mjhjmtu

it is a well know result that for two $ntimes n$ matrices $A$ and $B$ the eigenvalues of the product $AB$ are equal to those of the product $BA$.

Are there similar results for a product of $m>2$ matrices? E.g. does the product of four $ntimes n$ matrices $ABCD$ have the same eigenvalues as $ACBD$?

Thanks for your help!

Let check this on orthogonal matrices (e.g. rotations dim. $3 times 3$).

We have for example $R_1R_1^TR_2R_2^T=I$ with $1$ as eigenvalues. If we change the order of rotations for example $R_1R_2(R_2R_1)^T=R_1R_2R_1^TR_2^T$ certainly if $R_1$ and $R_2$ are not commutative we would not obtain identity matrix and consequently the eigenvalues are different.

In fact, $AB$ and $BA$ have the same characteristic polynomial.

The problem with expecting switches in between various other matrices to preserve eigenvalues, is that the characteristic polynomial does not behave well with multiplication of matrices from left/right.

In other words, if $A$ and $B$ have the same characteristic polynomial, then this may not be true of $CA$ and $CB$, where $C$ is some other matrix. This is because the trace(sum of eigenvalues) is not multiplicative in general, and we can use this to produce counterexamples.

To give an example, take $A = beginbmatrix1 0 \ 0 2 endbmatrix$ and $B = beginbmatrix 1 1 \ 0 2endbmatrix$. These matrices have the same characteristic polynomial. For yourself, find a $C$ such that $CA$ and $CB$ have different trace, and therefore different eigenvalues. Find a $D$ such that $AD$ and $BD$ have different trace. It follows that $CA,CB$ and $AD,BD$ don't have the same characteristic polynomials.

The stumbling block in trying to prove, say , that $CBA$ and $CAB$ have the same eigenvalues, is the inability to assure that multiplying by $C$, after knowing that the eigenvalues of $AB$ and $BA$ are the same, retains the equality of eigenvalues. This is why you can use the $C,A,B,D$ above to find counterexamples of the assertions you were making.

However, cyclic permutations retain eigenvalues, because we are just applying the "two-matrix" case to deduce this. For example, $ABCD = (AB)(CD)$ has the same eigenvalues as $(CD)(AB) = CDAB$ and so on. So while this is a generalization, it may not feel satisfactory.