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… ?
Best Answer
Hint For invertible $B$, the characteristic polynomial of $AB$ is $$\det(\lambda I - AB) = \det[(\lambda B^{-1} - A) B] = \det (\lambda B^{-1} - A)\det B.$$ Rewrite this to show that $$\det(\lambda I - AB) = \det(\lambda I - BA).$$
By applying the above argument inductively, we can also conclude that the eigenvalues of $A_1 \cdots A_r$ coincide with those of any product $A_s \cdots A_r A_1 \cdots A_{s-1}$ given by permuting the factors cyclically. On the other hand, this is not true for general permutations, that is, in general the eigenvalues of $A_{\sigma(1)} \cdots A_{\sigma(r)}$ are not those of $A_1 \cdots A_r$. For a minimal example, take $$A_1 = \pmatrix{1&0\\0&0}, \qquad A_2 = \pmatrix{0&1\\1&0}, \qquad A_3 = \pmatrix{0&1\\0&0} .$$ Then, $A_1 A_2 A_3 = 0$ but $A_1 A_3 A_2 = \pmatrix{1&0\\0&0}$.
You can also show using these ingredients that the trace of an $r$-fold product of square matrices is invariant under cyclic permutation, but we can again see from the above example that the same is not in general true for general permutations.