# topological proof of the Cayley-Hamilton theorem

Suppose then that the discriminant     of the characteristic polynomial is non-zero, and hence that $T:V\rightarrow V$ has $n=\dim V$ distinct eigenvalues     once we extend11Technically, this means that we must work with the vector space  $\bar{V}=V\otimes\bar{k}$, where $\bar{k}$ is the algebraic closure  of the original field of scalars, and with $\bar{T}:\bar{V}\to\bar{V}$ the extended automorphism with action $\bar{T}(v\otimes a)\to T(V)\otimes a,\quad v\in V,\;a\in\bar{k}.$ to the algebraic closure of the ground field. We can therefore choose a basis of eigenvectors    , call them $\mathbf{v}_{1},\ldots,\mathbf{v}_{n}$, with $\lambda_{1},\ldots,\lambda_{n}$ the corresponding eigenvalues. From the definition of characteristic polynomial we have that

 $c_{T}(x)=\prod_{i=1}^{n}(x-\lambda_{i}).$

The factors on the right commute, and hence

 $c_{T}(T)\mathbf{v}_{i}=0$

for all $i=1,\ldots,n$. Since $c_{T}(T)$ annihilates a basis, it must, in fact, be zero.

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

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

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

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

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

Title topological proof of the Cayley-Hamilton theorem  TopologicalProofOfTheCayleyHamiltonTheorem 2013-03-22 12:33:22 2013-03-22 12:33:22 rmilson (146) rmilson (146) 7 rmilson (146) Proof msc 15-00