I'm trying to calculate the matrix exponential $e^{At}$ for $$A=\frac{1}{2}\begin{bmatrix}-1&1&-1\\2&-2&0\\1&-1&-1\end{bmatrix}$$
I found the eigenvalues $\lambda_1=\lambda_2=-1, \lambda_3=0$ which give the eigenvectors $$v_{\lambda_{1,2}}=\begin{bmatrix}0\\1\\1\end{bmatrix},v_{\lambda_3}=\begin{bmatrix}1\\1\\0\end{bmatrix}$$
I now would like to find the matrices $S$ and $J$ so that $$A=SJS^{-1}$$
Constructing these matrices requires another column vector. I now try to find the generalized eigenvector for eigenvalue $-1$ as follows. $$(A-(-1)I)\mathbf{v}=\begin{bmatrix}1/2&1/2&-1/2\\1&0&0\\1/2&-1/2&1/2\end{bmatrix}\mathbf{v}=\begin{bmatrix}0\\1\\1\end{bmatrix}$$
But this has no solution, so I am stuck.
Best Answer
Wolfram alpha gives \begin{eqnarray} J&=& \begin{bmatrix} -1&1&0\\ 0&-1&0\\ 0&0&0 \end{bmatrix},\ \mbox{and}\\ S &=& \begin{bmatrix} 0&1&1\\ 1&-1&1\\ 1&0&0 \end{bmatrix}\ . \end{eqnarray} Since $$ J^n = \left(-1\right)^n\begin{bmatrix} 1&-n&0\\ 0&1&0\\ 0&0&0 \end{bmatrix}, $$ then $$ e^{tJ} = \sum_{n=0}^\infty\frac{t^n J^n}{n!}=\begin{bmatrix} \sum_{n=0}^\infty \frac{(-t)^n}{n!}&-\sum_{n=0}^\infty n\frac{(-t)^n}{n!}&0\\ 0&\sum_{n=0}^\infty \frac{(-t)^n}{n!}&0\\ 0&0&0 \end{bmatrix}\\ =\ \begin{bmatrix} e^{-t}&te^{-t}&0\\ 0&e^{-t}&0\\ 0&0&0 \end{bmatrix}\ . $$ So \begin{eqnarray} e^{tA} &=& S\, e^{tJ} S^{-1}\\ &=&\frac{1}{2} \begin{bmatrix} e^{-t}&-e^{-t}&e^{-t}\\ (t-1)\,e^{-t}&(1-t)\,e^{-t}&(1+t)\,e^{-t}\\ te^{-t}&-te^{-t}&(2+t)\,e^{-t} \end{bmatrix}\ . \end{eqnarray}