Determine $\mathit{e}^{At}$ by using generalized eigenvector method to find a fundamental matrix of $x'=Ax$ with $A=\begin{bmatrix} 5 &-4 &0 \\ 1&0 &2 \\ 0& 2 &5 \end{bmatrix}$.
I just want to know whether my solution is okay?
I found the eigenvalues to be $\lambda_{1}=0$ (multiplicity 1) and $\lambda_{2}=5$ (multiplicity 2).
I found the eigenvector of $\lambda_{1}$ to be $v_{1}=(-4,-5,2)$ and the eigenvector of $\lambda_{2}$ to be $v_{2}=(-2,0,1)$.
I used the generalized eigenvector property to find $v_{3}$, where $v_{2}=(A-\lambda_{2}I)v_{3}$. I got $v_{3}=(1/2,1/2,1)$.
So then $\mathit{e}^{At}=\begin{bmatrix} -4 &-2 &\frac{1}{2} \\ -5&0 &\frac{1}{2} \\ 2& 1 &1 \end{bmatrix}$. Is this okay?
Thanks for any help!
$e^{At} = P e^{Jt} P^{-1}$
and
$e^{Jt} = \begin{bmatrix} e^{J_1 t} && 0 \\ 0 && e^{J_2 t} \end{bmatrix}$
Firstly, as $J_1 = 0$, $e^{J_1t} = 1$ .
Now, if we open taylor series of exponential around $\lambda_2$
$e^{xt} = \sum \frac{e^{\lambda_2t}}{n!}(xt-\lambda_2t)^n $
$e^{J_2 t} = \sum \frac{e^{\lambda_2 t}}{n!}(J_2t-\lambda_2 I t)^n$
Now,
$J_2t-\lambda_2I t = \begin{bmatrix} 0 && 1t \\ 0 && 0 \end{bmatrix}$
Note that,
$\begin{bmatrix} 0 && 1t \\ 0 && 0 \end{bmatrix}^n = 0, for: n>1$
Then,
$e^{J_2 t} = e^{\lambda_2 t} \begin{bmatrix} 1 && 0 \\ 0 && 1 \end{bmatrix} + e^{\lambda_2 t} \begin{bmatrix} 0 && t \\ 0 && 0\end{bmatrix} = \begin{bmatrix} e^{\lambda_2 t} && te^{\lambda_2 t} \\ 0 && e^{\lambda_2 t} \end{bmatrix}$
And finally if your eigenvectors are correct, you have:
$e^{At} = P^{-1} e^{Jt} P = \begin{bmatrix} -4 && -2 && 1/2 \\ -5 && 0 && 1/2 \\ 2 && 1 && 1\end{bmatrix} \begin{bmatrix} 1 && 0 && 0 \\ 0 && e^{5 t} && te^{5t} \\ 0 && 0 && e^{5t} \end{bmatrix} \begin{bmatrix} -4 && -2 && 1/2 \\ -5 && 0 && 1/2 \\ 2 && 1 && 1\end{bmatrix}^{-1}$