I have been working through this useful article on calculating eigen values and vectors: https://medium.com/fintechexplained/what-are-eigenvalues-and-eigenvectors-a-must-know-concept-for-machine-learning-80d0fd330e47
At the bottom, the author provides a worked example, which is clear until this point:
Link to picture of the calculation
Where the eigen vector calculation result for (A - lambda * I) goes from -1/2-sqrt(15)/2, -1 on the top and 4, 1-sqrt(15)/2 on the bottom… to -1 + sqrt(15) on the top, and 8 on the bottom.
I don’t understand how it simplifies like this? If anyone could explain that would be great.
The eigenvalue-eigenvector pair $(\lambda,v) $ of a matrix $A$ is defined by the $$ Av = \lambda v $$ where $v\neq 0$. Rearranging terms yields $$ (A-\lambda I) v = 0. $$ This means that, knowing the eigenvalue, one has to solve $$ (A-\lambda I) x = 0 $$ for $x$ in order to get the eigenvector. In your example you have to solve $$\begin{bmatrix} -\frac12 - i \frac{\sqrt{15}}{2} & -1 \\ 4 & \frac12 - i \frac{\sqrt{15}}{2} \end{bmatrix} \begin{bmatrix} v_1 \\ v_2 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} $$ for $v_1$ and $v_2$. This yields $$ 4v_2 = (-\frac12 + i\frac{\sqrt{15} }{2})v_1, $$ hence by choosing $v_2 = 2$, we have $$ v=\begin{bmatrix} -1 + i\sqrt{15} \\ 8 \end{bmatrix}. $$ Note that if $v$ is an eigenvector, so is $cv$ for any nonzero constant $c$.