TITLE: Eigenvalues for a product of matrices
QUESTION [8 upvotes]: It was mentioned in one MSE answer that eigenvalues of products of square matrices are equal (see the answer of user1551 for Eigenvalues of Matrices and Eigenvalue of product of Matrices)          
Let's denote this fact: $ \ \ \ \ $ $\text{eig}(AB)=\text{eig}(BA)$.

However .. how can this be explained in the case where matrices don't
commute?   
Does some kind of geometrical interpretation of this    statement
exist - at least in the case of 3D orthogonal matrices where    it is
known that they usually don't commute ?
Can the statement be extended for a case of product of more number of matrices, for example:
$\text{eig}(A_1{A_2} ... A_n)=\text{eig}(A_n{A_{n-1}} ... A_1)=
   \text{eig}(A_{n-1}{A_{n-2}} ... A_n)=$  etc... ?

REPLY [3 votes]: On the spectra of cyclic permutation of matrix products
You can show that cyclic permutations of matrix produxts have the same spectra rather easily. Consider $$ A_1 \ldots A_n \vec v = \lambda \vec{v}.$$ Now multiply by $A_n$ on both sides. $$ A_n A_1 \ldots A_{n-1} (A_n \vec{v})  = \lambda  (A_n \vec{v}).$$ So $ A_n A_1 \ldots A_{n-1}$ has the same eigenvalues as $ A_1 \ldots A_n$ with corresponding eigenvectors given by the above expressions. You can do this procedure repeatedly to show that all cyclic permutations have the same spectrum. The two matrix case is a special case of this.
Geometric interpretation 
Sorry for the hand drawn picture. This gives an explanation for the case where the eigenvalue is $1$ or $-1$. The loci of vectors turned by the same amount due to a rotation matrix form a cone centred at the origin in 3D. When you combine two rotations the eigenvectors corresponding to $1$ lie in the intersection of two such cones, one for each rotation matrix. The picture shows such an intersection.  The cones intersect at two different vectors $\vec v$ and $A\vec{v}$ in the picture. When you reverse the order in which the rotation are applied the eigenvector changes from one of these vectors to the other. I cannot think of any geometric interpretation for the case of complex eigenvalues.