I have a question to the following problem:
"There is a linear, time-invariant System with the form $\frac{d\mathbf{x}}{dt}=\mathbf{A}x$. The Eigenvalues of the matrix $A$ are $s_1=-1$ and $s_2=-2$, as well as 2 corresponding Eigenvectors are $p_1=\begin{pmatrix}1 \\ 0\end{pmatrix}$ and $p_2=\begin{pmatrix}1 \\ 1\end{pmatrix}$.
Calculate the solution $\textbf{x}(t)$ of the system, if it gets started at the time $t=0$ at point $x(0)=\begin{pmatrix}3 \\ 1\end{pmatrix}$."
So as I suppose a state-transition matrix is needed, so I can calculate the solution by $x(t)=\Phi(t)x_0$. But how do I get $\Phi(t)$? There are no practical hints in my lecture notes to that.
Thank you very much already
It is a follow-up question to this thread: Eigenvectors Trajectories
Approach 1: You already have
$$\tag 1 x(t) = c_1e^{-t} \begin{bmatrix}1\\0\end{bmatrix} + c_2e^{-2t} \begin{bmatrix}1\\1\end{bmatrix}, x[0] = \begin{bmatrix}3\\1\end{bmatrix}$$
Substitute in $t = 0$, equate terms to the IC and find $c_1 = 2, c_2=1$.
Approach 2: Fundamental State Transition matrix
From $(1)$, we can write:
$$\Phi(t) = \begin{bmatrix}e^{-t}& e^{-2t}\\0&e^{-2t}\end{bmatrix}$$
Find:
$$x(t) = e^{A t} \cdot x[0]$$
Of course the answers are the same. Also, there are many approaches to solving these.