Two square matrices whose sum equals their product must commute

Question

Let $M$ and $N$ be square matrices satisfying $M + N = MN$ show that $M$ and $N$ commute.

Note that $\lambda = 1 \notin \text{spec}(N)$. To see this, note that $Nv = v \implies Mv + Nv = MNv \implies Mv + v = Mv \implies v = 0$ and so $v$ would not be an eigenvector.

Now, if $Nv = \lambda v$, then $Mv = \mu v$, with $\mu = \frac{\lambda }{\lambda -1}$

Thus, $M$ and $N$ have the same eigenspace but $\text{Spec}(N) \cap \text{Spec}(M) = \emptyset$.(*)

This precise condition (*)implies that $M$ and $N$ are simultaneously diagonalizable, that is, there exists an invertible matrix $P$ whose column vectors are the eigenvectors of $M$, which are also the eigenvectors of $N$ so that $P^{-1}MP = D_{M}$ and $P^{-1}N P = D_{N}$. The commutativity of $M$ and $N$ follows from the fact that $D_{M}$ and $D_{N}$, which are diagonal matrices with respective eigenvalues on the diagonals, commute.

If my argument doesn't contain an invalid claim, then invoking (*) may still be an overkill. I will appreciate seeing any simpler argument.

Answer 1

We can rewrite this equation as $$ M + N = MN \implies MN - M - N + I = I \implies\\ (M - I)(N - I) = I. $$ So, we have $(N - I) = (M - I)^{-1}$, which means that $N-I,M-I$ commute. This implies that $M,N$ commute, as was desired.

Answer 2

From the condition given, you get $$ MN+N^2 = MN^2, \quad NM + N^2 = NMN, $$ so subtracting the two gives $$ MN-NM = MN^2 - NMN = (MN-NM)N. $$ Now, if $NM-MN \neq 0$, take $v$ such that $w = v(NM-MN) \neq 0$. Then plugging this in the above equation tells you that $w$ is a left-eigenvector of $N$ for the eigenvalue 1. But you showed that 1 is not an eigenvalue.