Find the eigenvalues and a basis for each eigenspace in $C^2$
$\begin{bmatrix}1&5\\-2&3\end{bmatrix}$
$\lambda = 2 \pm 3i$
For the first eigenvalue, $2 + 3i$, I got this eigenvector:
$\begin{bmatrix}\frac{1-3i}{2}\\1\end{bmatrix}$
But the book says $\begin{bmatrix}{1-3i}\\2\end{bmatrix}$. This is just: $2\begin{bmatrix}\frac{1-3i}{2}\\1\end{bmatrix}$
So, where did I go wrong?
Best Answer
We have the matrix
$$\begin{bmatrix}1&5\\-2&3\end{bmatrix}$$
We find the eigenvalues from $|A - \lambda I| = 0$ as
$$\lambda_{1,2} = 2 ~ \pm 3i$$
We find the eigenvector of $\lambda_1 = 2 + 3i$, using the RREF of $[A - \lambda I]v_1 = 0$ as
$$\begin{bmatrix} 1 & \dfrac{-1 + 3 i}{2} \\ 0 & 0 \\ \end{bmatrix}v_1 = \begin{bmatrix} 0 \\ 0 \ \end{bmatrix}$$
Eigenvectors are not unique, and we can write (verify that each $v_1$ works)
$$v_1 = \begin{bmatrix} \dfrac{1 - 3 i}{2} \\ 1 \ \end{bmatrix}~~\text{or}~~ v_1 = \begin{bmatrix} 1 - 3 i \\ 2 \ \end{bmatrix}$$
For $\lambda_2's$ eigenvector, we just take the complex conjugate of $\lambda_1's$.
In other words, your result and the book's result are equivalent and both are correct.