Does this proof of "$AB$ and $BA$ have the same characteristic polynomial" assume $k$ algebraically closed?

Question

I am looking at Theorem 1.3.22 of this book. I was wondering whether it makes the assumption that the matrices are in an algebraically closed field $k$, and if so, where does it use such assumption.

Answer 1

When we say like "$n$ eigenvalues" you can understand that to mean:

• the eigenvalues exist in $k$
• the eigenvalues exist in some algebraic extension/closure of $k$

The proof itself doesn't reference the eigenvalues, just the matrices themselves.

Sometimes you'll see a proof begin with like: "fix an algebraic closure" or "fix an extension which contains all the eigenvalues." This is not necessary for this proof.

You can see similar statements to "$n$ eigenvalues." For instance: every quadratic has either 2 real roots, 1 repeated root with multiplicity, or 2 complex roots. There is an understanding that we know there are 2 roots even if they appear with multiplicity or appear not in $\mathbf{R}$ but in $\mathbf{R}[\sqrt{-1}]$.