Complex eigenvalues and rotations

Question

Suppose that $A$ is a $2\times 2$ real matrix with two complex eigenvalues $z_1,z_2$ such that $z_1z_2 = 1$. Prove that $A$ is a rotation by some angle $\theta$.

There are multiple ways to prove this. However, I am looking for the most "elementary" proof possible. Usually one can regard $A$ as a matrix with complex coefficients and find an eigenbasis consisting of complex vectors to conclude that $A^{-1} = A^t$, and the proof follows from there.

Suppose you are trying to justify this to someone who never seen complex vector spaces before, how do you derive it?

Answer 1

This is not true. Consider e.g. $$ A=\pmatrix{1&-2\\ 1&-1}. $$ Its two eigenvalues are $\pm i$ but $A$ is clearly not a rotation matrix.

Answer 2

That is false.

Since the eigenvalues of our matrix $A$ are conjugate, both have modulus $1$ and they are in the form $\cos(\theta)\pm i\sin(\theta)$. Then $tr(A)=2\cos(\theta), \det(A)=1$.

Finally, $A=\begin{pmatrix}a&b\\c&2\cos(\theta)-a\end{pmatrix}$, where $2a\cos(\theta)-a^2-bc=1$, depends on $2$ parameters (in addition to $\theta$) and, then, is not necessarily a rotation.

EDIT. You want an additional condition so that $A$ is orthogonal. This one must concern its eigenvectors; indeed, a rotation admits always $[1,\pm i]^T$ as a basis of eigenvectors; that is, it suffices to say that $A$ commutes with ONE rotation.

$\textbf{Proposition.}$ Let $A\in M_2(\mathbb{R})$. If $\det(A)=1$ and $A$ commutes with $Rot(\pi/2)$, then $A\in O^+(2)$ -and conversely-.