Method 2:

$e^{At}=Pe^{\Lambda t}P^{-1}$ where $e^{\Lambda t}=\begin{bmatrix}e^{(1+j)t}&0\\ 0 & e^{(1-j)t}\end{bmatrix}$ . Eigen values are $1\pm j$ . The corresponding eigenvectors are found by using equation $Av_i=\lambda_i v_i$ as follows:

$\displaystyle \begin{bmatrix}1 & 1\\ -1 & 1\end{bmatrix}\begin{bmatrix}v_1\\ v_2\end{bmatrix}=(1+j)\begin{bmatrix}v_1\\ v_2\end{bmatrix}$

Taking , we get . So, the eigenvector corresponding to is $\begin{bmatrix}1\\ j\end{bmatrix}$ and the one corresponding to is $\begin{bmatrix}1\\ -j\end{bmatrix}$ . The transformation matrix is given by

$\displaystyle P=[v_1\;\;v_2]=\begin{bmatrix}1 & 1 \\ j & -j \end{bmatrix} \Rightarrow P^{-1}= \frac{1}{2}\begin{bmatrix}1 & -j \\ 1 & j \end{bmatrix}$

Now,