If $A = LU$ and $B = UL$, the $A$ and $B$ have the same eigenvalues

Question

If $A = LU$ and $B = UL$, where $L$ is a unit lower triangular and $U$ is an upper triangular matrices, then $A$ and $B$ have the same eigenvalues.

The eigenvalues of $A$ are obtained by solving the equation: $\det(A - \lambda I) = 0$. We have

$$\det(A - \lambda I) = \det(LU - \lambda I)= \det(L(U - \lambda L^{-1})) = \det(L)\det(U - \lambda L^{-1}).$$ Now, from the previous equation, we have $$\det(L)\det(U - \lambda L^{-1}) = \det(U - \lambda L^{-1})\det(L) = \det(UL - \lambda I) = \det(B - \lambda I), $$ thus, we have $$\det(A - \lambda I) = \det(B - \lambda I).$$ Does this means that they have the same eigenvalues?

Answer 1

Your argument is fine, but it requires $L$ to be invertible.

In general, the eigenvalues of $AB$ are the same as the eigenvalues of $BA$ for any two square matrices $A,B$.