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?

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    $\begingroup$ Avoid demanding that answers have a certain form. If you don't allow for references or links, you'll miss out on other proofs. Also, there is no reason for demanding that an answer contain only one proof. What's the goal of this? $\endgroup$ – Pedro Tamaroff Apr 23 '16 at 14:57
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    $\begingroup$ That seems like an arbitrary thing to ask for. $\endgroup$ – Pedro Tamaroff Apr 23 '16 at 15:05
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    $\begingroup$ At any rate, do not expect people to comply with this demand. $\endgroup$ – Pedro Tamaroff Apr 23 '16 at 15:08
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    $\begingroup$ @PedroTamaroff you can delete your comments. $\endgroup$ – user217174 May 4 '16 at 22:16
  • $\begingroup$ Also, I would avoid asking moderators do delete their comments :p $\endgroup$ – fonini May 4 '16 at 23:10

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 $N\in M_n(k)$, $\chi_N(N) = \varphi_N(\chi_M(M)) = \varphi_N(0) = 0$.

  • $\begingroup$ What's a $k$-algebra morphism? Can you elaborate on that step? $\endgroup$ – littleO Apr 23 '16 at 23:28
  • $\begingroup$ @littleO A morphism that preserves operations : it is linear (where $A$ and $k$ are considered as vector spaces on $k$), and preserves multiplication and unit (in other words, that is a unital ring homomorphism between $A$ and $k$ seen as rings) $\endgroup$ – yago Aug 5 '16 at 13:42

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.

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    $\begingroup$ This proof also works over any integral domain by going up to the algebraic closure of the field of fractions and using that the diagonalizable matrices are dense in the Zariski topology (which is regular which is sufficient to emulate the Hausdorffness to make morphisms uniquely determined by their value on a dense subset) $\endgroup$ – Tobias Kildetoft Apr 23 '16 at 18:54

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.


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


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