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Find the General solution of $\textbf{x}^{'}=\begin{pmatrix} 2&2+i\\-1&-1-i\\ \end{pmatrix}\textbf{x}$

I started out by finding the eigenvalues.

$\begin{gather}(2-\lambda)(-1-i-\lambda)+2+i=0\\-2-2i-2\lambda +\lambda +\lambda i +\lambda^2+2+i=0\\ \lambda^2+\lambda(i-1)-i=0\\ \lambda_1=\dfrac{1-i +\sqrt{2i}}{2}\\ \lambda_2=\dfrac{1-i-\sqrt{2i}}{2} \end{gather}$

The book gets a really nice looking general solution of $$c_1\begin{pmatrix} 2+i\\-1 \end{pmatrix}e^t+c_2\begin{pmatrix} 1\\-1 \end{pmatrix}e^{it}$$

But this contradicts what I got for my eigenvalues. Where is my mistake?

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Looks like a simple algebra issue finding the roots of:

$$\lambda^2+\lambda(i-1)-i=0 \implies ~ \lambda_1 = 1, ~ \lambda_2 = i$$

Note: $\dfrac{1-i + \sqrt{2 i}}{2}$ can be nicely reduced as:

$$2 i = 1 + 2i -1 = 1 + 2i + i^2 = (1+i)^2$$

This gives:

$$\lambda_{1,2} = \dfrac{1-i \pm~\sqrt{(1+i)^2}}{2} = \dfrac{1-i \pm~(1+i)}{2} = 1, ~-i$$

You could have also done a similar reduction for the second one and get $\lambda_{1,2} = 1, ~-i$.

Update

When finding the eigenvectors for $\lambda_1 = 1$, we want the rref of $[A - I]v_1 = 0$. This produces:

$$\begin{pmatrix} 1 & 2+i\\0 & 0\\ \end{pmatrix}v_1 = 0$$

So we have $a = (-2 - i) b$. We are free to choose $b \ne 0$ (can not have a zero eigenvector).

For example, if we choose $b = 1$, we would get $v_1 = (-2 - i, 1)$.

If we choose $b = -1$, we would get $v_1 = (2 + i, -1)$.

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  • $\begingroup$ I did not know that nor would I have thought of writing it like that. I knew I was missing something this definitely helps $\endgroup$ – adam Mar 27 '14 at 0:40
  • $\begingroup$ By the way I get the eigenvector corresponding to 1 of $\begin{pmatrix} -2-i\\1 \end{pmatrix}$ which is the negative of the solution. Is this correct? $\endgroup$ – adam Mar 27 '14 at 1:09
  • $\begingroup$ but you need to write it as a basis though right? $\endgroup$ – adam Mar 27 '14 at 1:22

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