Prove $\det(A+BC)=\det(A+CB)$ if $AB=BA$

Question

Let $A$, $B$ and $C$ be three endomorphisms of a finite-dimensional vector space such that $AB=BA$. Prove that $$\det\left(A+BC\right)=\det\left(A+CB\right)$$

Answer 1

To see this, note that:

1) $\det(A+BC)=\det(A)\det(I+A^{-1}BC)$,

2) $AB = BA$ implies $BA^{-1}=A^{-1}B$,

3) $\det(I+ABC)=\det(I+BCA)$ (this is true for any cyclic permutation of $ABC$).

Combining these gives us: $$\det(A+BC)=\det(A)\det(I+BA^{-1}C)=\det(A)\det(I+A^{-1}CB)= \det(A+CB).$$

Note: we have implicitly assumed that $A$ is invertible.

Answer 2

First assume that $B$ is invertible. We then have $$A = B^{-1}AB$$ and

$$\det(A+BC) = \det(B^{-1}(A+BC)B) = \det(B^{-1}AB + CB) = \det(A + CB)$$

Since invertible matrices are dense and determinant is continuous, equality extends to non-invertible $B$'s by continuity.