How could I find the eigen-function of the following system. Consider the following ODE system $$\begin{bmatrix}\dot{x}\\\dot{y}\end{bmatrix} = \begin{bmatrix}0&1\\1&0\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}.$$ The eigen-value is easy to find here, $$\det(\lambda I - A) = \det\begin{bmatrix}\lambda&-1\\-1&\lambda\end{bmatrix} = \lambda^{2} + 1.$$ Thus we have $\lambda_{1} = i,\lambda_{2} = -i$, and plugging in we got eigen-vector of $A$ are $v_{1} = \begin{bmatrix}-1\\1\end{bmatrix}$, $v_{2} = \begin{bmatrix}1\\1\end{bmatrix}$. But what are the corresponding eigen-function here?
Are they $\exp(tv_{1}), \exp(tv_{2})$ respectively? Or am I missing something?
Ok, made a mistake, $$\det(\lambda I - A) = \det\begin{bmatrix}\lambda&-1\\-1&\lambda\end{bmatrix} = \lambda^{2} - (-1)^2 = \lambda^{2} - 1 .$$ Thus $\lambda_{1} = 1,\lambda_{2} = -1$, and plugging in we got eigen-vector of $A$ are $v_{1} = \begin{bmatrix}1\\1\end{bmatrix}$, $v_{2} = \begin{bmatrix}-1\\1\end{bmatrix}$.