### 数学代写|计算线性代数代写Computational Linear Algebra代考|MATH4076

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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|Determinant of a Matrix

Definition 2.20. Let us consider $n$ objects. We will call permutation every grouping of these objects. For example, if we consider three objects $a, b$, and $c$, we could group them as $a-b-c$ or $a-c-b$, or $c-b-a$ or $b-a-c$ or $c-a-b$ or $b-c-a$ . In this case, there are totally six possible permutations. More generally, it can be checked that for $n$ objects there are $n$ ! ( $n$ factorial) permutations where $n !=$ $(n)(n-1)(n-2) \ldots(2)(1)$ with $n \in \mathbb{N}$ and $(0) !=1$.

We could fix a reference sequence (e.g. $a-b-c$ ) and name it fundamental permutation. Every time two objects in a permutation follow each other in a reverse order with respect to the fundamental we will call it inversion. Let us define even class permutation a permutation undergone to an even number of inversions and odd class permutation a permutation undergone to an odd number of inversions, see also [1].

In other words, a sequence is an even class permutation if an even number of swaps is necessary to obtain the fundamental permutation. Analogously, a sequence is an odd class permutation if an odd number of swaps is necessary to obtain the fundamental permutation.

Example 2.19. Let us consider the fundamental permutation $a-b-c-d$ associated with the objects $a, b, c, d$. The permutation $d-a-c-b$ is of even class since two swaps are required to reconstruct the fundamental permutation. At first we swap $a$ and $d$ to obtain $a-d-c-b$ and then we swap $d$ and $b$ to obtain the fundamental permutation $a-b-c-d$.

On the contrary, the permutation $d-c-a-b$ is of odd class since three swaps are necessary to reconstruct the fundamental permutation. Let us reconstruct the fundamental permutation step-by-step. At first we swap $d$ and $b$ and obtain $b-c-$ $a-d$. Then, let us swap $b$ and $a$ to obtain $a-c-b-d$. Eventually, we swap $c$ and $b$ to obtain the fundamental permutation $a-b-c-d$.

## 数学代写|计算线性代数代写Computational Linear Algebra代考|Linear Dependence of Row and Column Vectors of a Matrix

Definition 2.24. Let A be a matrix. The $i^{t h}$ row is said linear combination of the other rows if each of its element $a_{i, j}$ can be expressed as weighted sum of the other elements of the $j^{t h}$ column by means of the same scalars $\lambda_{1}, \lambda_{2}, \ldots, \lambda_{i-1}, \lambda_{i+1}, \ldots \lambda_{n}$ :
$$\mathbf{a}{\mathbf{i}}=\lambda{1} \mathbf{a}{1}+\lambda{2} \mathbf{a}{2}+\cdots+\lambda{i-1} \mathbf{a}{\mathbf{i}-\mathbf{1}}+\lambda{i+1} \mathbf{a}{\mathbf{i}+\mathbf{1}}+\ldots+\lambda{n} \mathbf{a}{\mathbf{n}}$$ Equivalently, we may express the same concept by considering each row element: \begin{aligned} &\forall j: \exists \lambda{1}, \lambda_{2}, \ldots, \lambda_{i-1}, \lambda_{i+1}, \ldots \lambda_{n} \mid \ &a_{i, j}=\lambda_{1} a_{1, j}+\lambda_{2} a_{2, j}+\ldots \lambda_{i-1} a_{i-1, j}+\lambda_{i+1} a_{i+1, j}+\ldots \lambda_{n} a_{n, j} . \end{aligned}

Example 2.27. Let us consider the following matrix:
$$\mathbf{A}=\left(\begin{array}{lll} 0 & 1 & 1 \ 3 & 2 & 1 \ 6 & 5 & 3 \end{array}\right)$$
The third row is a linear combination of the first two by means of scalars $\lambda_{1}, \lambda_{2}=$ 1,2 , the third row is equal to the weighted sum obtained by multiplying the first row by 1 and summing to it the second row multiplied by 2 :
$$(6,5,3)=(0,1,1)+2(3,2,1)$$
that is
$$\mathbf{a}{3}=\mathbf{a}{1}+2 \mathbf{a}{2} .$$ Definition 2.25. Let A be a matrix. The $j^{t h}$ column is said linear combination of the other column if each of its element $a{i, j}$ can be expressed as weighted sum of the other elements of the $i^{t h}$ row by means of the same scalars $\lambda_{1}, \lambda_{2}, \ldots, \lambda_{j-1}$, $\lambda_{j+1}, \ldots \lambda_{n}$ :
$$\mathbf{a}^{\mathbf{j}}=\lambda_{1} \mathbf{a}^{\mathbf{1}}+\lambda_{2} \mathbf{a}^{2}+\cdots+\lambda_{j-1} \mathbf{a}^{\mathbf{j}-\mathbf{1}}+\lambda_{j+1} \mathbf{a}^{\mathbf{j}+\mathbf{1}}+\ldots+\lambda_{n} \mathbf{a}^{\mathbf{n}}$$
Equivalently, we may express the same concept by considering each row element:
\begin{aligned} &\forall i: \exists \lambda_{1}, \lambda_{2}, \ldots, \lambda_{j-1}, \lambda_{j+1}, \ldots \lambda_{n} \mid \ &a_{i, j}=\lambda_{1} a_{i, 1}+\lambda_{2} a_{i, 2}+\ldots \lambda_{i-1} a_{i, j-1}+\lambda_{i+1} a_{i, j+1}+\ldots \lambda_{n} a_{i, n} \end{aligned}

## 数学代写|计算线性代数代写Computational Linear Algebra代考|Laplace Theorems on Determinants

Theorem 2.2. I Laplace Theorem Let $\mathbf{A} \in \mathbb{R}{n, n}$. The determinant of $\mathbf{A}$ can be computed as the sum of each row (element) multiplied by the corresponding cofactor: $\operatorname{det} \mathbf{A}=\sum{j=1}^{n} a_{i, j} A_{i, j}$ for any arbitrary $i$ and
$\operatorname{det} \mathbf{A}=\sum_{i=1}^{n} a_{i, j} A_{i, j}$ for any arbitrary $j$.
The I Laplace Theorem can be expressed in the equivalent form: the determinant of a matrix is equal to scalar product of a row (column) vector by the corresponding vector of cofactors.
Example 2.46. Let us consider the following $\mathbf{A} \in \mathbb{R}{3,3}$ : $$\mathbf{A}=\left(\begin{array}{ccc} 2 & -1 & 3 \ 1 & 2 & -1 \ -1 & -2 & 1 \end{array}\right)$$ The determinant of this matrix is $\operatorname{det} \mathbf{A}=4-1-6+6+1-4=0$. Hence, the matrix is singular. Let us now calculate the determinant by applying the I Laplace Theorem. If we consider the first row, it follows that det $\mathbf{A}=a{1,1} A_{1,1}+a_{1,2}(-1) A_{1,2}+$ $a_{1,3} A_{1,3}$, $\operatorname{det} \mathbf{A}=2(0)+1(0)+3(0)=0$. We arrive to the same conclusion.
Example 2.47. Let us consider the following $\mathbf{A} \in \mathbb{R}{3,3}$ : $$\mathbf{A}=\left(\begin{array}{lll} 1 & 2 & 1 \ 0 & 1 & 1 \ 4 & 2 & 0 \end{array}\right)$$ The determinant of this matrix is $\operatorname{det} \mathbf{A}=8-4-2=2$. Hence, the matrix is nonsingular. Let us now calculate the determinant by applying the I Laplace Theorem. If we consider the second row, it follows that $\operatorname{det} \mathbf{A}=a{2,1}(-1) A_{2,1}+a_{2,2} A_{2,2}+$ $a_{2,3}(-1) A_{2,3}$, $\operatorname{det} \mathbf{A}=0(-1)(-2)+1(-4)+1(-1)(-6)=2$. The result is the same.

## 数学代写|计算线性代数代写Computational Linear Algebra代考|Linear Dependence of Row and Column Vectors of a Matrix

$$\mathbf{a i}=\lambda 1 \mathbf{a} 1+\lambda 2 \mathbf{a} 2+\cdots+\lambda i-1 \mathbf{a} \mathbf{i}-\mathbf{1}+\lambda i+1 \mathbf{a} \mathbf{i}+\mathbf{1}+\ldots+\lambda n \mathbf{a n}$$

$$\forall j: \exists \lambda 1, \lambda_{2}, \ldots, \lambda_{i-1}, \lambda_{i+1}, \ldots \lambda_{n} \mid \quad a_{i, j}=\lambda_{1} a_{1, j}+\lambda_{2} a_{2, j}+\ldots \lambda_{i-1} a_{i-1, j}+\lambda_{i+1} a_{i+1, j}+\ldots \lambda_{n} a_{n, j}$$

$$\mathbf{A}=\left(\begin{array}{lllllll} 0 & 1 & 13 & 2 & 16 & 5 & 3 \end{array}\right)$$

$$(6,5,3)=(0,1,1)+2(3,2,1)$$

$$\mathbf{a} 3=\mathbf{a} 1+2 \mathbf{a} 2 .$$

$$\mathbf{a}^{\mathbf{j}}=\lambda_{1} \mathbf{a}^{\mathbf{1}}+\lambda_{2} \mathbf{a}^{2}+\cdots+\lambda_{j-1} \mathbf{a}^{\mathbf{j}-\mathbf{1}}+\lambda_{j+1} \mathbf{a}^{\mathbf{j}+\mathbf{1}}+\ldots+\lambda_{n} \mathbf{a}^{\mathbf{n}}$$

$$\forall i: \exists \lambda_{1}, \lambda_{2}, \ldots, \lambda_{j-1}, \lambda_{j+1}, \ldots \lambda_{n} \mid \quad a_{i, j}=\lambda_{1} a_{i, 1}+\lambda_{2} a_{i, 2}+\ldots \lambda_{i-1} a_{i, j-1}+\lambda_{i+1} a_{i, j+1}+\ldots \lambda_{n} a_{i, n}$$

## 数学代写|计算线性代数代写Computational Linear Algebra代考|Laplace Theorems on Determinants

$\operatorname{det} \mathbf{A}=\sum j=1^{n} a_{i, j} A_{i, j}$ 对于任何任意 $i$ 和
$\operatorname{det} \mathbf{A}=\sum_{i=1}^{n} a_{i, j} A_{i, j}$ 对于任何任意 $j$.
। 拉普拉斯定理可以表示为等价形式: 矩阵的行列式等于行 (列) 向量与相应的辅因子向量的标量积。 例 2.46。让我们考虑以下内容 $\mathbf{A} \in \mathbb{R} 3,3$ :

$$\mathbf{A}=\left(\begin{array}{llllllll} 1 & 2 & 10 & 1 & 14 & 2 & 0 \end{array}\right)$$

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