### 数学代写|计算线性代数代写Computational Linear Algebra代考|Block Multiplication and Triangular Matrices

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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

## 数学代写|计算线性代数代写Computational Linear Algebra代考|Block Multiplication

A rectangular matrix $A$ can be partitioned into submatrices by drawing horizontal lines between selected rows and vertical lines between selected columns. For example, the matrix
$$A=\left[\begin{array}{lll} 1 & 2 & 3 \ 4 & 5 & 6 \ 7 & 8 & 9 \end{array}\right]$$ can be partitioned as
(i) $\left[\begin{array}{ll}A_{11} & A_{12} \ A_{21} & A_{22}\end{array}\right]=\left[\begin{array}{l|ll}1 & 2 & 3 \ \hline 4 & 5 & 6 \ 7 & 8 & 9\end{array}\right]$,
(ii) $\left[\boldsymbol{a}: 1, \boldsymbol{a}{: 2}, \boldsymbol{a}: 3\right]=\left[\begin{array}{c|c|c}1 & 2 & 3 \ 4 & 5 & 6 \ 7 & 8 & 9\end{array}\right]$, (iii) $\left[\begin{array}{c}a{1}^{T} \ a_{2:}^{T} \ a_{3:}^{T}\end{array}\right]=\left[\begin{array}{lll}\frac{1}{2} & 3 \ \frac{4}{}\end{array}\right.$
(iv) $\left[A_{11}, A_{12}\right]=\left[\begin{array}{c|ll}1 & 2 & 3 \ 4 & 5 & 6 \ 7 & 8 & 9\end{array}\right]$.
In ( $i$ ) the matrix $A$ is divided into four submatrices
$$A_{11}=[1], \quad A_{12}=[2,3], A_{21}=\left[\begin{array}{l} 4 \ 7 \end{array}\right] \text {, and } A_{22}=\left[\begin{array}{ll} 5 & 6 \ 8 & 9 \end{array}\right] \text {, }$$
while in (ii) and (iii) A has been partitioned into columns and rows, respectively. The submatrices in a partition are often referred to as blocks and a partitioned matrix is sometimes called a block matrix.

In the following we assume that $\boldsymbol{A} \in \mathbb{C}^{m \times p}$ and $\boldsymbol{B} \in \mathbb{C}^{p \times n}$. Here are some rules and observations for block multiplication.

1. If $\boldsymbol{B}=\left[\boldsymbol{b}{: 1}, \ldots, \boldsymbol{b}{: n}\right]$ is partitioned into columns then the partition of the product $\boldsymbol{A B}$ into columns is
$$A B=\left[A b_{: 1}, A b_{: 2}, \ldots, A b_{: n}\right]$$
In particular, if $\boldsymbol{I}$ is the identity matrix of order $p$ then
$$A=A I=A\left[e_{1}, e_{2}, \ldots, e_{p}\right]=\left[A e_{1}, A e_{2}, \ldots, A e_{p}\right]$$
and we see that column $j$ of $A$ can be written $A e_{j}$ for $j=1, \ldots, p$.
2. Similarly, if $\boldsymbol{A}$ is partitioned into rows then
$$\boldsymbol{A} \boldsymbol{B}=\left[\begin{array}{c} a_{1:}^{T} \ a_{2:}^{T} \ \vdots \ a_{m:}^{T} \end{array}\right] \boldsymbol{B}=\left[\begin{array}{c} a_{1:}^{T} \boldsymbol{B} \ a_{2:}^{T} \boldsymbol{B} \ \vdots \ a_{m:}^{T} \boldsymbol{B} \end{array}\right]$$
and taking $\boldsymbol{A}=\boldsymbol{I}$ it follows that row $i$ of $\boldsymbol{B}$ can be written $\boldsymbol{e}_{i}^{T} \boldsymbol{B}$ for $i=1, \ldots, m$.

## 数学代写|计算线性代数代写Computational Linear Algebra代考|Triangular Matrices

Lemma $2.4$ (Inverse of a Block Triangular Matrix) Suppose
$$A=\left[\begin{array}{cc} \boldsymbol{A}{11} & \boldsymbol{A}{12} \ \mathbf{0} & \boldsymbol{A}{22} \end{array}\right]$$ where $\boldsymbol{A}, \boldsymbol{A}{11}$ and $\boldsymbol{A}{22}$ are square matrices. Then $\boldsymbol{A}$ is nonsingular if and only if both $A{11}$ and $A_{22}$ are nonsingular. In that case
$$\boldsymbol{A}^{-1}=\left[\begin{array}{cc} \boldsymbol{A}{11}^{-1} & \boldsymbol{C} \ \mathbf{0} & \boldsymbol{A}{22}^{-1} \end{array}\right]$$
for some matrix $\boldsymbol{C}$.
Proof Suppose $A$ is nonsingular. We partition $B:=A^{-1}$ conformally with $A$ and have
$$B A=\left[\begin{array}{ll} B_{11} & B_{12} \ B_{21} & B_{22} \end{array}\right]\left[\begin{array}{cc} A_{11} & A_{12} \ \mathbf{0} & A_{22} \end{array}\right]=\left[\begin{array}{ll} I & 0 \ \mathbf{0} & I \end{array}\right]=I$$
Using block-multiplication we find
$$B_{11} A_{11}=I, B_{21} A_{11}=\mathbf{0}, B_{21} A_{12}+B_{22} A_{22}=I, \quad B_{11} A_{12}+B_{12} A_{22}=\mathbf{0}$$

The first equation implies that $A_{11}$ is nonsingular, this in turn implies that $\boldsymbol{B}{21}=$ $\mathbf{0} \boldsymbol{A}{11}^{-1}=\mathbf{0}$ in the second equation, and then the third equation simplifies to $\boldsymbol{B}{22} \boldsymbol{A}{22}=\boldsymbol{I}$. We conclude that also $\boldsymbol{A}{22}$ is nonsingular. From the fourth equation we find $$B{12}=C=-A_{11}^{-1} A_{12} A_{22}^{-1}$$
Conversely, if $\boldsymbol{A}{11}$ and $\boldsymbol{A}{22}$ are nonsingular then
$$\left[\begin{array}{cc} A_{11}^{-1} & -A_{11}^{-1} A_{12} A_{22}^{-1} \ \mathbf{0} & A_{22}^{-1} \end{array}\right]\left[\begin{array}{cc} A_{11} & A_{12} \ \mathbf{0} & A_{22} \end{array}\right]=\left[\begin{array}{ll} I & 0 \ 0 & I \end{array}\right]=I$$
and $A$ is nonsingular with the indicated inverse.
Consider now a triangular matrix.

## 数学代写|计算线性代数代写Computational Linear Algebra代考|3 by 3 Example

Gaussian elimination with row interchanges is the classical method for solving $n$ linear equations in $n$ unknowns. ${ }^{1}$ We first recall how it works on a $3 \times 3$ system.
Example $3.1$ (Gaussian Elimination on a $3 \times 3$ System) Consider a nonsingular system of three equations in three unknowns:
$a_{11}^{(1)} x_{1}+a_{12}^{(1)} x_{2}+a_{13}^{(1)} x_{3}=b_{1}^{(1)}, \quad \mathbf{I}$
$a_{21}^{(1)} x_{1}+a_{22}^{(1)} x_{2}+a_{23}^{(1)} x_{3}=b_{2}^{(1)}, \quad$ II
$a_{31}^{(1)} x_{1}+a_{32}^{(1)} x_{2}+a_{33}^{(1)} x_{3}=b_{3}^{(1)}$. III.

To solve this system by Gaussian elimination suppose $a_{11}^{(1)} \neq 0$. We subtract $l_{21}^{(1)}:=$ $a_{21}^{(1)} / a_{11}^{(1)}$ times equation I from equation II and $l_{31}^{(1)}:=a_{31}^{(1)} / a_{11}^{(1)}$ times equation I from equation III. The result is
$a_{11}^{(1)} x_{1}+a_{12}^{(1)} x_{2}+a_{13}^{(1)} x_{3}=b_{1}^{(1)}, \quad \mathrm{I}$
$a_{22}^{(2)} x_{2}+a_{23}^{(2)} x_{3}=b_{2}^{(2)}, \quad \mathbf{I I}^{\prime}$
$a_{32}^{(2)} x_{2}+a_{33}^{(2)} x_{3}=b_{3}^{(2)}, \quad \mathrm{III}^{\prime}$,
where $b_{i}^{(2)}=b_{i}^{(1)}-l_{i 1}^{(1)} b_{i}^{(1)}$ for $i=2,3$ and $a_{i j}^{(2)}=a_{i j}^{(1)}-l_{i, 1}^{(1)} a_{1 j}^{(1)}$ for $i, j=2,3$. If $a_{11}^{(1)}=0$ and $a_{21}^{(1)} \neq 0$ we first interchange equation I and equation II. If $a_{11}^{(1)}=$ $a_{21}^{(1)}=0$ we interchange equation I and III. Since the system is nonsingular the first column cannot be zero and an interchange is always possible.

If $a_{22}^{(2)} \neq 0$ we subtract $l_{32}^{(2)}:=a_{32}^{(2)} / a_{22}^{(2)}$ times equation $\mathrm{II}^{\prime}$ from equation $\mathrm{III}^{\prime}$ to obtain
$a_{11}^{(1)} x_{1}+a_{12}^{(1)} x_{2}+a_{13}^{(1)} x_{3}=b_{1}^{(1)}, \quad \mathbf{I}$
$a_{22}^{(2)} x_{2}+a_{23}^{(2)} x_{3}=b_{2}^{(2)}, \quad$ II $^{\prime}$
$a_{33}^{(3)} x_{3}=b_{3}^{(3)}, \quad \mathrm{III}^{\prime \prime}$,
where $a_{33}^{(3)}=a_{33}^{(2)}-l_{32}^{(2)} a_{23}^{(2)}$ and $b_{3}^{(3)}=b_{3}^{(2)}-l_{32}^{(2)} b_{2}^{(2)}$. If $a_{22}^{(2)}=0$ then $a_{32}^{(2)} \neq 0$ (cf. Sect. 3.4) and we first interchange equation $\mathrm{II}^{\prime}$ and equation $\mathrm{III}^{\prime}$. The reduced system is easy to solve since it is upper triangular. Starting from the bottom and moving upwards we find
\begin{aligned} &x_{3}=b_{3}^{(3)} / a_{33}^{(3)} \ &x_{2}=\left(b_{2}^{(2)}-a_{23}^{(2)} x_{3}\right) / a_{22}^{(2)} \ &x_{1}=\left(b_{1}^{(1)}-a_{12}^{(1)} x_{2}-a_{13}^{(1)} x_{3}\right) / a_{11}^{(1)} \end{aligned}

## 数学代写|计算线性代数代写Computational Linear Algebra代考|Block Multiplication

(i)\left[\begin{array}{ll}A_{11} & A_{12} \A_{21} & A_{22}\end{array}\right]=\left[\begin{array}{l| ll}1 & 2 & 3 \ \hline 4 & 5 & 6 \ 7 & 8 & 9\end{array}\right]\left[\begin{array}{ll}A_{11} & A_{12} \A_{21} & A_{22}\end{array}\right]=\left[\begin{array}{l| ll}1 & 2 & 3 \ \hline 4 & 5 & 6 \ 7 & 8 & 9\end{array}\right],
(ii)[一个:1,一个:2,一个:3]=[123 456 789], (iii)[一个1吨 一个2:吨 一个3:吨]=[123 4
(四)[一个11,一个12]=[123 456 789].

1. 如果乙=[b:1,…,b:n]被划分为列，然后是产品的划分一个乙成列是
一个乙=[一个b:1,一个b:2,…,一个b:n]
特别是，如果我是阶单位矩阵p然后
一个=一个我=一个[和1,和2,…,和p]=[一个和1,一个和2,…,一个和p]
我们看到那一栏j的一个可以写一个和j为了j=1,…,p.
2. 同样，如果一个然后被划分为行
一个乙=[一个1:吨 一个2:吨 ⋮ 一个米:吨]乙=[一个1:吨乙 一个2:吨乙 ⋮ 一个米:吨乙]
并采取一个=我它跟随那一行一世的乙可以写和一世吨乙为了一世=1,…,米.

## 数学代写|计算线性代数代写Computational Linear Algebra代考|Triangular Matrices

[一个11−1−一个11−1一个12一个22−1 0一个22−1][一个11一个12 0一个22]=[我0 0我]=我

## 数学代写|计算线性代数代写Computational Linear Algebra代考|3 by 3 Example

X3=b3(3)/一个33(3) X2=(b2(2)−一个23(2)X3)/一个22(2) X1=(b1(1)−一个12(1)X2−一个13(1)X3)/一个11(1)

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