Section: 对对等化 | 5.3 对等化

Section outline

In middle school and elementary school you have probably factored numbers or even factored polynomials and similarly you can "factor" matrices. Using eigenvalues you can factor some matrix $\begin{aligned} A = P D P^{- 1} \end{aligned}$ where D is a diagonal matrix of the eigenvalues and P is the matrix of the linearly independent eigenvectors.
::在初中和小学中,您可能已经设定了因数或甚至因数多语种,同样您也可以使用“因数”矩阵。使用 eigenvalues,您可以选择一些矩阵 A= PDP-1,其中D 是egenvalies的对数矩阵,P 是线性独立源数的矩阵。

Let's first look $\begin{aligned} A = [\begin{matrix} 1 & 2 \\ 3 & 7 \end{matrix}] \end{aligned}$ .
::让我们首先看A=[1237]。

Now, find the characteristic equation of $\begin{aligned} det (A - λ I) = det ([\begin{matrix} 1 - λ & 2 \\ 3 & 7 - λ \end{matrix}]) = (1 - λ) (7 - λ) - 6 = λ^{2} - 8 λ + 1 = 0 \end{aligned}$
::现在, 找到 det( AI) =det( [1\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\0\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\0\\\\\\\\\

Now applying the quadratic formula $\begin{aligned} λ = \frac{8 \pm \sqrt{60}}{2} = 4 \pm \sqrt{15} \end{aligned}$
::现在应用二次公式 8602=415

Hence $\begin{aligned} D = [\begin{matrix} 4 + \sqrt{15} & 0 \\ 0 & 4 - \sqrt{15} \end{matrix}] \end{aligned}$
::因此D=[4+15004-15]

Now we find the eigenvectors by finding $\begin{aligned} Nul ([\begin{matrix} - 3 - \sqrt{15} & 2 \\ 3 & 3 - \sqrt{15} \end{matrix}]) \end{aligned}$ which can be done by solving for $\begin{aligned} x_{1} and x_{2} where [\begin{matrix} - 3 - \sqrt{15} & 2 \\ 3 & 3 - \sqrt{15} \end{matrix}] \cdot [\begin{matrix} x_{1} \\ x_{2} \end{matrix}] = \vec{0} = [\begin{matrix} 0 \\ 0 \end{matrix}] \end{aligned}$
::现在我们通过找到Nul([-3- 15233- 15] ) 来找到源子,可以通过在 [-3- 15233- 15] 和 [x1x2] =0\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\

From this we get that $\begin{aligned} x_{1} = (\frac{\sqrt{15}}{3} - 1) x_{2} \end{aligned}$

Hence, the eigenvector is $\begin{aligned} [\begin{matrix} \sqrt{15} - 3 \\ 3 \end{matrix}] \end{aligned}$

Then, to find the other eigenvector we have to calculate

$\begin{aligned} Nul ([\begin{matrix} - 3 + \sqrt{15} & 2 \\ 3 & 3 + \sqrt{15} \end{matrix}]) \end{aligned}$

Hence, we have to solve the equation

$\begin{aligned} [\begin{matrix} - 3 + \sqrt{15} & 2 \\ 3 & 3 + \sqrt{15} \end{matrix}] \cdot [\begin{matrix} x_{1} \\ x_{2} \end{matrix}] = [\begin{matrix} 0 \\ 0 \end{matrix}] \\ (- 3 + \sqrt{15}) x_{1} + 2 x_{2} = 0 \\ 3 x_{1} + (3 + \sqrt{15}) x_{2} = 0 \\ x_{1} = (1 - \frac{\sqrt{15}}{3}) x_{2} \\ The eigenvector is \\ [\begin{matrix} 3 - \sqrt{15} \\ 3 \end{matrix}] \end{aligned}$

Hence, we get that

$\begin{aligned} P = [\begin{matrix} \sqrt{15} - 3 & 3 - \sqrt{15} \\ 3 & 3 \end{matrix}] \\ P^{- 1} = [\begin{matrix} \frac{\sqrt{15} + 3}{12} & \frac{1}{6} \\ - \frac{\sqrt{15} + 3}{12} & \frac{1}{6} \end{matrix}] \\ A = P D P^{- 1} \\ A = [\begin{matrix} \sqrt{15} - 3 & 3 - \sqrt{15} \\ 3 & 3 \end{matrix}] \cdot [\begin{matrix} 4 + \sqrt{15} & 0 \\ 0 & 4 - \sqrt{15} \end{matrix}] \cdot [\begin{matrix} \frac{\sqrt{15} + 3}{12} & \frac{1}{6} \\ - \frac{\sqrt{15} + 3}{12} & \frac{1}{6} \end{matrix}] \end{aligned}$

I'll start one more example and I'd like you all to finish it off
::我再举一个例子,请你们全部完成

Let
::Le 让我们

$\begin{aligned} A = [\begin{matrix} 4 & 3 & - 2 \\ 1 & 7 & 6 \\ - 1 & 4 & - 2 \end{matrix}] \\ det (A - λ \cdot I) = det ([\begin{matrix} 4 - λ & 3 & - 2 \\ 1 & 7 - λ & 6 \\ - 1 & 4 & - 2 - λ \end{matrix}]) \\ = (4 - λ) ((7 - λ) (- 2 - λ) - 24) - 3 ((- 2 - λ) + 6) - 2 (4 + 7 - λ) \\ = () \end{aligned}$
::A=[43-2176-14-2]det(AI)=det([43-217]6-14-2])=(4(7)(-2)-24)-3((-2)+6)-2(4+7)=()

Finally, this video is good to review from MIT open courseware:
::这段影片在麻省理工学院公开课程软件中,