MHB How can you diagonalize the given matrix and use it to solve a system of ODE's?

[Ackbach]

Here's this week's problem.

-----

Diagonalize the matrix
$$A=\begin{bmatrix}
0 &1 &0 \\
0 &0 &1 \\
30 &1 &-6
\end{bmatrix}.$$
Note that this procedure can be used to compute $e^{At}$, and thus
solve the system of ODE's $x'=Ax$, the solution being $e^{At}x_{0}$.

-----

Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!

 

[Ackbach]

No one solved this week's POTW. You can read my solution below.

Spoiler

We find the eigenvalues and eigenvectors thus:
$$\det(A-\lambda I)=\left|
\begin{matrix}
-\lambda &1 &0 \\
0 &-\lambda &1 \\
30 &1 &-6-\lambda
\end{matrix}
\right|
=-\lambda(-\lambda(-6-\lambda)-1)-1(-30)=0,$$
or
$$\lambda^2(-6-\lambda)+\lambda+30=0\quad\implies\quad
-\lambda^3-6\lambda^2+\lambda+30=0,$$
or
$$\lambda^3+6\lambda^2-\lambda-30=0.$$
We can see that $\lambda=2$ is a solution, making $\lambda-2$ a factor.
The other factors are $x+5$ and $x+3$. That is, the eigenvalues of
this matrix are $2, -3,$ and $-5$. Let us take the $\lambda=2$
eigenvalue. To find the corresponding eigenvector, we solve the system
$$Ax=2x,\quad\implies\quad
\begin{bmatrix}
-2 &1 &0 \\
0 &-2 &1 \\
30 &1 &-8
\end{bmatrix}x=0.$$
One solution is
$$\chi_2=\begin{bmatrix} 1\\ 2\\ 4\end{bmatrix}.$$
Similarly, solutions for the other eigenvalues are as follows:
$$\chi_{-3}=\begin{bmatrix}1 \\ -3 \\ 9 \end{bmatrix}\quad\text{and}
\quad \chi_{-5}=\begin{bmatrix}1 \\ -5 \\ 25 \end{bmatrix}.$$
Therefore, if we have
$$P=\begin{bmatrix}1 &1 &1 \\ -5 &-3 &2 \\ 25 &9 &4\end{bmatrix}\quad
\text{and}\quad D=\begin{bmatrix}-5 &0 &0 \\ 0 &-3 &0 \\ 0 &0 &2
\end{bmatrix},$$
then $A=PDP^{-1}$.
Note that
$$P^{-1}=\begin{bmatrix}-3/7 &1/14 &1/14 \\ 1 &-3/10 &-1/10 \\
3/7 &8/35 &1/35 \end{bmatrix}.$$