I am trying to find eigenvectors of the matrix
$$
A=
\begin{bmatrix}
5 & 10 \\
-1 & -1 \\
\end{bmatrix}
$$
I have solved the characteristic polynomial and obtained eigenvalues of $$\lambda = 2 \pm i$$
In the real-valued eigenvalue case, this is straightforward, since I can easily compute the nullspace of $A-\lambda I$. I suspect that a similar technique is required for complex eigenvalues, however I cannot seem to find a simple way to row-reduce and find the nullspace of
$$A-\lambda I =
\begin{bmatrix}
3-i & 10 \\
-1 & -3-i \\
\end{bmatrix}
$$
I know that eigenvectors will be complex conjugates of one another, so I am only computing the eigenvector corresponding to $2+i$.
Can anyone explain how to proceed?
Much thanks in advance.
Best Answer
By definition, this linear system has rank $1$, so you really have only one equation in two variables to solve, say $$x+(3+i)y=0,\quad\text{whence}\quad x=-(3+i)y.$$ One obtains at once by conjugation the two independent eigenvectors: $$\begin{pmatrix}3+i\\-1\end{pmatrix},\quad \begin{pmatrix}3-i\\-1\end{pmatrix}. $$