### 机器学习代写|主成分分析作业代写PCA代考| Rayleigh Quotient and Its Characteristics

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• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

If the negative direction of RQ gradient is regarded as the gradient flow of vector $\boldsymbol{x}$, e.g.’,
$$\dot{\boldsymbol{x}}=-[\boldsymbol{C}-r(\boldsymbol{x}) \boldsymbol{I}] \boldsymbol{x}$$
then vector $x$ can be computed iteratively by the following gradient algorithm:
$$\boldsymbol{x}(k+1)=\boldsymbol{x}(k)+\mu \dot{\boldsymbol{x}}=\boldsymbol{x}(k)-\mu[\boldsymbol{C}-r(\boldsymbol{x}) \boldsymbol{I}] \boldsymbol{x} .$$
It is worth noting that the gradient algorithm of RQ has faster convergence speed than the iterative algorithm of standard RQ.

In the following, the conjugate gradient algorithm for RQ will be introduced, where $\boldsymbol{A}$ in the RQ is a real symmetric matrix.

Starting from some initial vector, the conjugate gradient algorithm uses the iterative equation, e.g.,
$$\boldsymbol{x}{k+1}=\boldsymbol{x}{k}+\alpha_{k} \boldsymbol{P}{k}$$ to update and approach the eigenvector, associated with the minimal or maximal eigenvalue of a symmetric matrix. The real coefficient $\alpha{k}$ is
$$\alpha_{k}=\pm \frac{1}{2 D}\left(-B+\sqrt{B^{2}-4 C D}\right),$$
where ” $+$ ” is used in the updating of the eigenvector associated with the minimal eigenvalue, and “-” is used in the updating of the eigenvector associated with the maximal eigenvalue. The formulae for parameters D, B, C in the above equations are

$$\left{\begin{array}{c} D=P_{b}(k) P_{c}(k)-P_{a}(k) P_{d}(k) \ B=P_{b}(k)-\lambda_{k} P_{d}(k) \ C=P_{a}(k)-\lambda_{k} P_{c}(k) \ P_{a}(k)=\boldsymbol{P}{k}^{\mathrm{T}} \boldsymbol{A} \boldsymbol{x}{k} /\left(\boldsymbol{x}{k}^{\mathrm{T}} \boldsymbol{x}{k}\right) \ P_{b}(k)=\boldsymbol{p}{k}^{\mathrm{T}} \boldsymbol{A} \boldsymbol{p}{k} /\left(\boldsymbol{x}{k}^{\mathrm{T}} \boldsymbol{x}{k}\right) \ P_{c}(k)=\boldsymbol{p}{k}^{\mathrm{T}} \boldsymbol{x}{k} /\left(\boldsymbol{x}{k}^{\mathrm{T}} \boldsymbol{x}{k}\right) \ P_{d}(k)=\boldsymbol{p}{k}^{\mathrm{T}} \boldsymbol{p}{k} /\left(\boldsymbol{x}{k}^{\mathrm{T}} \boldsymbol{x}{k}\right) \ \lambda_{k}=r\left(\boldsymbol{x}{k}\right)=\boldsymbol{x}{k}^{\mathrm{T}} \boldsymbol{A} \boldsymbol{x}{k} /\left(\boldsymbol{x}{k}^{\mathrm{T}} \boldsymbol{x}{k}\right) . \end{array}\right.$$ At the $k+1$ th iteration, the search direction can be selected as $$\boldsymbol{p}{k+1}=\boldsymbol{r}{k+1}+b(k) \boldsymbol{p}{k}$$
where $b(-1)=0$ and $\boldsymbol{r}{k+1}$ is the residual vector at the $k+1$ th iteration. $\boldsymbol{r}{k+1}$ and $b(k)$ can be computed, respectively, as
$$\boldsymbol{r}{k+1}=-\frac{1}{2} \nabla{x} r\left(\boldsymbol{x}{k+1}\right)=\left(\lambda{k+1} \boldsymbol{x}{k+1}-\boldsymbol{A} \boldsymbol{x}{k+1}\right) /\left(\boldsymbol{x}{k=1}^{\mathrm{T}} \boldsymbol{x}{k+1}\right)$$
and
$$b(k)=-\frac{\boldsymbol{r}{k+1}^{\mathrm{T}} \boldsymbol{A} \boldsymbol{p}{k}+\left(\boldsymbol{r}{k+1}^{\mathrm{T}} \boldsymbol{r}{k+1}\right)\left(\boldsymbol{x}{k+1}^{\mathrm{T}} \boldsymbol{p}{k}\right)}{\boldsymbol{p}{k}^{\mathrm{T}}\left(\boldsymbol{A} \boldsymbol{p}{k}-\lambda_{k+1} \boldsymbol{I}\right) \boldsymbol{p}{k}}$$ Equations (2.5)-(2.9) constitute the conjugate gradient algorithm for RQ, which was proposed in [11]. If the updated $x{k}$ is normalized to one and “t” (or “-“) is selected in Eq. (2.6), the above algorithm will obtain the minimal (or maximal) eigenvalue of matrix $A$ and its associated eigenvectors.

## 机器学习代写|主成分分析作业代写PCA代考|Generalized Rayleigh Quotient

Definition $2.3$ Assume that $A \in \mathbb{C}^{n \times n}, \mathbf{B} \in \mathbb{C}^{n \times n}$ are both Hermitian matrices, and $\boldsymbol{B}$ is positive definite. The generalized RQ or generalized Rayleigh-Ritz of the matrix pencil $(\boldsymbol{A}, \boldsymbol{B})$ is a scalar function, e.g.,
$$r(\boldsymbol{x})=\frac{\boldsymbol{x}^{H} \boldsymbol{A} \boldsymbol{x}}{\boldsymbol{x}^{H} \boldsymbol{B} \boldsymbol{x}},$$
where $x$ is a quantity to be selected, and the objective is to maximize or minimize the generalized RQ.

In order to solve for the generalized RQ, define a new vector $\tilde{\boldsymbol{x}}=\boldsymbol{B}^{1 / 2} \boldsymbol{x}$, where $\boldsymbol{B}^{1 / 2}$ is the square root of the positive definite $\boldsymbol{B}$. Replace $\boldsymbol{x}$ by $\boldsymbol{B}^{-1 / 2} \tilde{\boldsymbol{x}}$ in (2.43). Then it holds that
$$r(\tilde{\boldsymbol{x}})=\frac{\tilde{\boldsymbol{x}}^{H}\left(\boldsymbol{B}^{-1 / 2}\right)^{H} \boldsymbol{A}\left(\boldsymbol{B}^{-1 / 2}\right)^{H} \tilde{\boldsymbol{x}}}{\tilde{\boldsymbol{x}}^{H} \tilde{\boldsymbol{x}}},$$
which shows that the generalized RQ of matrix pencil $(\boldsymbol{A}, \boldsymbol{B})$ is equivalent to the RQ of matrix product $\left(\boldsymbol{B}^{-1 / 2}\right)^{H} \boldsymbol{A}\left(\boldsymbol{B}^{-1 / 2}\right)^{H}$. From the Rayleigh-Ritz theorem, it is clear that when vector $\tilde{x}$ is the eigenvector associated with the smallest eigenvalue $\lambda_{\min }$ of matrix product $\left(\boldsymbol{B}^{-1 / 2}\right)^{H} \boldsymbol{A}\left(\boldsymbol{B}^{-1 / 2}\right)^{H}$, the generalized RQ obtains $\lambda_{\min }$. And if vector $\tilde{\boldsymbol{x}}$ is the eigenvector associated with the largest eigenvalue $\lambda_{\max }$ of matrix product $\left(\boldsymbol{B}^{-1 / 2}\right)^{H} \boldsymbol{A}\left(\boldsymbol{B}^{-1 / 2}\right)^{H}$, the generalized RQ obtains $\lambda_{\max }$.

In the following, we review the eigen decomposition of matrix product $\left(\boldsymbol{B}^{-1 / 2}\right)^{H} \boldsymbol{A}\left(\boldsymbol{B}^{-1 / 2}\right)^{H}$, e.g.,
$$\left(\boldsymbol{B}^{-1 / 2}\right)^{H} \boldsymbol{A}\left(\boldsymbol{B}^{-1 / 2}\right)^{H} \tilde{\boldsymbol{x}}=i \tilde{\boldsymbol{x}} .$$
If $\mathbf{B}=\sum_{i=1}^{n} \beta_{i} v_{i} v_{i}^{H}$ is an eigen decomposition of matrix $\boldsymbol{B}$, then
$$\mathbf{B}^{1 / 2}=\sum_{i=1}^{n} \sqrt{\beta_{i}} v_{i} v_{i}^{H}$$
and $\boldsymbol{B}^{1 / 2} \boldsymbol{B}^{1 / 2}=\boldsymbol{B}$. Since matrix $\boldsymbol{B}^{1 / 2}$ and $\boldsymbol{B}^{-1 / 2}$ have the same eigenvectors and their eigenvalues are reciprocals to each other, then it follows that
$$\mathbf{B}^{-1 / 2}=\sum_{i=1}^{n} \frac{1}{\sqrt{\beta_{i}}} v_{i} v_{i}^{H},$$
which shows that $\boldsymbol{B}^{-1 / 2}$ is also an Hermitian matrix, e.g., $\left(\boldsymbol{B}^{-1 / 2}\right)^{H}=\boldsymbol{B}^{-1 / 2}$.

## 机器学习代写|主成分分析作业代写PCA代考|Differential and Integral of Matrix with Respect to Scalar

If $\boldsymbol{A}(t)=\left{a_{i j}(t)\right}_{m \times n}$ is a real matrix function of scalar $t$, then its differential and integral are, respectively, defined as
\left{\begin{aligned} \frac{\mathrm{d}}{\mathrm{d} t} \boldsymbol{A}(t) &=\left{\frac{\mathrm{d}}{\mathrm{d}{t}} a{i j}(t)\right}_{m \times n} \ \int A(t) \mathrm{d} t &=\left{\int a_{i j}(t) \mathrm{d} t\right}_{m \times n} \end{aligned}\right.
If $\boldsymbol{A}(\mathrm{t})$ and $\boldsymbol{B}(\mathrm{t})$ are, respectively, $m \times n$ and $n \times r$ matrices, then
$$\frac{\mathrm{d}}{\mathrm{d} t}[\boldsymbol{A}(t) \boldsymbol{B}(t)]=\left[\frac{\mathrm{d} \boldsymbol{A}(t)}{\mathrm{d} t}\right] \boldsymbol{B}(t)+\boldsymbol{A}(t)\left[\frac{\mathrm{d} \boldsymbol{B}(t)}{\mathrm{d} t}\right] .$$
If $\boldsymbol{A}(\mathrm{t})$ and $\boldsymbol{B}(\mathrm{t})$ are both $m \times n$ matrices, then
$$\frac{\mathrm{d}}{\mathrm{d} t}[\boldsymbol{A}(t)+\boldsymbol{B}(t)]=\frac{\mathrm{d} \boldsymbol{A}(t)}{\mathrm{d} t}+\frac{\mathrm{d} \boldsymbol{B}(t)}{\mathrm{d} t}$$
If $A(\mathrm{t})$ is a rank- $n$ invertible square matrix, then
$$\frac{\mathrm{d} \boldsymbol{A}^{-1}(t)}{\mathrm{d} t}=-\boldsymbol{A}^{-1}(t) \frac{\mathrm{d} \boldsymbol{A}(t)}{\mathrm{d} t} \boldsymbol{A}^{-1}(t) .$$

## 主成分分析代写

X˙=−[C−r(X)一世]X

X(ķ+1)=X(ķ)+μX˙=X(ķ)−μ[C−r(X)一世]X.

Xķ+1=Xķ+一种ķ磷ķ更新和逼近与对称矩阵的最小或最大特征值相关的特征向量。实际系数一种ķ是

$$\左{D=磷b(ķ)磷C(ķ)−磷一种(ķ)磷d(ķ) 乙=磷b(ķ)−λķ磷d(ķ) C=磷一种(ķ)−λķ磷C(ķ) 磷一种(ķ)=磷ķ吨一种Xķ/(Xķ吨Xķ) 磷b(ķ)=pķ吨一种pķ/(Xķ吨Xķ) 磷C(ķ)=pķ吨Xķ/(Xķ吨Xķ) 磷d(ķ)=pķ吨pķ/(Xķ吨Xķ) λķ=r(Xķ)=Xķ吨一种Xķ/(Xķ吨Xķ).\对。一种吨吨H和ķ+1吨H一世吨和r一种吨一世这n,吨H和s和一种rCHd一世r和C吨一世这nC一种nb和s和l和C吨和d一种s\boldsymbol{p}{k+1}=\boldsymbol{r}{k+1}+b(k) \boldsymbol{p}{k} 在H和r和b(−1)=0一种ndrķ+1一世s吨H和r和s一世d在一种l在和C吨这r一种吨吨H和ķ+1吨H一世吨和r一种吨一世这n.rķ+1一种ndb(ķ)C一种nb和C这米p在吨和d,r和sp和C吨一世在和l是,一种s \boldsymbol{r}{k+1}=-\frac{1}{2} \nabla{x} r\left(\boldsymbol{x}{k+1}\right)=\left(\lambda{k +1} \boldsymbol{x}{k+1}-\boldsymbol{A} \boldsymbol{x}{k+1}\right) /\left(\boldsymbol{x}{k=1}^{\mathrm {T}} \boldsymbol{x}{k+1}\right) 一种nd b(k)=-\frac{\boldsymbol{r}{k+1}^{\mathrm{T}} \boldsymbol{A} \boldsymbol{p}{k}+\left(\boldsymbol{r}{ k+1}^{\mathrm{T}} \boldsymbol{r}{k+1}\right)\left(\boldsymbol{x}{k+1}^{\mathrm{T}} \boldsymbol{p }{k}\right)}{\boldsymbol{p}{k}^{\mathrm{T}}\left(\boldsymbol{A} \boldsymbol{p}{k}-\lambda_{k+1} \ boldsymbol{I}\right) \boldsymbol{p}{k}}$$ 方程 (2.5)-(2.9) 构成了 RQ 的共轭梯度算法，该算法在 [11] 中提出。如果更新Xķ被归一化为 1 并且在方程式中选择“t”（或“-”）。(2.6)，上述算法将得到矩阵的最小（或最大）特征值一种及其相关的特征向量。

## 机器学习代写|主成分分析作业代写PCA代考|Generalized Rayleigh Quotient

r(X)=XH一种XXH乙X,

r(X~)=X~H(乙−1/2)H一种(乙−1/2)HX~X~HX~,

(乙−1/2)H一种(乙−1/2)HX~=一世X~.

## 机器学习代写|主成分分析作业代写PCA代考|Differential and Integral of Matrix with Respect to Scalar

\left{\begin{aligned} \frac{\mathrm{d}}{\mathrm{d} t} \boldsymbol{A}(t) &=\左{\frac{\mathrm{d}}{\mathrm{d} {t}} a {ij}(t)\right}_{m \times n} \ \int A(t) \mathrm{d} t &=\left{\int a_{ij}(t) \mathrm{d} t\right}_{m \times n} \end{aligned}\right. 一世F一种(吨)一种nd乙(吨)一种r和,r和sp和C吨一世在和l是,米×n一种ndn×r米一种吨r一世C和s,吨H和n \frac{\mathrm{d}}{\mathrm{d} t}[\boldsymbol{A}(t) \boldsymbol{B}(t)]=\left[\frac{\mathrm{d} \boldsymbol{ A}(t)}{\mathrm{d} t}\right] \boldsymbol{B}(t)+\boldsymbol{A}(t)\left[\frac{\mathrm{d} \boldsymbol{B} (t)}{\mathrm{d} t}\right] 。 一世F一种(吨)一种nd乙(吨)一种r和b这吨H米×n米一种吨r一世C和s,吨H和n \frac{\mathrm{d}}{\mathrm{d} t}[\boldsymbol{A}(t)+\boldsymbol{B}(t)]=\frac{\mathrm{d} \boldsymbol{A} (t)}{\mathrm{d} t}+\frac{\mathrm{d} \boldsymbol{B}(t)}{\mathrm{d} t} 一世F一种(吨)一世s一种r一种nķ−n一世n在和r吨一世bl和sq在一种r和米一种吨r一世X,吨H和n \frac{\mathrm{d} \boldsymbol{A}^{-1}(t)}{\mathrm{d} t}=-\boldsymbol{A}^{-1}(t) \frac{\mathrm {d} \boldsymbol{A}(t)}{\mathrm{d} t} \boldsymbol{A}^{-1}(t) 。

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