1. ## "Eigenpair" Fundamental Matrix

I came across this problem in my class notes. I can't solve this one and I was wondering how to solve it.

Use the "eigenpair" method to find a fundamental matrix for the differential equation system. Use your answer to find e^At

x'=Ax [0 1]
[-2 -3] x

2. Originally Posted by meks08999
I came across this problem in my class notes. I can't solve this one and I was wondering how to solve it.

Use the "eigenpair" method to find a fundamental matrix for the differential equation system. Use your answer to find e^At

x'=Ax [0 1]
[-2 -3] x
You notation is confusing. I know you mean $\begin{bmatrix}0 & 1 \\ -2 & -3\end{bmatrix}$ but what do you mean by " $A x\begin{bmatrix}0 & 1 \\ -2 & -3\end{bmatrix}x$"? I suspect you have combined two notations and really mean that "Ax" is $\begin{bmatrix}0 & 1 \\ -2 & -3\end{bmatrix}x$.

If that is the case, then you need to find all eigenvalues and a corresponding eigenvector for each eigenvalue (the "eigenpairs").

Once you have that you can write $A= PDP^{-1}$ where "P" is the matrix having eigevectors as columns and D is a diagonal matrix with the eigenvalues on the diagonal.

It is easy to find powers of a diagonal matrix: $D^n$ is just the diagonal matrix having the nth power of the eigenvalues on the diagonal. Further, $A^2= (PDP^{-1})^2= (PDP^{-1})(PDP^{-1})= (PD)(P^{-1}P)Dp^{-1}= PD^2P^{-1}$, $A^3= A^2A= (PD^2P^{-1})(PDP^{-1})= PD^3P^{-1}$, etc.

$e^{At}= I+ At+ \frac{1}{2!}A^2t^2+ \frac{1}{3!}A^3t^3+ \cdot\cdot\cdot= PP^{-1}+ PDtP^{-1}+ \frac{1}{2!}PD^2t^2P^{-1}+ \frac{1}{3!}PD^3t^3+ \cdot\cdot\cdot$ $= P(I+ Dt+ \frac{1}{2}(Dt)^2+ \frac{1}{3!}(Dt)^3+ \cdot\cdot\cdot)P^{-1}= Pe^{Dt}P^{-1}$ where $e^{Dt}$ is just the diagonal matrix having $e^{\lambda t}$ on the diagonal with $\lambda$ being the eigenvalues.

3. Ya. Sorry about the notation. It is Ax= the matrix. So thanks for the help.