An example:
$$ \begin{pmatrix} 1 & 2 & 0 \\ 2 & -3 & 4 \\ 4 & -8 & 7 \\ \end{pmatrix} $$
Has eigenvalues $3$, $1+2i$, $1-2i$. How would you find the eigenvectors for this matrix without annoying row-reduction algebra with complex numbers?
For example, if I choose $\lambda = 1+2i$, I will have a matrix as such:
$$ \begin{pmatrix} -2i & 2 & 0 \\ 2 & -4-2i & 4 \\ 4 & -8 & 6-2i\end{pmatrix}$$
Now, if this were a $2 \times 2$ matrix, I would know that there has to be kernel with dimension $1$, so I could simply set one of the two rows to $0, 0$. But in a $3 \times 3$ matrix, setting any row to $0, 0, 0$ might result in my destroying a linearly independent row (although the odds are $1/3$). Is there an easy trick to find eigenvectors in such a situation without complex algebra?