About the Eigenvalues of Orthogonal Matrix plus Perturbation Let $O$ be an orthogonal matrix, $O^T O = I$, thus its eigenvalues lie on the unit circle, $\lambda(O)=e^{i\theta}$. Furthermore, assume the form 
$O = X Y$, where both matrices satisfy $X^2 = I$ and $X=X^T$, and also $Y^2=I$ and $Y=Y^T$, thus $\lambda(X),\lambda(Y) \in \{-1,1\}$.
(These matrices do not commute $[X,Y] \ne 0$, $[O,X]\ne 0$.)
What can be said about the eigenvalues of the matrix $P = O + \delta X = X(Y + \delta I)$? Numerically it is possible to verify that up to some value of $\delta$, the complex eigenvalues of $P$ still falls on a circle whose radius now depends on $\delta$. Is it possible to prove this?
Any ideas or techniques on how to approach this problem?
 A: The reason is very simple, actually. Consider the following tautological identity, for some $0\leq z\in \mathbb{R}$,
$$
P^2-2\frac{P+z P^{-1}}{2}P+z I=0.
$$
Denote $\frac{P+z P^{-1}}{2}=B$, then find
$$
P^2-2BP+z I=0.
$$
Note that $[B,P]=0$ and thus if we can diagonalize $P$ and $B$, we can do so simulatneosly, and then for respective eigenvalues we will find
$$
\lambda_P=\lambda_B\pm i \sqrt{z-\lambda_B^2}.
$$
Assume that $\lambda_B^2\in\mathbb R$  and $\lambda_B^2<z$, then obviously
$$
|\lambda_P|=\sqrt z\quad\text{and}\quad \mathrm{Re}\,\lambda_P=\lambda_B
$$
If $\lambda_B^2\in\mathbb R$ and $\lambda_B^2\geq z$, then 
$$\lambda_P\in \mathbb{R}.$$
Now the idea is that if we pick $z=1-\delta^2\in \mathbb{R}$, then $B$ is real symmetric and thus diagonalizable with real eigenvalues. Indeed, 
$P^{-1}=\frac{Y-\delta I}{1-\delta^2}X$, and so
$$
2B=P+(1-\delta^2)P^{-1}=XY+YX=O+O^T.
$$
Even more, $B$ does not depend on $\delta$, and its spectrum is just the real part of the spectrum of $O$, so one can write down an explicit form for the spectrum of $P$ in terms of that of $O$. For example, the OP observation that the real parts of the complex eigenvalues are those of $O$ is immediately explained.
Edit: corrected minor errors as pointed out by the OP
