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For simplicity, consider $M_2(\mathbb{R})$, the set of $2\times 2$ real matrices.

Fact: For any $A,B\in M_2(\mathbb{R})$, $AB$ and $BA$ have same characteristic polynomials.

The wiki proof is as follows:

(1) $M_2(\mathbb{R})$ can be identified topologically with Euclidean topological space $\mathbb{R}^4$

(2) $GL_2(\mathbb{R})$ is an open subset of $M_2(\mathbb{R})$.

(3) The fact is true for $A,B$ in this open subset; hence true for $M_2(\mathbb{R}).$

Question 1: Can one clarify a little the argument (3)?

Question 2. How should the proof be modified for arbitrary field $K$ instead of $\mathbb{R}$?

p Groups
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1 Answers1

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It's not just that it's an open subset. The important thing is that it's a dense open subset. In the Zariski topology this should work over arbitrary fields, as the general linear group is a dense open subset in the Zariski topology. (I haven't really checked the details of this; there may be some caveat that I missed).

Matt Samuel
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