How does rotating an operator affect its eigenvalues and eigenvectors?

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Well, in general the to an eigenvalue ##a## there are several linearly independent eigenvectors (in QT one talks about "degeneracy"), and then I label them as ##|a,\beta\rangle##. You can always choose them as orthonormal sets. For a pure state represented by a normalized ##|\psi \rangle## the probability to measure the value ##a## of the observable ##A## then is
$$P(a)=\sum_{\beta} |\langle a,\beta|\psi \rangle|^2.$$
That's of course the same as
$$P(a)=\sum_{\beta} \langle a,\beta|\hat{\rho}|a,\beta \rangle$$
with
$$\hat{\rho}=|\psi \rangle \langle \psi|.$$
 
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