Donde, l son los autovalores (o eigenvalores) de la matriz de estabilidad, y se conocen como funciones de Morse.

Question

Deje $A$ ser una matriz con valores propios $\lambda_1,...\lambda_n$. A continuación, se $e^{\lambda_1},...e^{\lambda_n}$ autovalores de a $e^A$? Aquí, $$e^A = I + A + \dfrac{A^2}{2!}+...\dfrac{A^k}{k!}+...$$

Answer 1

Si $A$ es diagonalizable, entonces puede ser escrito $A = SDS^{-1}$, luego $$A^k = (SDS^{-1})^k = (SDS^{-1})^{k-2}(SDS^{-1})(SDS^{-1})$$ where we see that $DS^{-1}SD$ in the middle can be simplified to $D^2$ and then by induction we can show that $^k = SD^kS^{-1}$ and as $D$ contains the $\lambda_i$ on the diagonal then they themselves will be powered by k. Then you see that for each eigenvalue each term will become $\frac{{\lambda_i}^k}{k!}$ and that series must be $e^{\lambda_i}$.