节: 对对等化 | 5.3 对等化

章节大纲

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,