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• Statistical Inference 统计推断
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• (Generalized) Linear Models 广义线性模型
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• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

## 数学代写|微分几何代写Differential Geometry代考|Bilinear Forms on V

Many applications of bilinear forms involve a bilinear form $\langle\cdot, \cdot\rangle$ on $V$.
When we consider the components of a bilinear form on $V$ with respect to bases, we always assume that $\mathcal{A}=\mathcal{B}$. The components $\left(c_{i j}\right)$ described in Proposition $4.2 .3$ can be written as an $n \times n$ matrix. In Proposition 4.2.6, we also suppose that $\mathcal{A}^{\prime}=\mathcal{B}^{\prime}$ so that change of coordinate matrices are equal, $P=Q$. Then
$$\bar{c}{k \ell}=\bar{p}{k}^{i} \tilde{p}{\ell}^{j} c{i j} .$$
The matrices of components $\left(c_{i j}\right)$ and $\left(\bar{c}{k \ell}\right)$ are not necessarily similar. If they were, they would satisfy (4.7). Consequently, though we do depict the components of a bilinear form according to Proposition 4.2.3, the matrix does not behave under coordinate changes like a matrix that represents a linear transformation. We leave it as an exercise for the reader to prove that $$\bar{C}=\left(P^{-1}\right)^{\top} C P^{-1},$$ where $C$ is the matrix $\left(c{i j}\right)$ and $\bar{C}$ is the matrix with components $\left(\bar{c}_{k \ell}\right)$. Hence, $C$ and $\bar{C}$ are similar only if $P$ is orthogonal.
Definition 4.2.8. A bilinear form $\langle, \cdot\rangle$ on a vector space $V$ is called

1. symmetric if $\langle y, x\rangle=\langle x, y\rangle$ for all $x, y \in V$; and
2. antisymmetric (or skew-symmetric) if $\langle y, x\rangle=-\langle x, y\rangle$ for all $x, y \in V$.

## 数学代写|微分几何代写Differential Geometry代考|Signature of a Symmetric Bilinear Form

Theorem 4.2.14 (Sylvester’s Law of Inertia). Let $\langle\cdot, \cdot\rangle$ be a symmetric bilinear form on $V$ with $\operatorname{dim} V=n$. Setting $r=n-(p+q)$, the triple of nonnegative integers $(p, q, r)$ arising in (4.20) is independent of the basis.

Proof. Let $\mathcal{B}=\left(e_{1}, e_{2}, \ldots, e_{n}\right)$ be an ordered basis of $V$ with respect to which the component matrix of $\langle\cdot, \cdot\rangle$ is given in (4.20). The rank of $\langle\cdot, \cdot\rangle$, which is independent of any basis, is $p+q$.

Let $V_{1}$ be a subspace of $V$ of maximal dimension such that $\langle\cdot, \cdot\rangle$ restricted to $V_{1}$ is positive-definite. Then $\operatorname{dim} V_{1}=p^{\prime} \geq p$ because $\langle\cdot,$,$rangle is positive-definite over$ $\operatorname{Span}\left(e_{1}, \ldots, e_{p}\right)$. By Proposition $4.2 .10$, there is a basis $\mathcal{B}{1}$ of $V{1}$ with respect to which that matrix of the form is $I_{p^{\prime}}$.

Let $V_{2}$ be a subspace of $V$ of maximal dimension such that $\left.\langle\cdot,\rangle\right$, restricted to $V_{1}$ is is negative-definite, i.e., for all $v \in V_{2}-{0},\langle v, v\rangle<0$. Over $\operatorname{Span}\left(e_{p+1}, \ldots, e_{p+q}\right)$, the form is positive-definite, so $\operatorname{dim} V_{2}=p^{\prime} \geq p$. By Proposition 4.2.10, there is a basis $\mathcal{B}{2}$ of $V{2}$ with respect to which that matrix of the form is $-I_{q^{\prime}}$.

Clearly, $V_{1} \cap V_{2}={0}$ so the subspace $V_{1}+V_{2}$ has dimension $p^{\prime}+q^{\prime}$. Assume that $p^{\prime}>p$ or $q^{\prime}>q$. Then with respect to $\mathcal{B}{1} \cup \mathcal{B}{2}$, the restriction of $\langle, \cdot\rangle$ to $V_{1}+V_{2}$ is
$$\left(\begin{array}{cc} I_{p^{\prime}} & 0 \ 0 & -I_{q^{\prime}} \end{array}\right),$$
which implies that the rank of $\langle, y\rangle$ on $V$ is greater than $p+q$. This is a contradiction. We deduce that $\operatorname{dim} V_{1}=p$ and $\operatorname{dim} V_{2}=q$. Since $V_{1}$ and $V_{2}$ were defined without reference to a basis, the theorem follows.

# 微分几何代考

## 数学代写|微分几何代写Differential Geometry代考| Bilinear Forms on V

$$\bar{c} k \ell=\bar{p} k^{i} \tilde{p} \ell^{j} c i j .$$

$$\bar{C}=\left(P^{-1}\right)^{\top} C P^{-1},$$

1. 对称如果 $\langle y, x\rangle=\langle x, y\rangle$ 面向所有人 $x, y \in V$;和
2. 反对称 (或偏对称) 如果 $\langle y, x\rangle=-\langle x, y\rangle$ 面向所有人 $x, y \in V$.

## 数学代写|微分几何代写Differential Geometry代考| Signature of a Symmetric Bilinear Form

$$\left(\begin{array}{llll} I_{p^{\prime}} & 0 & 0 & -I_{q^{\prime}} \end{array}\right),$$

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## MATLAB代写

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assignmentutor™您的专属作业导师
assignmentutor™您的专属作业导师