### 统计代写|数值分析和优化代写numerical analysis and optimazation代考|Conjugate Gradients statistics-lab™ 为您的留学生涯保驾护航 在代写数值分析和优化numerical analysis and optimazation方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写数值分析和优化numerical analysis and optimazation方面经验极为丰富，各种代写数值分析和优化numerical analysis and optimazation相关的作业也就用不着说。

• 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

## 统计代写|数值分析和优化代写numerical analysis and optimazation代考|Conjugate Gradients

We have already seen how the positive definiteness of $A$ can be used to define a norm. This can be extended to have a concept similar to orthogonality.
Definition 2.8. The vectors $\mathbf{u}$ and $\mathbf{v}$ are conjugate with respect to the positive definite matrix $A$, if they are nonzero and satisfy $\mathbf{u}^{T} A \mathbf{v}=0$.

Conjugacy plays such an important role, because through it search directions can be constructed such that the new gradient $\mathbf{g}^{(k+1)}$ is not just orthogonal to the current search direction $\mathbf{d}^{(k)}$ but to all previous search directions. This avoids revisiting search directions as in Figure 2.1.

Specifically, the first search direction is chosen as $\mathbf{d}^{(0)}=-\mathbf{g}^{(0)}$. The following search directions are then constructed by
$$\mathbf{d}^{(k)}=-\mathbf{g}^{(k)}+\beta^{(k)} \mathbf{d}^{(k-1)}, \quad k=1,2, \ldots,$$
where $\beta^{(k)}$ is chosen such that the conjugacy condition $\mathbf{d}^{(k)^{T}} A \mathbf{d}^{(k-1)}=0$ is satisfied. This yields
$$\beta^{(k)}=\frac{\mathbf{g}^{(k)^{T}} A \mathbf{d}^{(k-1)}}{\mathbf{d}^{(k-1)^{T}} A \mathbf{d}^{(k-1)}}, \quad k=1,2, \ldots$$
This gives the conjugate gradient method.
The values of $\mathbf{x}^{(k+1)}$ and $\omega^{(k)}$ are calculated as before in (2.7) and $(2.9)$.
These search directions satisfy the descent condition. From Equation (2.8) we have seen that the descent condition is equivalent to $\mathbf{d}^{(k)^{T}} \mathbf{g}^{(k)}<0$. Using the formula for $\mathbf{d}^{(k)}$ above and the fact that the new gradient is orthogonal to the previous search direction, i.e., $\mathbf{d}^{(k-1)^{T}} \mathbf{g}^{(k)}=0$, we see that
$$\mathbf{d}^{(k)^{T}} \mathbf{g}^{(k)}=-\left|\mathbf{g}^{(k)}\right|^{2}<0$$

## 统计代写|数值分析和优化代写numerical analysis and optimazation代考|Krylov Subspaces and Pre-Conditioning

Definition 2.9. Let $A$ be an $n \times n$ matrix, $\mathbf{b} \in \mathbb{R}^{n}$ a non-zero vector, then for a number $m$ the space spanned by $A^{j} \mathbf{b}, j=0, \ldots, m-1$ is the $m^{\text {th }}$ Krylov subspace of $\mathbb{R}^{n}$ and is denoted by $K_{m}(A, \mathbf{b})$.

In our analysis of the conjugate gradient method we saw that in the $k^{\text {th }}$ iteration the space spanned by the search directions $\mathbf{d}^{(j)}$ and the space spanned by the gradients $\mathbf{g}^{(j)}, j=0, \ldots, k$, are the same.

Lemma 2.3. The space spanned by $\mathbf{g}^{(j)}\left(\right.$ or $\left.\mathbf{d}^{(j)}\right), j=0, \ldots, k$, is the same as the $k+1^{\text {th }}$ Krylov subspace.
Proof. For $k=0$ we have $\mathbf{d}^{(0)}=-\mathbf{g}^{(0)}=\mathbf{b} \in K_{1}(A, \mathbf{b})$.
We assume that the space spanned by $\mathbf{g}^{(j)}, j=0, \ldots, k$, is the same as the space spanned by $\mathbf{b}, A \mathbf{b}, \ldots, A^{k} \mathbf{b}$ and increase $k$ by one.

In the formula $\mathbf{g}^{(k+1)}=\mathbf{g}^{(k)}+\omega^{(k)} A \mathbf{d}^{(k)}$ both $\mathbf{g}^{(k)}$ and $\mathbf{d}^{(k)}$ can be expressed as linear combinations of $\mathbf{b}, A \mathbf{b}, \ldots, A^{k} \mathbf{b}$ by the inductive hypothesis. Thus $\mathbf{g}^{(k+1)}$ can be expressed as a linear combination of $\mathbf{b}, A \mathbf{b}, \ldots, A^{k+1} \mathbf{b}$. Equally, using $\mathbf{d}^{(k+1)}=-\mathbf{g}^{(k+1)}+\beta^{(k)} \mathbf{d}^{(k)}$ we show that $\mathbf{d}^{(k+1)}$ lies in the space spanned by b, $A \mathbf{b}, \ldots, A^{k+1} \mathbf{b}$, which is $K_{k+2}(A, \mathbf{b})$. This completes the proof.

In the following lemma we show some properties of the Krylov subspaces.
Lemma 2.4. Given $A$ and b. Then $K_{m}(A, \mathbf{b})$ is a subspace of $K_{m+1}(A, \mathbf{b})$ and there exists a positive integer $s \leq n$ such that for every $m \geq s K_{m}(A, \mathbf{b})=$ $K_{s}(A, \mathbf{b})$. Furthermore, if we express $\mathbf{b}$ as $\mathbf{b}=\sum_{i=1}^{t} c_{i} \mathbf{v}{i}$, where $\mathbf{v}{1}, \ldots, \mathbf{v}{t}$ are eigenvectors of $A$ corresponding to distinct eigenvalues and all coefficients $c{i}, i=1, \ldots, t$, are non-zero, then $s=t$.

Proof. Clearly, $K_{m}(A, \mathbf{b}) \subseteq K_{m+1}(A, \mathbf{b})$. The dimension of $K_{m}(A, \mathbf{b})$ is less than or equal to $m$, since it is spanned by $m$ vectors. It is also at most $n$ since

$K_{m}(A, \mathbf{b})$ is a subspace of $\mathbb{R}^{n}$. The first Krylov subspace has dimension 1 . We let $s$ be the greatest integer such that the dimension of $K_{s}(A, \mathbf{b})$ is $s$. Then the dimension of $K_{s+1}(A, \mathbf{b})$ cannot be $s+1$ by the choice of $s$. It has to be $s$, since $K_{s}(A, \mathbf{b}) \subseteq K_{s+1}(A, \mathbf{b})$. Hence the two spaces are the same. This means that $A^{s} \mathbf{b} \in K_{s}(A, \mathbf{b})$, i.e., $A^{s} \mathbf{b}=\sum_{j=0}^{s-1} a_{j} A^{j} \mathbf{b}$. But then
$$A^{s+r} \mathbf{b}=\sum_{j=0}^{s-1} a_{j} A^{j+r} \mathbf{b}$$
for any positive $r$. This means that also the spaces $K_{s+r+1}(A$, b $)$ and $K_{s+r}(A, \mathbf{b})$ are the same. Therefore, for every $m \geq s K_{m}(A, \mathbf{b})=K_{s}(A, \mathbf{b})$.
Suppose now that $\mathbf{b}=\sum_{i=1}^{t} c_{i} \mathbf{v}{i}$, where $\mathbf{v}{1}, \ldots, \mathbf{v}{t}$ are eigenvectors of $A$ corresponding to distinct eigenvalues $\lambda{i}$. Then for every $j=1, \ldots, s$
$$A^{j} \mathbf{b}=\sum_{i=1}^{t} c_{i} \lambda_{i}^{j} \mathbf{v}_{i}$$

## 统计代写|数值分析和优化代写numerical analysis and optimazation代考|Eigenvalues and Eigenvectors

So far we only considered eigenvalues when analyzing the properties of numerical methods. In the following sections we look at how to determine eigenvalues and eigenvectors. Let $A$ be a real $n \times n$ matrix. The eigenvalue equation is given by
$$A \mathbf{v}=\lambda \mathbf{v} \text {, }$$
where $\lambda$ is scalar. It may be complex if $A$ is not symmetric. There exists $\mathbf{v} \in \mathbb{R}^{n}$ satisfying the eigenvalue equation if and only if the determinant $\operatorname{det}(A-\lambda I)$ is zero. The function $p(\lambda)=\operatorname{det}(A-\lambda I), \lambda \in \mathbb{C}$, is a polynomial of degree $n$. However, calculating the eigenvalues by finding the roots of $p$ is generally unsuitable because of loss of accuracy due to rounding errors. In Chapter 1 , Fundamentals, we have seen how even finding the roots of a quadratic can be difficult due to loss of significance.

If the polynomial has some multiple roots and if $A$ is not symmetric, then $A$ might have less than $n$ linearly independent eigenvalues. However, there are always $n$ mutually orthogonal real eigenvectors when $A$ is symmetric. In the following we assume that $A$ has $n$ linearly independent eigenvectors $\mathbf{v}{j}$ for each eigenvalue $\lambda{j}, j=1, \ldots, n$. This can be achieved by perturbing $A$ slightly if necessary. The task is now to find $\mathbf{v}{j}$ and $\lambda{j}, j=1, \ldots, n$.

## 统计代写|数值分析和优化代写numerical analysis and optimazation代考|Conjugate Gradients

We have already seen how the positive definiteness of A can be used to define a norm. This can be extended to have a concept similar to orthogonality.
Definition 2.8. The vectors u and v are conjugate with respect to the positive definite matrix A, if they are nonzero and satisfy uTAv=0.

Conjugacy plays such an important role, because through it search directions can be constructed such that the new gradient g(k+1) is not just orthogonal to the current search direction d(k) but to all previous search directions. This avoids revisiting search directions as in Figure 2.1.

Specifically, the first search direction is chosen as d(0)=−g(0). The following search directions are then constructed by
d(k)=−g(k)+β(k)d(k−1),k=1,2,…,
where β(k) is chosen such that the conjugacy condition d(k)TAd(k−1)=0 is satisfied. This yields
This gives the conjugate gradient method.
The values of x(k+1) and ω(k) are calculated as before in (2.7) and (2.9).
These search directions satisfy the descent condition. From Equation (2.8) we have seen that the descent condition is equivalent to d(k)Tg(k)<0. Using the formula for d(k) above and the fact that the new gradient is orthogonal to the previous search direction, i.e., d(k−1)Tg(k)=0, we see that
d(k)Tg(k)=−|g(k)|2<0

## 统计代写|数值分析和优化代写numerical analysis and optimazation代考|Krylov Subspaces and Pre-Conditioning

Definition 2.9. Let A be an n×n matrix, b∈Rn a non-zero vector, then for a number m the space spanned by Ajb,j=0,…,m−1 is the mth  Krylov subspace of Rn and is denoted by Km(A,b).

In our analysis of the conjugate gradient method we saw that in the kth  iteration the space spanned by the search directions d(j) and the space spanned by the gradients g(j),j=0,…,k, are the same.

Lemma 2.3. The space spanned by g(j)( or d(j)),j=0,…,k, is the same as the k+1th  Krylov subspace.
Proof. For k=0 we have d(0)=−g(0)=b∈K1(A,b).
We assume that the space spanned by g(j),j=0,…,k, is the same as the space spanned by b,Ab,…,Akb and increase k by one.

In the formula g(k+1)=g(k)+ω(k)Ad(k) both g(k) and d(k) can be expressed as linear combinations of b,Ab,…,Akb by the inductive hypothesis. Thus g(k+1) can be expressed as a linear combination of b,Ab,…,Ak+1b. Equally, using d(k+1)=−g(k+1)+β(k)d(k) we show that d(k+1) lies in the space spanned by b, Ab,…,Ak+1b, which is Kk+2(A,b). This completes the proof.

In the following lemma we show some properties of the Krylov subspaces.
Lemma 2.4. Given A and b. Then Km(A,b) is a subspace of Km+1(A,b) and there exists a positive integer s≤n such that for every m≥sKm(A,b)= Ks(A,b). Furthermore, if we express b as b=∑i=1tcivi, where v1,…,vt are eigenvectors of A corresponding to distinct eigenvalues and all coefficients ci,i=1,…,t, are non-zero, then s=t.

Proof. Clearly, Km(A,b)⊆Km+1(A,b). The dimension of Km(A,b) is less than or equal to m, since it is spanned by m vectors. It is also at most n since

Km(A,b) is a subspace of Rn. The first Krylov subspace has dimension 1 . We let s be the greatest integer such that the dimension of Ks(A,b) is s. Then the dimension of Ks+1(A,b) cannot be s+1 by the choice of s. It has to be s, since Ks(A,b)⊆Ks+1(A,b). Hence the two spaces are the same. This means that Asb∈Ks(A,b), i.e., Asb=∑j=0s−1ajAjb. But then
As+rb=∑j=0s−1ajAj+rb
for any positive r. This means that also the spaces Ks+r+1(A, b ) and Ks+r(A,b) are the same. Therefore, for every m≥sKm(A,b)=Ks(A,b).
Suppose now that b=∑i=1tcivi, where v1,…,vt are eigenvectors of A corresponding to distinct eigenvalues λi. Then for every j=1,…,s
Ajb=∑i=1tciλijvi

## 统计代写|数值分析和优化代写numerical analysis and optimazation代考|Eigenvalues and Eigenvectors

So far we only considered eigenvalues when analyzing the properties of numerical methods. In the following sections we look at how to determine eigenvalues and eigenvectors. Let A be a real n×n matrix. The eigenvalue equation is given by
Av=λv,
where λ is scalar. It may be complex if A is not symmetric. There exists v∈Rn satisfying the eigenvalue equation if and only if the determinant det⁡(A−λI) is zero. The function p(λ)=det⁡(A−λI),λ∈C, is a polynomial of degree n. However, calculating the eigenvalues by finding the roots of p is generally unsuitable because of loss of accuracy due to rounding errors. In Chapter 1 , Fundamentals, we have seen how even finding the roots of a quadratic can be difficult due to loss of significance.

If the polynomial has some multiple roots and if A is not symmetric, then A might have less than n linearly independent eigenvalues. However, there are always n mutually orthogonal real eigenvectors when A is symmetric. In the following we assume that A has n linearly independent eigenvectors vj for each eigenvalue λj,j=1,…,n. This can be achieved by perturbing A slightly if necessary. The task is now to find vj and λj,j=1,…,n.

## 广义线性模型代考

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