举例矩阵在$ℂ$和$\Bbb F_5$可对角化,但在$ℝ$上不可

hbghlyj

$$A=\left(\begin{array}{cc} 1 & -1 \\ 1 & 1 \end{array}\right) $$ $$\chi_{A}(x)=(1-x)(1-x)+1=x^{2}-2 x+2 . $$ \begin{aligned} \mathbb{F}=\mathbb{R}& \Rightarrow m_{A}(x)=\chi_{A}(x) \text { has no roots, } \\ &\Rightarrow A \text { is not triangularisable, nor diagonalisable; } \\ \mathbb{F}=\mathbb{C}& \Rightarrow m_{A}(x)=\chi_{A}(x)=(x-(1+i))(x-(1-i)) \\ & \Rightarrow A \text { is diagonalisable; } \\ \mathbb{F}=\mathbb{F}_{5}& \Rightarrow m_{A}(x)=\chi_{A}(x)=(x-3)(x-4) \\ &\Rightarrow A \text { is diagonalisable. } \end{aligned}

hbghlyj

Blackboard bold这里有一个Unicode表格,可以发现Unicode Code Point (hex)这一列是四位的只有ℂ,ⅅ,ⅆ,ⅇ,ℍ,ⅈ,ⅉ,ℕ,ℙ,ℚ,ℝ,ℤ,ℾ,ℽ,ℿ,ℼ,⅀.剩下的都是五位的,在这个论坛因为MySQL数据库编码是utf8所以无法保存(新版论坛会变成4个问号,旧版会切断后面的).

hbghlyj

$$B=\left(\begin{array}{ccc} 0 & 0 & -1 \\ 0 & -1 & 0 \\ 1 & 0 & 1 \end{array}\right)$$$χ_B(λ)=-λ^3 - 1⇒m_B(λ)=λ^3 + 1$
$m_B(λ)=(1+λ)(1-λ+λ^2)⇒B$ is not diagonalisable over $ℝ$.
$m_B(λ)=(1+λ)((-1)^{1/3} - λ)((-1)^{2/3} - λ) ⇒B$ is diagonalisable over $ℂ$.
$m_B(λ)= (1 +λ) (1 -λ + λ^2)⇒B$ is not diagonalisable over $\Bbb F_5$.

hbghlyj

hbghlyj 发表于 2022-7-7 19:23
Blackboard bold这里有一个Unicode表格,可以发现Unicode Code Point (hex)这一列是四位的只有ℂ,ⅅ,ⅆ,ⅇ, ...

kuing修改Server charset后可以直接存入数据库
如𝔼、𝔽