I know this question has already been asked, but

I need a proof that for $A,B \in M_n(K)$, $$AB=I_n \Rightarrow BA=I_n$$ using Fitting's lemma.

I thought of using the fact that $K^n$ is a simple $M_n(K)$-module, but I don't see what endomorphism of $K^n$ I can use Fitting's lemma on, since multiplying by $BA$ is not $M_n(K)$-linear.

  • $\begingroup$ Do you have to use Fitting's lemma? Because there's a pretty easy solution without it; if you want to see, leave a comment and I'll answer. Cheers! $\endgroup$ Jun 21, 2015 at 18:30
  • $\begingroup$ Thank you but I know how to prove it without Fitting's lemma, but this is an exercise where it is explicitly mentioned that we have to use Fitting's lemma. $\endgroup$
    – Nitrogen
    Jun 21, 2015 at 18:33
  • 1
    $\begingroup$ Well, that's what I figured; if I come up with anything, I'll let you know. Best of luck with it! $\endgroup$ Jun 21, 2015 at 18:40

1 Answer 1


For a given matrix $N\in M_n(K)$ the map $\phi_N:K^n\to K^n$ defined by $\phi_N(x)=Nx$ is a homomorphism of $K$-vector spaces. By Fitting's Lemma there are two subspaces $V,W\subset K^n$ such that $K^n=V\oplus W$, $\phi_N^k(V)=0$ for $k\ge n$, and $\phi_N$ is an automorphism of $W$.

We apply the Fitting's Lemma to the map $\phi_{BA}$. From $AB=I$ we get $(BA)^2=BA$, that is, $BA$ is idempotent. Since $(BA)^nx=0\ \forall x\in V$ we get $(BA)x=0\ \forall x\in V$. Moreover, by multiplying (on the left) with $A$ we get $Ax=0\ \forall x\in V$. This shows that $V\subseteq\ker\phi_A$. From $\phi_{AB}=1_{K^n}$, equivalently, $\phi_A\circ\phi_B=1_{K^n}$ we know that $\phi_A$ is surjective, and since $K^n$ is a finite dimensional vector space we conclude that $\phi_A$ is bijective. It follows that $\ker\phi_A=0$, so $V=0$, and therefore $\phi_{BA}$ is an automorphism of $K^n$, that is, $\phi_{BA}$ is bijective. This entails that $BA$ is invertible (why?), and from $(BA)^2=BA$ we get $BA=I$.


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