Finding a real $3\times3$ matrix with eigenvalues $1$, $i$ and $-i$ geometrically

Question

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?

Answer 1

The $2\times2$ matrix $\pmatrix{0&1\\-1&0}$, which corresponds geometrically to $90^\circ$ rotation

(multiplication by $i$), has eigenvalues $i$ and $-i$.

Put a $\color{blue}1$ on the diagonal, and you're done: $\pmatrix{\color{blue}1&0&0\\0&0&1\\0&-1&0}$.