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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?

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    $\begingroup$ The fact that the eigenvalues varies continuously doesn't imply that $\lambda(P)$ is still on a circle right? Otherwise this would be true for any arbitrary perturbation. Numerically I can see some interesting properties about the eigenvalues of $P$, for instance the real part of its eigenvalues are the same as the eigenvalues of $XY$ alone. Thus the perturbation $\delta X$ is only affecting the imaginary part. I don't know how to prove these things. You can think about $\delta$ being small but this is true even for a not so small $\delta$. $\endgroup$
    – ModularFunction
    Commented May 17, 2016 at 4:18
  • $\begingroup$ I've run some tests in Mathematica, with random $X, Y$ and it doesn't seem that your conjecture holds. Note however that it holds for pure imaginary $\delta$, since in that case $P$ is proportional to unitary matrix. $\endgroup$
    – Peter Kravchuk
    Commented May 17, 2016 at 4:21
  • $\begingroup$ I was not totally clear, sorry for that. What I means is that the complex eigenvalues fall on a circle. I'll include an example in the posting. $\endgroup$
    – ModularFunction
    Commented May 17, 2016 at 4:28
  • $\begingroup$ Indeed, it seems to hold for complex eigenvalues. $\endgroup$
    – Peter Kravchuk
    Commented May 17, 2016 at 4:30
  • $\begingroup$ I find that the complex eigenvalues fall on a circle of radius $r = (4 - 4 \delta^2)^{1/2}$, and the real ones which does not fall on the circle are $2\pm 2\delta$. If you vary $\delta$ eventually all eigenvalues start to collapse on the real line. What I'm interested is in understanding why the complex eigenvalues always have the form $\lambda(P) = 2(1-\delta^2) e^{i\theta}$. $\endgroup$
    – ModularFunction
    Commented May 17, 2016 at 4:33

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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

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  • $\begingroup$ Thanks a lot Peter! Very nice answer!! Just for future reference there is a typo on $z^2 \to z$ inside the square root, and $|\lambda_P| =\sqrt{ z } $ with $z > 0$. Your answer explains everything. $\endgroup$
    – ModularFunction
    Commented May 17, 2016 at 7:13

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