Algebro-geometric proof of Cayley-Hamilton

Here is a sketch of proof of the Cayley-Hamilton theorem via classical algebraic geometry.

The set of n x n matrices over an algebraically closed field can be identified with the affine space $\mathbb{A}^{n^2}$ . Let $V$ be the subset of matrices that satisfy their own characteristic polynomial. We will prove that $V$ is in fact all of $\mathbb{A}^{n^2}$ . Since affine space is irreducible, it suffices to show that $V$ is closed and $V$ contains a non-empty open set.

Fix a matrix $M$ . First, observe that the coefficients of the characteristic polynomial are polynomials in the entries in $M$ . In particular, the condition that a matrix satisfy its own characteristic polynomial amounts to a collection of polynomials in the entries of $M$ vanishing. This establishes that $V$ is closed.

Let $U$ be the set of matrices that have $n$ distinct eigenvalues. A matrix has $n$ distinct eigenvalues if and only if its characteristic polynomial has no double roots when it splits. This occurs if and only if the discriminant of the characteristic polynomial is nonzero. The discriminant is a polynomial in the coefficients of the characteristic polynomial. Thus the condition that a matrix have $n$ distinct eigenvalues amounts to a polynomial in the entries of $M$ not vanishing. Thus $U$ is open.

Finally, we have to show $U \subseteq V$ . It is easy to check this for $U$ a diagonal matrix. The general result follows from the fact that the determinant and thus the characteristic polynomial is basis-invariant.

I learned this from https://aergus.net/blog/posts/using-zariski-topology-to-prove-the-cayley-hamilton-theorem.html/

Arithmetic variety

Algebro-geometric proof of Cayley-Hamilton

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