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

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

Bepaal eerst de eigenwaarden en eigenvectoren (voor de stappen, zie rekenmachine voor eigenwaarden en eigenvectoren).

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

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

Stel de matrix $$$P$$$ op, waarvan kolom $$$i$$$ eigenvector nr. $$$i$$$ is: $$$P = \left[\begin{array}{cc}- \frac{3 + \sqrt{33}}{6} & \frac{-3 + \sqrt{33}}{6}\\1 & 1\end{array}\right]$$$.

Vorm de diagonaalmatrix $$$D$$$ waarbij het element in rij $$$i$$$, kolom $$$i$$$ gelijk is aan eigenwaarde nr. $$$i$$$: $$$D = \left[\begin{array}{cc}- \frac{-5 + \sqrt{33}}{2} & 0\\0 & \frac{5 + \sqrt{33}}{2}\end{array}\right]$$$.

De matrices $$$P$$$ en $$$D$$$ zijn zodanig dat voor de oorspronkelijke matrix $$$\left[\begin{array}{cc}1 & 2\\3 & 4\end{array}\right] = P D P^{-1}$$$ geldt.

$$$P^{-1} = \left[\begin{array}{cc}- \frac{\sqrt{33}}{11} & - \frac{-11 + \sqrt{33}}{22}\\\frac{\sqrt{33}}{11} & \frac{\sqrt{33} + 11}{22}\end{array}\right]$$$ (voor de stappen, zie inverse matrix-calculator.)

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