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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|Complex Vectors, Matrices and Systems of Linear Equation

All the theory and examples in the previous chapters refer to real numbers. However, it is worth mentioning that the algebra analysed up to this point can be straightforwardly extended to the complex case. This section elaborates on this statement by showing several examples.

Definition 5.3. Let $\mathbb{C}$ be the complex set and $\mathbb{C}^{n}=\mathbb{C} \times \mathbb{C} \times \ldots \times \mathbb{C}$ the Cartesian product obtained by the composition of the complex set calculated $n$ times.

A generic element $\mathbf{u} \in \mathbb{C}^{n}$ is named complex vector and is an $n$-tuple of the type
$$\mathbf{u}=\left(a_{1}+j b_{1}, a_{2}+j b_{2}, \ldots, a_{n}+j b_{n}\right)$$
where each component $a_{k}+j b_{k}$ is a complex number.
Example 5.15. The following is a complex vector of $\mathbb{C}^{3}$ :
$$\mathbf{u}=(3-j 2,4,1+j 7) .$$
By using the operation of sum and product of complex numbers we can define the scalar product of complex vectors.

Definition 5.4. Let $\mathbf{u}=\left(u_{1}, u_{2}, \ldots, u_{n}\right)$ and $\mathbf{v}=\left(v_{1}, v_{2}, \ldots, v_{n}\right)$ be two vectors of $\mathbb{C}^{n}$. The scalar product uv is
$$\mathbf{u v}=\sum_{j=1}^{n} u_{j} v_{j} .$$
Example 5.16. Let us consider the following complex vectors of $\mathbb{C}^{3}$ :
\begin{aligned} &\mathbf{u}=(1,1+j 2,0) \ &\mathbf{v}=(3-j, j 5,6-j 2) \end{aligned}
The scalar product is
$$\mathbf{u v}=3-j+-10+j 5+0=-7+j 4$$
Similarly, we may think about a matrix whose elements are complex numbers.

## 数学代写|计算线性代数代写Computational Linear Algebra代考|Operations of Polynomials

Definition 5.8. Let $n \in \mathbb{N}$ and $a_{0}, a_{1}, \ldots, a_{n} \in \mathbb{C}$. The function $p(z)$ in the complex variable $z \in \mathbb{C}$ defined as
$$p(z)=a_{0}+a_{1} z+a_{2} z^{2}+\ldots+\ldots a_{n} z^{n}=\sum_{k=0}^{n} a_{k} z^{k}$$
is said complex polynomial in the coefficients $a_{k}$ and complex variable $z$. The order $n$ of the polynomial is the maximum value of $k$ corresponding to a non-null coefficient $a_{k}$
Example 5.26. The following function
$$p(z)=4 z^{4}-5 z^{3}+z^{2}-6$$
is a polynomial.
Definition 5.9. Let $p(z)=\sum_{k=0}^{n} a_{k} z^{k}$ be a polynomial. If $\forall k \in \mathbb{N}$ with $k \leq n: a_{k}=0$, the polynomial is said null polynomial.

Definition 5.10. Let $p(z)=\sum_{k=0}^{n} a_{k} z^{k}$ be a polynomial. If $\forall k \in \mathbb{N}$ with $0<k \leq n$ : $a_{k}=0$ and $a_{0} \neq 0$, the polynomial is said constant polynomial.

Definition 5.11. Identity Principle. Let $p_{1}(z)=\sum_{k=0}^{n} a_{k} z^{k}$ and $p_{2}(z)=\sum_{k=0}^{n} b_{k} z^{k}$ be two complex polynomials. The two polynomials are said identical $p_{1}(z)=p_{2}(z)$ if and only if the following two conditions are both satisfied:

• the order $n$ of the two polynomials is the same
• $\forall k \in \mathbb{N}$ with $k \leq n: a_{k}=b_{k}$.
Example 5.27. Let $p_{1}(z)=\sum_{k=0}^{n} a_{k} z^{k}$ and $p_{2}(z)=\sum_{k=0}^{m} b_{k} z^{k}$ be two complex polynomials with $m<n$. The two polynomials are identical if and only if
• $\forall k \in \mathbb{N}$ with $k \leq m: a_{k}=b_{k}$
• $\forall k \in \mathbb{N}$ with $m<k \leq n: a_{k}=0$

## 数学代写|计算线性代数代写Computational Linear Algebra代考|Roots of Polynomials

Definition 5.15. Let $p(z)$ be a polynomial. The values of $z$ such that $p(z)=0$ are said roots or solutions of the polynomial.

Corollary 5.1. Ruffini’s Theorem. Let $p(z)=\sum_{k=0}^{n} a_{k} z^{k}$ be a complex polynomial having order $n \geq 1$. The polynomial $p(z)$ is divisible by $(z-\alpha)$ if and only if $p(\alpha)=$ $0(\alpha$ is a root of the polynomial).
Proof. If $p(z)$ is divisible by $(z-\alpha)$ then we may write
$$p(z)=(z-\alpha) q(z)$$

Thus, for $z=\alpha$ we have
$$p(\alpha)=(\alpha-\alpha) q(\alpha)=0 . \square$$
If $\alpha$ is a root of the polynomial, then $p(\alpha)=0$. Considering that
$$p(z)=(z-\alpha) q(z)+r(z)$$
and for the little Bézout’s Theorem $p(\alpha)=r$, it follows that $r=0$ and that
$$p(z)=(z-\alpha) q(z)$$
that is $p(z)$ is divisible by $(z-\alpha)$.
Example 5.33. Let us consider the division of polynomials
$$\frac{\left(-z^{4}+3 z^{2}-5\right)}{(z+2)} \text {. }$$
It can be easily verified that the polynomial reminder of this division is
$$r=p(-2)=-9$$

## 数学代写|计算线性代数代写Computational Linear Algebra代考|Complex Vectors, Matrices and Systems of Linear Equation

$$\mathbf{u}=\left(a_{1}+j b_{1}, a_{2}+j b_{2}, \ldots, a_{n}+j b_{n}\right)$$

$$\mathbf{u}=(3-j 2,4,1+j 7)$$

$$\mathbf{u v}=\sum_{j=1}^{n} u_{j} v_{j}$$

$$\mathbf{u}=(1,1+j 2,0) \quad \mathbf{v}=(3-j, j 5,6-j 2)$$

$$\mathbf{u v}=3-j+-10+j 5+0=-7+j 4$$

## 数学代写|计算线性代数代写Computational Linear Algebra代考|Operations of Polynomials

$$p(z)=a_{0}+a_{1} z+a_{2} z^{2}+\ldots+\ldots a_{n} z^{n}=\sum_{k=0}^{n} a_{k} z^{k}$$

$$p(z)=4 z^{4}-5 z^{3}+z^{2}-6$$

• 命令 $n$ 两个多项式相同
• $\forall k \in \mathbb{N}$ 和 $k \leq n: a_{k}=b_{k}$.
例 5.27。让 $p_{1}(z)=\sum_{k=0}^{n} a_{k} z^{k}$ 和 $p_{2}(z)=\sum_{k=0}^{m} b_{k} z^{k}$ 是两个复多项式 $m<n$. 两个多项式相同当且仅当
• $\forall k \in \mathbb{N}$ 和 $k \leq m: a_{k}=b_{k}$
• $\forall k \in \mathbb{N}$ 和 $m<k \leq n: a_{k}=0$

## 数学代写|计算线性代数代写Computational Linear Algebra代考|Roots of Polynomials

$$p(z)=(z-\alpha) q(z)$$

$$p(\alpha)=(\alpha-\alpha) q(\alpha)=0$$

$$p(z)=(z-\alpha) q(z)+r(z)$$

$$p(z)=(z-\alpha) q(z)$$

$$\frac{\left(-z^{4}+3 z^{2}-5\right)}{(z+2)}$$

$$r=p(-2)=-9$$

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