Diagonalizar $$$\left[\begin{array}{cc}1 & 2\\3 & 4\end{array}\right]$$$

Diagonaliza $$$\left[\begin{array}{cc}1 & 2\\3 & 4\end{array}\right]$$$.

Primero, encuentra los valores propios y los vectores propios (para los pasos, consulta calculadora de valores propios y vectores propios).

Valor propio: $$$- \frac{-5 + \sqrt{33}}{2}$$$, vector propio: $$$\left[\begin{array}{c}- \frac{3 + \sqrt{33}}{6}\\1\end{array}\right]$$$.

Valor propio: $$$\frac{5 + \sqrt{33}}{2}$$$, vector propio: $$$\left[\begin{array}{c}\frac{-3 + \sqrt{33}}{6}\\1\end{array}\right]$$$.

Forme la matriz $$$P$$$, cuya columna $$$i$$$ es el vector propio número $$$i$$$: $$$P = \left[\begin{array}{cc}- \frac{3 + \sqrt{33}}{6} & \frac{-3 + \sqrt{33}}{6}\\1 & 1\end{array}\right]$$$.

Forme la matriz diagonal $$$D$$$ cuyo elemento en la fila $$$i$$$, columna $$$i$$$ es el valor propio n.º $$$i$$$: $$$D = \left[\begin{array}{cc}- \frac{-5 + \sqrt{33}}{2} & 0\\0 & \frac{5 + \sqrt{33}}{2}\end{array}\right]$$$.

Las matrices $$$P$$$ y $$$D$$$ son tales que la matriz inicial $$$\left[\begin{array}{cc}1 & 2\\3 & 4\end{array}\right] = P D P^{-1}$$$.

$$$P^{-1} = \left[\begin{array}{cc}- \frac{\sqrt{33}}{11} & - \frac{-11 + \sqrt{33}}{22}\\\frac{\sqrt{33}}{11} & \frac{\sqrt{33} + 11}{22}\end{array}\right]$$$ (para los pasos, consulte calculadora de matriz inversa).

$$$P = \left[\begin{array}{cc}- \frac{3 + \sqrt{33}}{6} & \frac{-3 + \sqrt{33}}{6}\\1 & 1\end{array}\right]\approx \left[\begin{array}{cc}-1.457427107756338 & 0.457427107756338\\1 & 1\end{array}\right]$$$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