Find the general solution of x'=(2, -1, 2+i, -1-i)x. (this is a matrix, 2 and -1 on the left, 2+i, -1-i on the right, and I know that the 2 roots are -i, 1. but I don't know how to find a, b for the first root.)

Question

anonymous · Answer

Sorry about the wait.
$$x'=\begin{pmatrix}2&2+i\-1&-1-i\end{pmatrix}x$$
This is your equation, right?

anonymous · Answer

Yes.

anonymous · Answer

$$\begin{vmatrix}2-\lambda&2+i\-1&-1-i-\lambda\end{vmatrix}=\lambda^2-(1-i)\lambda-i=0~~\Rightarrow~~\lambda_1=1,\lambda_2=-i$$
And you're having trouble with the eigenvectors, right?

anonymous · Answer

Yes, continue.

anonymous · Answer

For $\lambda_1=1$, you have the following equation:
$$\begin{pmatrix}1&2+i\-1&-2-i\end{pmatrix}\begin{pmatrix}a_1\a_2\end{pmatrix}=\begin{pmatrix}0\0\end{pmatrix}$$
where $\vec{a}=\begin{pmatrix}a_1\a_2\end{pmatrix}$ is the eigenvectors for $\lambda_1$.

The second row is (-1) times the first, so you're left with the following equation with two unknowns:
$$a_1+(2+i)a_2=0$$
Letting $a_2=1$, you get $a_1=-2-i$, so
$$\vec{a}=\begin{pmatrix}-2-i\1\end{pmatrix}$$

anonymous · Answer

For $\lambda_2=-i$, you have the following equation:
$$\begin{pmatrix}2+i&2+i\-1&-1\end{pmatrix}\begin{pmatrix}b_1\b_2\end{pmatrix}=\begin{pmatrix}0\0\end{pmatrix}$$
where $\vec{b}=\begin{pmatrix}b_1\b_2\end{pmatrix}$ is the eigenvectors for $\lambda_2$.

The two rows are the same, so you have
$$-b_1-b_2=0$$
Letting $b_2=1$, you get $b_1=-1$, so
$$\vec{b}=\begin{pmatrix}-1\1\end{pmatrix}$$

anonymous · Answer

But the first vector is 2+i, -1. How do I get that? How do I find a, b for the first one?

anonymous · Answer

Are you saying that's what your answer key says?

This usually comes up as an issue, but it's really not. If you let $a_2=-1$, you'll get $a_1=2+i$. It all depends on the values you choose. They're really the same eigenvector.

anonymous · Answer

But how do you get that? a+b(2+i)=0, and -a-b(2-i)=0, how do you solve for a and b?

anonymous · Answer

You fix one of the values. Pick an easy one to work with.

anonymous · Answer

I get it, thanks.

anonymous · Answer

Your welcome. Check out Paul's online notes for this topic. It helped me out when I first learned this stuff.