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

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?

• How about $\pmatrix{1&0&0\\0&0&1\\0&-1&0}$? Commented Nov 18, 2020 at 23:04

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: $$\pmatrix{\color{blue}1&0&0\\0&0&1\\0&-1&0}$$.