Ytukyg

I see many proofs for the Cayley-Hamilton Theorem in textbooks and net, so I want to know how many proofs are there for this important and applicable theorem?

My favorite : let $k$ be your ground field, and let $A = k[X_{ij}]_{1leqslant i,jleqslant n}$ be the ring of polynomials in $n^2$ indeterminates over $k$, and $K = Frac(A)$.

Then put $M = (X_{ij})_{ij}in M_n(A)$ the "generic matrix".

For any $N=(a_{ij})_{ij}in M_n(k)$, there is a unique $k$-algebra morphism $varphi_N:Ato k$ defined by $varphi(X_{ij}) = a_{ij}$ that satisfies $varphi(M)=N$.

Then the characteristic polynomial of $M$ is separable (ie $M$ has $n$ distinct eingenvalues in an algebraic closure $widetilde{K}$ of $K$). Indeed, otherwise its resultant $Res(chi_M)$ is zero, so for any $Nin M_n(k)$, $Res(chi_N) = Res(chi_{varphi_N(M)})= varphi_N(Res(chi_M)) = 0$, so no matrix $Nin M_n(k)$ would have distinct eigenvalues (but obviously some do, just take a diagonal matrix).

It's easy to show that matrices with separable characteristic polynomial satisfy Cayley-Hamilton (because they are diagonalizable in an algebraic closure), so $M$ satisfies Cayley-Hamilton.

Now for any $Nin M_n(k)$, $chi_N(N) = varphi_N(chi_M(M)) = varphi_N(0) = 0$.

Here is a neat proof from Qiaochu Yuan's answer to this question:

If $L$ is diagonalizable with eigenvalues $lambda_1, dots lambda_n$, then it's clear that $(L - lambda_1) dots (L - lambda_n) = 0$, which is the Cayley-Hamilton theorem for $L$. But the Cayley-Hamilton theorem is a "continuous" fact: for an $n times n$ matrix it asserts that $n^2$ polynomial functions of the $n^2$ entries of $L$ vanish. And the diagonalizable matrices are dense (over $mathbb{C}$). Hence we get Cayley-Hamilton in general.

One can prove this theorem by use of the fact that the matrix representation of all linear map on a complex vector space, is Triangularisable with respect to a basis ${v_1,...,v_n}$.

So if $T$ be a linear map there are ${lambda_1,...,lambda_n}$ s.t

$$T(v_1)=lambda_1 v_1 $$
$$T(v_2)=a_{11} v_1+lambda_2 v_2 $$
$$.$$
$$.$$
$$.$$
$$T(v_n)=a_{n1}v_1+a_{n2}v_2+...+lambda_n v_n $$

And by computation we can find that the matrix $S=(T-lambda_1)(T-lambda_2)...(T-lambda_n)$ vanishes all $v_i$, and so $Sequiv 0$.

For more details you can see Herstein's Topics in Algebra.