topological proof of the Cayley-Hamilton theorem

We begin by showing that the theorem is true if the characteristicpolynomial does not have repeated roots, and then prove the general case.

Suppose then that the discriminant of the characteristic polynomial isnon-zero, and hence that $T:V\to V$ has $n=dimV$ distincteigenvalues once we extend to the algebraic closure of the groundfield.We can therefore choose a basis of eigenvectors, call them${𝐯}_1,\mathrm{\dots },{𝐯}_n$, with ${\lambda }_1,\mathrm{\dots },{\lambda }_n$ thecorresponding eigenvalues. From the definition of characteristicpolynomial we have that

$$c_T(x)=\prod _{i=1}^n(x-{\lambda }_i).$$

The factors on the right commute, and hence

$$c_T(T){𝐯}_i=0$$

for all $i=1,\mathrm{\dots },n$. Since $c_T(T)$ annihilates a basis, it must, infact, be zero.

To prove the general case, let $\delta (p)$ denote the discriminant ofa polynomial $p$, and let us remark that the discriminant mapping

$$T\mapsto \delta (c_T),T\in \mathrm{End}(V)$$

is polynomial on$\mathrm{End}(V)$. Hence the set of $T$ with distinct eigenvalues is a denseopen subset of $\mathrm{End}(V)$ relative to the Zariskitopology. Now the characteristic polynomial map

$$T\mapsto c_T(T),T\in \mathrm{End}(V)$$

is a polynomial map on the vectorspace $\mathrm{End}(V)$. Since it vanishes on a denseopen subset, it must vanish identically. Q.E.D.