\begin{align} A = \begin{bmatrix} 0.5&-0.6\\0.75&1.1 \end{bmatrix} \end{align}

Find eigenvalues and a basis for the eigenspace

\begin{align} (1.1-\lambda)(0.5-\lambda)+(0.6)(0.75) = 0\\ \lambda^2 -1.6\lambda +1 \\ \frac{1.6 \pm \sqrt{-1.44}}{2} = 0.8 \pm 0.6\imath \end{align}

Substituting our new $\lambda$ back into our A matrix,

\begin{align} \begin{bmatrix} 0.5-(0.8+0.6\imath)&-0.6\\0.75&1.1-(0.8+0.6\imath) \end{bmatrix} = \begin{bmatrix} -0.3+0.6\imath&-0.6\\0.75&0.3+0.6\imath \end{bmatrix}. \end{align}

Row reduction here doesn't quite work the way we understand it, but we can notice that each equation in the matrix should have the same nontrivial solution. This was given by the book and I have no justification for it beyond "the book said so." Because this example is based on preliminary reading, I'm not certain as to why, just that it is. I hope to have this question answered in class. Continuing!

\begin{align} 0.75x_1 = (-0.3 -0.6\imath)x_2\\ x_1 = (-0.4 - 0.8\imath)x_2 \end{align}

If we choose a nice $x_2$ (like 5), we eliminate the need for any ugly decimals or even worse…fractions…*shudder*

$\vec{v}_1 = \begin{bmatrix} -2-4\imath\\5 \end{bmatrix}$
$\vec{v}_2 = \begin{bmatrix} -2+4\imath\\5 \end{bmatrix}$

Complex Eigenvalues ALWAYS occur in complex pairs. This is again stated in the book as a fact. If one plugs in our other eigenvalue (which is the complex conjugate of the first), we do in fact find our second eigenvalue to be the complex conjugate of the first, but I have not seen it generally proved that this is the case.

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