• 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|高等线性代数代写Advanced Linear Algebra代考|Jordan canonical form for nilpotent matrices

We will see that the Cayley-Hamilton theorem (Theorem 4.1.1) plays a crucial role in obtaining the Jordan canonical of a matrix. In this section we focus on the case when $p_A(\lambda)=\lambda^n$. Thus $A^n=0$. A matrix with this property is called nilpotent.
Given a matrix $A$, we introduce the following quantities:
$$w_k(A, \lambda)=\operatorname{dim} \operatorname{Ker}\left(A-\lambda I_n\right)^k-\operatorname{dim} \operatorname{Ker}\left(A-\lambda I_n\right)^{k-1}, k=1, \ldots, n .$$
Here $\left(A-\lambda I_n\right)^0=I_n$, so $w_1(A, \lambda)=\operatorname{dim} \operatorname{Ker}\left(A-\lambda I_n\right)$. The numbers $w_k(A, \lambda)$ are collectively called the Weyr characteristics of $A$. The spaces $\operatorname{Ker}\left(A-\lambda I_n\right)^k$ are called generalized eigenspaces of $A$ at $\lambda$.

We also introduce the Jordan block $J_k(\lambda)$ of size $k$ at $\lambda$, as being the $k \times k$ upper triangular matrix
$$J_k(\lambda)=\left(\begin{array}{ccccc} \lambda & 1 & 0 & \cdots & 0 \ 0 & \lambda & 1 & \cdots & 0 \ \vdots & & \ddots & \ddots & \vdots \ 0 & 0 & \cdots & \lambda & 1 \ 0 & 0 & \cdots & 0 & \lambda \end{array}\right)$$

We write $\oplus_{k=1}^p A_k$ for the block diagonal matrix
$$\oplus_{k=1}^p A_k=\left(\begin{array}{cccc} A_1 & 0 & \cdots & 0 \ 0 & A_2 & \cdots & 0 \ \vdots & \vdots & \ddots & \vdots \ 0 & 0 & \cdots & A_p \end{array}\right)$$
When we have a block diagonal matrix with $p$ copies of the same matrix $B$, we write $\oplus_{k=1}^p B$

## 数学代写|高等线性代数代写Advanced Linear Algebra代考|The minimal polynomial

As we have seen in Theorem 4.1.1, the characteristic polynomial $p_A(t)$ of a matrix $A$ has the property that $p_A(A)=0$. There are many other monic polynomials $p(t)$ that also satisfy $p(A)=0$. Of particular interest is the one of lowest possible degree. This so-called “minimal polynomial” of $A$ captures some essential features of the Jordan canonical form of the matrix $A$.
Given $A \in \mathbb{F}^{n \times n}$ we define its minimal polynomial $m_A(t)$ to be the lowest-degree monic polynomial so that $m_A(A)=0$.
Example 4.5.1 Let
$$A=\left(\begin{array}{lll} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 2 \end{array}\right) \text {. }$$
Then $m_A(t)=(t-1)(t-2)$. Indeed,
$$m_A(A)=\left(A-I_3\right)\left(A-2 I_3\right)=0,$$
and any monic degree- 1 polynomial has the form $t-\lambda$, but $A-\lambda I_3 \neq 0$ for all $\lambda$.

Proposition 4.5.2 Every $A \in \mathbb{F}^n$ has a unique minimal polynomial $m_A(t)$, and every eigenvalue of $A$ is a root of $m_A(t)$. Moreover, if $p(A)=0$, then $m_A(t)$ divides $p(t)$. In particular, $m_A(t)$ divides $p_A(t)$.

Proof. As $p_A(A)=0$, there certainly exists a degree- $n$ polynomial satisfying $p(A)=0$, and thus there exists also a nonzero polynomial of lowest degree which can always be made monic by multiplying by a nonzero element of $\mathbb{F}$. Next suppose that $m_1(t)$ and $m_2(t)$ are both monic polynomials of lowest possible degree $k$ so that $m_1(A)=0=m_2(A)$. Then by Proposition 4.3.1 there exists $q(t)$ and $r(t)$ with $\operatorname{deg} r<k$ so that
$$m_1(t)=q(t) m_2(t)+r(t)$$

# 高等线性代数代考

$$w_k(A, \lambda)=\operatorname{dim} \operatorname{Ker}\left(A-\lambda I_n\right)^k-\operatorname{dim} \operatorname{Ker}\left(A-\lambda I_n\right)^{k-1}, k=1, \ldots, n .$$

$$J_k(\lambda)=\left(\begin{array}{ccccc} \lambda & 1 & 0 & \cdots & 0 \ 0 & \lambda & 1 & \cdots & 0 \ \vdots & & \ddots & \ddots & \vdots \ 0 & 0 & \cdots & \lambda & 1 \ 0 & 0 & \cdots & 0 & \lambda \end{array}\right)$$

$$\oplus_{k=1}^p A_k=\left(\begin{array}{cccc} A_1 & 0 & \cdots & 0 \ 0 & A_2 & \cdots & 0 \ \vdots & \vdots & \ddots & \vdots \ 0 & 0 & \cdots & A_p \end{array}\right)$$

. <

$$A=\left(\begin{array}{lll} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 2 \end{array}\right) \text {. }$$

$$m_A(A)=\left(A-I_3\right)\left(A-2 I_3\right)=0,$$

$$m_1(t)=q(t) m_2(t)+r(t)$$

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