Let K be an algebraically closed field and a,b n×n matrices over K. Then ab and ba have the same characteristics polynomial.

There are a lot of proof of this fact however here we want to gave a proof that use the machinery of Algebraic Geometry.

Recall that we define the affine n space over K as Aⁿ=Kⁿ . This set can be endowed with a nice topology, namely the Zariski topology. This is the topology which closed subset are the set of the form Z(I) for some subset I of K[x1,…,xn] where Z(I)={P∈Aⁿ : f(P)=0 ∀ f ∈ I}. Using the Nullstellensatz it can be shown that Aⁿ is an irreducible topological space (i.e if Z₁,Z₂ are closed subset of Aⁿ such that Aⁿ=Z₁∪Z₂then Aⁿ=Z₁ or Aⁿ=Z₂) .

We first observe this:

Lemma:

Let x,y be n×n matrices over K such that x is invertible.Then xy and yx have the same characteristics polynomial.

Proof:

Since x is invertible we can write yx=x⁻¹(xy)x so that xy and yx are similar. □

In the following for a matrix c we write Pc to indicate the characteristics polynomial of c. Then, identifying the n×n matrices over K with A^(n²) , we can define:

Z₁={(x,y)∈A^(2n²) : Pxy=Pyx}

Z₂={(x,y)∈A^(2n²) : det(A)=0}

Clearly Z₁ and Z₂ are closed subset of A^(2n²) since they are the zero set of some polynomial. Moreover:

A^(2n²)=Z₁ ∪ Z₂

since, if (x,y)∈A^(2n²) , if x is not singular then for the lemma xy han yx have the same characteristics polynomial, so that (x,y)∈Z₁.

Now since A^(2n²) is irreducible and clearly A^(2n²)≠Z₂ we get A^(2n²)=Z₁ that is the fact that we wanted to prove. □