# Quod Erat Demonstrandum

## 2016/07/26

### 相同特徵值及凱萊哈密頓

Filed under: NSS,Pure Mathematics — johnmayhk @ 10:29 上午
Tags: , ,

https://johnmayhk.wordpress.com/2016/07/22/flf-and-matrix/

$M=\left(\begin{array}{rcl}a& b\\c& d\\\end{array}\right)$　的特徵方程為 $\det(M-\lambda I)=0$，即

$\lambda^2-(a+d)\lambda+(ad-bc)=0$

$(a+d)$ 就是矩陣 $M$ 的跡（trace），即對角元的和，也是特徵值的和（sum of roots）；而

$(ad-bc)$ 就是矩陣 $M$ 的行列式（determinant），也是特徵值的積（product of roots）。

$\left(\begin{array}{rcl}2& 0\\0& 8\\\end{array}\right)$

trace = 2+8 = 10，所以隨便寫兩個，總和是 10 的兩個數做元，比如是 6,4：

$\left(\begin{array}{rcl}6& b\\c& 4\\\end{array}\right)$

$\left(\begin{array}{rcl}6& 4\\2& 4\\\end{array}\right)$

$\left(\begin{array}{rcl}5& 1\\9& 5\\\end{array}\right)$

$\left(\begin{array}{rcl}2& 0\\0& 2\\\end{array}\right)$

$M=I\left(\begin{array}{rcl}2& 0\\0& 2\\\end{array}\right)I^{-1}$

$\left(\begin{array}{rcl}1& 1\\-1& 3\\\end{array}\right)$

https://www.symbolab.com/solver/matrix-diagonalization-calculator/diagonalize%20%5Cbegin%7Bpmatrix%7D1%261%5C%5C%20-1%263%5Cend%7Bpmatrix%7D

$M=SJS^{-1}$

(1) $J=\left(\begin{array}{rcl}\lambda_1& 0\\0& \lambda_2\\\end{array}\right)$，或

(2) $J=\left(\begin{array}{rcl}\lambda & 1\\0& \lambda \\\end{array}\right)$

$\left(\begin{array}{rcl}1& 1\\-1& 3\\\end{array}\right)$

http://www.wolframalpha.com/input/?i=jordan+decomposition+%7B%7B1,1%7D,%7B-1,3%7D%7D

$\left(\begin{array}{rcl}0& 0\\1& 0\\\end{array}\right)$

$\left(\begin{array}{rcl}0& 1\\1& 0\\\end{array}\right)\left(\begin{array}{rcl}0& 1\\0& 0\\\end{array}\right)$

$\left(\begin{array}{rcl}0& 1\\1& 0\\\end{array}\right)\left(\begin{array}{rcl}0& 1\\0& 0\\\end{array}\right)\left(\begin{array}{rcl}0& 1\\1& 0\\\end{array}\right)$，故

$\left(\begin{array}{rcl}0& 0\\1& 0\\\end{array}\right)=\left(\begin{array}{rcl}0& 1\\1& 0\\\end{array}\right)\left(\begin{array}{rcl}0& 1\\0& 0\\\end{array}\right)\left(\begin{array}{rcl}0& 1\\1& 0\\\end{array}\right)=\left(\begin{array}{rcl}0& 1\\1& 0\\\end{array}\right)\left(\begin{array}{rcl}0& 1\\0& 0\\\end{array}\right)\left(\begin{array}{rcl}0& 1\\1& 0\\\end{array}\right)^{-1}$

http://www.wolframalpha.com/input/?i=jordan+decomposition+%7B%7B0,0%7D,%7B1,0%7D%7D

Jordan decomposition 是更一般的手法分解方陣（square matrix），無論它可被對角化與否。

$\left(\begin{array}{rcl}2& -1\\-1& 2\\\end{array}\right)$

http://www.wolframalpha.com/input/?i=jordan+decomposition+%7B%7B2,-1%7D,%7B-1,2%7D%7D

$\lambda =4$$\left(\begin{array}{rcl}2\\1\\\end{array}\right)$

$(M-\lambda I)\left(\begin{array}{rcl}h\\k\\\end{array}\right)=\left(\begin{array}{rcl}2\\1\\\end{array}\right)$

$\left(\begin{array}{rcl}-2& 4\\-1& 2\\\end{array}\right)\left(\begin{array}{rcl}h\\k\\\end{array}\right)=\left(\begin{array}{rcl}2\\1\\\end{array}\right)$

$\left(\begin{array}{rcl}h\\k\\\end{array}\right)=\left(\begin{array}{rcl}-1\\0\\\end{array}\right)$

$M=\left(\begin{array}{rcl}2& 4\\-1& 6\\\end{array}\right)=\left(\begin{array}{rcl}2& -1\\1& 0\\\end{array}\right)\left(\begin{array}{rcl}4& 1\\0& 4\\\end{array}\right)\left(\begin{array}{rcl}2& -1\\1& 0\\\end{array}\right)^{-1}$

http://www.wolframalpha.com/input/?i=%7B%7B2,-1%7D,%7B1,0%7D%7D%7B%7B4,1%7D,%7B0,4%7D%7D%7B%7B2,-1%7D,%7B1,0%7D%7D%5E(-1)

$(X+Y)^n=\displaystyle \sum_{r=0}^nC^n_rX^{n-r}Y^r$

（其中 $X^0=X$$Y^0=Y$

$\left(\begin{array}{rcl}\lambda & 1\\0& \lambda \\\end{array}\right)^n$

$=(\left(\begin{array}{rcl}\lambda & 0\\0& \lambda\\\end{array}\right)+\left(\begin{array}{rcl}0 & 1\\0& 0\\\end{array}\right))^n$

$=\displaystyle \sum_{r=0}^nC^n_r\left(\begin{array}{rcl}\lambda & 0\\0& \lambda\\\end{array}\right)^{n-r}\left(\begin{array}{rcl}0 & 1\\0& 0\\\end{array}\right)^r$

（這是因為　$\left(\begin{array}{rcl}\lambda & 0\\0& \lambda\\\end{array}\right)\left(\begin{array}{rcl}0 & 1\\0& 0\\\end{array}\right)=\left(\begin{array}{rcl}0 & 1\\0& 0\\\end{array}\right)\left(\begin{array}{rcl}\lambda & 0\\0& \lambda\\\end{array}\right)$

$=\left(\begin{array}{rcl}\lambda & 0\\0& \lambda\\\end{array}\right)^n+C^n_1\left(\begin{array}{rcl}\lambda & 0\\0& \lambda\\\end{array}\right)^{n-1}\left(\begin{array}{rcl}0 & 1\\0& 0\\\end{array}\right)$

（這是因為　$\left(\begin{array}{rcl}0 & 1\\0& 0\\\end{array}\right)^r=\left(\begin{array}{rcl}0 & 0\\0& 0\\\end{array}\right)$ for $r\ge 2$

$=\left(\begin{array}{rcl}\lambda^n & 0\\0& \lambda^n\\\end{array}\right)+n\left(\begin{array}{rcl}\lambda^{n-1} & 0\\0& \lambda^{n-1}\\\end{array}\right)\left(\begin{array}{rcl}0 & 1\\0& 0\\\end{array}\right)$

$=\left(\begin{array}{rcl}\lambda^n & n\lambda^{n-1}\\0& \lambda^n\\\end{array}\right)=\lambda^{n-1}\left(\begin{array}{rcl}\lambda & n\\0& \lambda\\\end{array}\right)$

$\left(\begin{array}{rcl}2& 4\\-1& 6\\\end{array}\right)=\left(\begin{array}{rcl}2& -1\\1& 0\\\end{array}\right)\left(\begin{array}{rcl}4& 1\\0& 4\\\end{array}\right)\left(\begin{array}{rcl}2& -1\\1& 0\\\end{array}\right)^{-1}$

$\left(\begin{array}{rcl}2& 4\\-1& 6\\\end{array}\right)^n=\left(\begin{array}{rcl}2& -1\\1& 0\\\end{array}\right)\left(\begin{array}{rcl}4& 1\\0& 4\\\end{array}\right)^n\left(\begin{array}{rcl}2& -1\\1& 0\\\end{array}\right)^{-1}$

$=\left(\begin{array}{rcl}2& -1\\1& 0\\\end{array}\right)4^{n-1}\left(\begin{array}{rcl}4& n\\0& 4\\\end{array}\right)\left(\begin{array}{rcl}2& -1\\1& 0\\\end{array}\right)^{-1}$

$= 4^{n-1}\left(\begin{array}{rcl}-2n+4& 4n\\-n& 2n+4\\\end{array}\right)$

https://johnmayhk.wordpress.com/2013/06/19/ch

$M$ 為 3×3 矩陣，$p(t)$$M$ 的特徵方程，即

$p(t)=\det(M-tI)$

$p(t)=a_0+a_1t+a_2t^2+a_3t^3$

$N=M-tI$，得

$\det(N)=a_0+a_1t+a_2t^2+a_3t^3$

$N^{-1}=\frac{adj(N)}{\det(N)}$

$\Rightarrow \det(N)I=adj(N)N$ ………. (*)

$adj(N)$ 的每個元寫為 $b_0+b_1t+b_2t^2$ 這種形式，不難想像，整個 $adj(N)$ 一定能寫成以下形式：

$adj(N)=B_0+B_1t+B_2t^2$

$(a_0+a_1t+a_2t^2+a_3t^3)I=(B_0+B_1t+B_2t^2)(M-tI)$

$a_0I=B_0M$
$a_1I=-B_0I+B_1M$
$a_2I=-B_1I+B_2M$
$a_3I=-B_2$

$a_0I=B_0M$
$a_1M=-B_0M+B_1M^2$
$a_2M^2=-B_1M^2+B_2M^3$
$a_3M^3=-B_2M^3$

$a_0I+a_1M+a_2M^2+a_3M^3=0\Rightarrow p(M)=0$（零矩陣）

Q.E.D.