$$$\left[\begin{array}{cc}1 & 2\\3 & 4\end{array}\right]$$$을 대각화하라
사용자 입력
$$$\left[\begin{array}{cc}1 & 2\\3 & 4\end{array}\right]$$$을(를) 대각화하시오.
풀이
먼저 고유값과 고유벡터를 구합니다(자세한 단계는 고유값 및 고유벡터 계산기를 참조하세요).
고유값: $$$- \frac{-5 + \sqrt{33}}{2}$$$, 고유벡터: $$$\left[\begin{array}{c}- \frac{3 + \sqrt{33}}{6}\\1\end{array}\right]$$$.
고유값: $$$\frac{5 + \sqrt{33}}{2}$$$, 고유벡터: $$$\left[\begin{array}{c}\frac{-3 + \sqrt{33}}{6}\\1\end{array}\right]$$$.
행렬 $$$P$$$를 구성하되, $$$i$$$번째 열이 $$$i$$$번째 고유벡터가 되도록 하라: $$$P = \left[\begin{array}{cc}- \frac{3 + \sqrt{33}}{6} & \frac{-3 + \sqrt{33}}{6}\\1 & 1\end{array}\right]$$$.
$$$i$$$행 $$$i$$$열의 원소가 $$$i$$$번째 고유값인 대각행렬 $$$D$$$를 구성하십시오: $$$D = \left[\begin{array}{cc}- \frac{-5 + \sqrt{33}}{2} & 0\\0 & \frac{5 + \sqrt{33}}{2}\end{array}\right]$$$
행렬 $$$P$$$와 $$$D$$$는 처음 주어진 행렬 등식 $$$\left[\begin{array}{cc}1 & 2\\3 & 4\end{array}\right] = P D P^{-1}$$$를 만족한다.
$$$P^{-1} = \left[\begin{array}{cc}- \frac{\sqrt{33}}{11} & - \frac{-11 + \sqrt{33}}{22}\\\frac{\sqrt{33}}{11} & \frac{\sqrt{33} + 11}{22}\end{array}\right]$$$ (단계별 풀이에 대해서는 역행렬 계산기를 참조하세요.)
정답
$$$P = \left[\begin{array}{cc}- \frac{3 + \sqrt{33}}{6} & \frac{-3 + \sqrt{33}}{6}\\1 & 1\end{array}\right]\approx \left[\begin{array}{cc}-1.457427107756338 & 0.457427107756338\\1 & 1\end{array}\right]$$$A
$$$D = \left[\begin{array}{cc}- \frac{-5 + \sqrt{33}}{2} & 0\\0 & \frac{5 + \sqrt{33}}{2}\end{array}\right]\approx \left[\begin{array}{cc}-0.372281323269014 & 0\\0 & 5.372281323269014\end{array}\right]$$$A
$$$P^{-1} = \left[\begin{array}{cc}- \frac{\sqrt{33}}{11} & - \frac{-11 + \sqrt{33}}{22}\\\frac{\sqrt{33}}{11} & \frac{\sqrt{33} + 11}{22}\end{array}\right]\approx \left[\begin{array}{cc}-0.522232967867094 & 0.238883516066453\\0.522232967867094 & 0.761116483933547\end{array}\right]$$$A