Given the matrix: $A=\begin{pmatrix}-2&0&0\\4&-2&0\\1&0&-2\end{pmatrix}$, find $e^{At}$.
I found the eigenvalues to be $-2,-2,-2$. I need to use the Jordan form to solve it. I'm practicing for exam in a couple of hrs. Thank you. I found the eigenvectors, $$v_1=(0,1,0), v_2=(0,1,1), V_3=(1,0,0).$$
I think the Jordan form is $e^{Jt}=\begin{pmatrix}e^{-2t}&1&0\\0&e^{-2t}&1\\0&0&e^{-2t}\end{pmatrix}$,Is this right?
Next I need to calculate $$Ve^{Jt}V^{-1}$$ Actually I do know the answer Im just practicing the way to do it. Can anybody tell me whats wrong with V_3? I found that $(A+2I)^3=0$ and I thought any eigenvector satisfies $0\cdot v_3=0$?!
The form of $e^{At}$ can be guessed by computing a few powers of $A$ by hand. One sees that these powers are all of the form $$ \begin{pmatrix} a &0 & 0 \\ b & a & 0 \\ c & 0 & a \end{pmatrix} $$ where $a$ is a power of (-2). Therefore one can guess that $$ e^{At} = \begin{pmatrix} e^{-2t} &0 & 0 \\ f(t) & e^{-2t} & 0 \\ g(t) & 0 & e^{-2t} \end{pmatrix} $$ with some functions $f(t), \, g(t)$ that need to be determined. Since $e^{0 \cdot A}$ is the identity matrix, we know that $f(0) = g(0) = 0$.
Using the differential equation $\frac{d}{dt} e^{At} = A e^{At}$ one obtains differential equations for $f$ and $g$ which can be easily solved.