Row reducing to find complex eigenvectors

Question

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.

Answer 1

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: $$\begin{pmatrix}3+i\\-1\end{pmatrix},\quad \begin{pmatrix}3-i\\-1\end{pmatrix}. $$

Answer 2

In order to row-reduce this, we want to get the second row to have the same first component as the first row. The first row has a first component of $3-i$ while the second row has a component of $-1$, so we multiply the second row by $\frac{3-i}{-1}=i-3$. $$\begin{bmatrix}3-i & 10 \\ 3-i & 10\end{bmatrix}$$ Then, we subtract the second row by the first row to eliminate that row: $$\begin{bmatrix}3-i & 10 \\ 0 & 0\end{bmatrix}$$ Finally, we divide the first row by $3-i$ to get the matrix into RREF: $$\begin{bmatrix}1 & 3+i \\ 0 & 0\end{bmatrix}$$ Finally, from the RREF, we get the vector for the null space: $$\begin{bmatrix}-(3+i) \\ 1\end{bmatrix}=\begin{bmatrix}3+i \\ -1\end{bmatrix}$$ The other eigenvector for the other eigenvalue can be found by conjugation.