How many ways are there to prove Cayley-Hamilton Theorem?

Solution 1:

My favorite : let $k$ be your ground field, and let $A = k[X_{ij}]_{1\leqslant i,j\leqslant n}$ 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 $\varphi_N:A\to k$ 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 $N\in M_n(k)$, $Res(\chi_N) = Res(\chi_{\varphi_N(M)})= \varphi_N(Res(\chi_M)) = 0$, so no matrix $N\in M_n(k)$ 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 $N\in M_n(k)$, $\chi_N(N) = \varphi_N(\chi_M(M)) = \varphi_N(0) = 0$.

Solution 2:

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.

Solution 3:

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 $S\equiv 0$.
For more details you can see Herstein's Topics in Algebra.

Solution 4:

A few years ago I gave an "ugly" and long proof (more than 4 pages), which is purely computational.

Basically, I took an $n\times n$ matrix $A=(a_{i,j})_{i,j}$, where $a_{i,j}$ are variables. Then I wrote the characteristic polynomial as $P_A(X)=c_nX^n+\cdots +c_0$, where each $c_k$ is an explicit polynomial in the variables $a_{i,j}$. Then I wrote explicitly the $n^2$ entries of each $A^k$ with $0\leq k\leq n$ as polynomials in the variables $a_{i,j}$. Finally, I proved that each of the $n^2$ entries of $P_A(A)=c_nA^n+\cdots +c_0I_n$ is $0$. It's just playing with sums of monomials and proving that all of them reduce in the end.

You can find the proof in the 3-4/2014 issue of Gazeta Matematica, Seria A, pages 32-36. Available online at this address: https://ssmr.ro/gazeta/gma/2014/gma3-4-2014-continut.pdf