In a problem I'm trying to find a real-valued $3\times3$ matrix that has the eigenvalues $1$, $i$ and $-i$ (which must mean that the corresponding eigenvectors for $i$ and $-i$ must be complex). I already know that one way to approach this is to write out a general $3\times3$ matrix of the form
$$A= \begin{pmatrix} a & b & c \\ d & e & f \\ g & h & j \end{pmatrix}$$
then evaluate the determinant of $A-\lambda I$ and set it equal to $0$ to form a characteristic polynomial for the eigenvalues and then choose $a,b,c,d,e,f,g,h,j$ such that the polynomial becomes $(\lambda-1)(\lambda^2+1)$ to give the required eigenvalues as roots.
However, I'm looking for a geometric way to find such a matrix (which is the approach hinted at in the problem). I thought maybe some kind of complex plane transformation might work, but I wasn't sure how a $3\times3$ matrix would apply in such a situation.
How can I find such a matrix geometrically, without having to do lots of algebra as in the method outlined above?