.

We can restate this definition to say that for any fixed $w_{0} \in W$, the function $x \mapsto\left\langle x, w_{0}\right\rangle$ is in $V^{}$ (corresponding to $(4.15)$ ) and that for any fixed $v_{0} \in V$, the function $x \mapsto\left\langle v_{0}, x\right\rangle$ is in $W^{}$ (corresponding to (4.16)).

The notation used for a bilinear form varies widely in the literature because of the many areas in which it is used. In terms of function notation, we might encounter the functional notation $f: V \times W \rightarrow K$ or perhaps $\omega: V \times W \rightarrow K$ for a bilinear form and $\langle\cdot, \cdot\rangle$ or $(\cdot, \cdot)$ for the “product” notation. If $V=W$, we sometimes write the pair $(V, f)$ to denote the vector space $V$ equipped with the bilinear form $f$.

Example 4.2.2. In elementary linear algebra, the most commonly known example of a bilinear form on $\mathbb{R}^{n}$ is the dot product between two vectors defined in terms of standard coordinates by
$$\vec{v} \cdot \vec{w}=v^{1} w^{1}+v^{2} w^{2}+\cdots+v^{n} w^{n} .$$
The following functions $\mathbb{R}^{n} \times \mathbb{R}^{n} \rightarrow \mathbb{R}$ are also bilinear forms:
\begin{aligned} &\langle\vec{v}, \vec{w}\rangle_{1}=v_{1} w_{2}+v_{2} w_{1}+v_{3} w_{3} \cdots+v_{n} w_{n} \ &\langle\vec{v}, \vec{w}\rangle_{2}=2 v_{1} w_{1}+v_{2} w_{2}+\cdots+v_{n} w_{n}+v_{1} w_{n} \ &\langle\vec{v}, \vec{w}\rangle_{3}=v_{1} w_{2}-v_{2} w_{1} \end{aligned}
Despite the variety depicted in the above example, bilinear forms on finite dimensional vector spaces can be completely characterized by a single matrix.

# 微分几何代考

## 数学代写|微分几何代写Differential Geometry代考| Hom Space and Dual

$$\left(T_{1}+T_{2}\right)(v) \stackrel{\text { def }}{=} T_{1}(v)+T_{1}(v) \quad \text { for all } v \in V$$

$$(\lambda T)(v) \stackrel{\text { def }}{=} \lambda(T(v)) \quad \text { for all } v \in V$$

$$T_{i j}\left(e_{k}\right)=\left{f_{i}, \quad \text { if } j=k, 0, \quad \text { if } j \neq k,\right.$$

$$L\left(e_{j}\right)=\sum_{i=1}^{n} a_{j}^{i} f_{i} .$$
$$L=\sum_{i=1}^{n} \sum_{j=1}^{m} a_{j}^{i} T_{i j},$$

$$\sum_{i=1}^{n} \sum_{j=1}^{m} c_{i j} T_{i j}=\mathbf{0}$$

$$\sum_{i=1}^{n} \sum_{j=1}^{m} c_{i j} T_{i j}\left(e_{k}\right)=\mathbf{0} \Longleftrightarrow \sum_{i=1}^{n} c_{i k} f_{i}=\mathbf{0} .$$

## 数学代写|微分几何代写Differential Geometry代考| Bilinear Forms and Inner Products

$$\left\langle v_{1}+v_{2}, w\right\rangle=\left\langle v_{1}, w\right\rangle+\left\langle v_{2}, w\right\rangle,\langle\lambda v, w\rangle=\lambda\langle v, w\rangle\left\langle v, w_{1}+w_{2}\right\rangle=\left\langle v, w_{1}\right\rangle+\left\langle v, w_{2}\right\rangle \quad\langle v, \lambda w\rangle=\lambda\langle v, w\rangle .$$

$$\vec{v} \cdot \vec{w}=v^{1} w^{1}+v^{2} w^{2}+\cdots+v^{n} w^{n} .$$

$$\langle\vec{v}, \vec{w}\rangle_{1}=v_{1} w_{2}+v_{2} w_{1}+v_{3} w_{3} \cdots+v_{n} w_{n} \quad\langle\vec{v}, \vec{w}\rangle_{2}=2 v_{1} w_{1}+v_{2} w_{2}+\cdots+v_{n} w_{n}+v_{1} w_{n}\langle\vec{v}, \vec{w}\rangle_{3}=v_{1} w_{2}-v_{2} w_{1}$$

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assignmentutor™您的专属作业导师
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