Summary of where we are: In middle of solving the state equation. We did LTV. In the case of LTI, we end up with e^{At}. We found it using using the first method. When has has distinct eigenvalues then we write e^{At}=V\begin{pmatrix} e^{\lambda _{1}t} & 0 & 0\\ 0 & \ddots & 0\\ 0 & 0 & e^{\lambda _{n}t}\end{pmatrix} V^{-1} Where V is called the modal matrix. (it has as its columns the eigenvectors of A). If we scale the eigenvectors, they still remain eigenvectors.
Reader: Show e^{At} is invariant under scaling of V.
Example: x_{1}^{\prime }=2x_{1} and x_{2}^{\prime }=-3x_{1}-3x_{2}. We want to find e^{At}. Hence A=\begin{pmatrix} 2 & 0\\ -3 & -3 \end{pmatrix} . The eigenvalues are \lambda _{1}=2,\lambda _{2}=-3 and the corresponding eigenvectors are v_{1}=\begin{pmatrix} 1\\ -\frac{3}{5}\end{pmatrix} ,v_{1}=\begin{pmatrix} 0\\ 1 \end{pmatrix} , hence V=\begin{pmatrix} 1 & 0\\ -\frac{3}{5} & 1 \end{pmatrix} ,therefore \begin{align*} e^{At} & =\begin{pmatrix} 1 & 0\\ -\frac{3}{5} & 1 \end{pmatrix}\begin{pmatrix} e^{2t} & 0\\ 0 & e^{-3t}\end{pmatrix}\begin{pmatrix} 1 & 0\\ -\frac{3}{5} & 1 \end{pmatrix} ^{-1}\\ & =\begin{pmatrix} e^{2t} & 0\\ -\frac{3}{5}e^{2t}+\frac{3}{5}e^{-3t} & e^{-3t}\end{pmatrix} \end{align*}
Notice at t=0\,then e^{At}=\begin{pmatrix} 1 & 0\\ 0 & 1 \end{pmatrix}
Reader: Show that \frac{d}{dt}e^{At}=A
What if we take the k^{th} derivative? then \frac{d^{k}}{dt^{k}}e^{At}=A^{k}
We will now do another example with complex eigenvalues. Let A=\begin{pmatrix} 0 & 1\\ -1 & -1 \end{pmatrix} , eigenvalues are \lambda _{1}=\frac{1}{2}i\sqrt{3}-\frac{1}{2},\lambda _{2}=-\frac{1}{2}i\sqrt{3}-\frac{1}{2}
Reader: Find e^{At} for the above.
We will find the eigenvectors to make the modal matrix. The eigenvectors are \begin{pmatrix} \frac{1}{2}i\sqrt{3}-\frac{1}{2}\\ 1 \end{pmatrix} and \begin{pmatrix} -\frac{1}{2}i\sqrt{3}-\frac{1}{2}\\ 1 \end{pmatrix} , hence V=\begin{pmatrix} \frac{1}{2}i\sqrt{3}-\frac{1}{2} & -\frac{1}{2}i\sqrt{3}-\frac{1}{2}\\ 1 & 1 \end{pmatrix} , therefore \begin{align*} e^{At} & =\begin{pmatrix} \frac{1}{2}i\sqrt{3}-\frac{1}{2} & -\frac{1}{2}i\sqrt{3}-\frac{1}{2}\\ 1 & 1 \end{pmatrix}\begin{pmatrix} e^{\left ( -\frac{1}{2}i\sqrt{3}-\frac{1}{2}\right ) t} & 0\\ 0 & e^{\left ( \frac{1}{2}i\sqrt{3}-\frac{1}{2}\right ) t}\end{pmatrix}\begin{pmatrix} \frac{1}{2}i\sqrt{3}-\frac{1}{2} & -\frac{1}{2}i\sqrt{3}-\frac{1}{2}\\ 1 & 1 \end{pmatrix} ^{-1}\\ & =\begin{pmatrix} \text{\tiny $ \frac{1}{2}e^{-\frac{1}{2}t-\frac{1}{2}i\sqrt{3}t}+\frac{1}{2}e^{\frac{1}{2}i\sqrt{3}t-\frac{1}{2}t}+\frac{1}{6}i\sqrt{3}e^{-\frac{1}{2}t-\frac{1}{2}i\sqrt{3}t}-\frac{1}{6}i\sqrt{3}e^{\frac{1}{2}i\sqrt{3}t-\frac{1}{2}t}$} & \text{\tiny $ \frac{1}{3}i\sqrt{3}\exp \left ( -t\left ( \frac{1}{2}i\sqrt{3}+\frac{1}{2}\right ) \right ) -\frac{1}{3}i\sqrt{3}e^{t\left ( \frac{1}{2}i\sqrt{3}-\frac{1}{2}\right ) } $}\\ \text{\tiny $ \frac{1}{3}i\sqrt{3}e^{t\left ( \frac{1}{2}i\sqrt{3}-\frac{1}{2}\right ) }-\frac{1}{3}i\sqrt{3}\exp \left ( -t\left ( \frac{1}{2}i\sqrt{3}+\frac{1}{2}\right ) \right ) $} & \text{\tiny $ \frac{1}{2}e^{-\frac{1}{2}t-\frac{1}{2}i\sqrt{3}t}+\frac{1}{2}e^{\frac{1}{2}i\sqrt{3}t-\frac{1}{2}t}-\frac{1}{6}i\sqrt{3}e^{-\frac{1}{2}t-\frac{1}{2}i\sqrt{3}t}+\frac{1}{6}i\sqrt{3}e^{\frac{1}{2}i\sqrt{3}t-\frac{1}{2}t} $} \end{pmatrix} \end{align*}
Now we will show another method to find e^{At}. This is using Laplace transform.
Reader: Show that e^{At}=\mathcal{L}^{-1}\left ( sI-A\right ) ^{-1}. Why is this true? \begin{align*} x^{\prime } & =Ax\\ sX\left ( s\right ) -x\left ( 0\right ) & =AX\left ( s\right ) \\ X\left ( s\right ) \left ( sI-A\right ) & =x\left ( 0\right ) \\ X\left ( s\right ) & =\left ( sI-A\right ) ^{-1}x\left ( 0\right ) \\ x\left ( t\right ) & =\mathcal{L}^{-1}\left ( sI-A\right ) ^{-1}x\left ( 0\right ) \end{align*}
Compare to x\left ( t\right ) =e^{At}x\left ( 0\right ) we see that e^{At}=\mathcal{L}^{-1}\left ( sI-A\right ) ^{-1} for any x\left ( 0\right )
Now we will given the third method to find e^{At}. This is called expansion of natural frequencies method. Here we allow repeated eigenvalues. In the first method (using modal matrix) the eigenvalues has to be distinct. Let the eigenvalues be \lambda _{1},\lambda _{1},\cdots ,\lambda _{m} with corresponding multiplies n_{1},n_{1},\cdots ,n_{m}. We will propose a form for e^{At} with some unknowns, then solve for these unknowns. Since all solution must have \exp and t multipliers (for repeated eigenvalues), let
\begin{equation} e^{At}={\displaystyle \sum \limits _{i=1}^{m}}{\displaystyle \sum \limits _{k=0}^{n_{i}-1}} Y_{k,i}t^{k}e^{\lambda _{i}t} \tag{1} \end{equation}
Where Y\left ( k,i\right ) are the unknowns. To find Y_{k,i}, we use \frac{d^{k}}{dt^{k}}e^{At}=A^{k}. Let us implement this on the first example we did above A=\begin{pmatrix} 2 & 0\\ -3 & -3 \end{pmatrix} \lambda _{1}=2,\lambda _{2}=-3, hence m=2, and the multiplies are n_{1}=1,n_{2}=1, hence using (1) gives \begin{equation} e^{At}=Y_{0,1}e^{2t}+Y_{0,2}e^{-3t} \tag{2} \end{equation}
\begin{align*} \left . e^{At}\right \vert _{t=0} & =I=\begin{pmatrix} 1 & 0\\ 0 & 1 \end{pmatrix} =Y_{0,1}+Y_{0,2}\\ \left . \frac{d}{dt}e^{At}\right \vert _{t=0} & =A=\begin{pmatrix} 2 & 0\\ -3 & -3 \end{pmatrix} =2Y_{0,1}-3Y_{0,2} \end{align*}
We have 2 equations above in 2 unknowns Y_{0,1},Y_{0,2}. We solve for these, then using (2) gives e^{At}. Solving gives Y_{0,1}=\begin{pmatrix} 1 & 0\\ \frac{-3}{5} & 0 \end{pmatrix} ,Y_{0,2}=\begin{pmatrix} 0 & 0\\ \frac{3}{5} & 1 \end{pmatrix} , hence (2) becomes\begin{align*} e^{At} & =\begin{pmatrix} 1 & 0\\ \frac{-3}{5} & 0 \end{pmatrix} e^{2t}+\begin{pmatrix} 0 & 0\\ \frac{3}{5} & 1 \end{pmatrix} e^{-3t}\\ & =\begin{pmatrix} e^{2t} & 0\\ \frac{3}{5}e^{-3t}-\frac{3}{5}e^{2t} & e^{-3t}\end{pmatrix} \end{align*}
Lets now do repeated eigenvalues. See my expansion of natural frequencies method notes for this larger example and more examples using this method using a symbolic function written to process this method.