8 added 329 characters in body

EDIT OF AUG. 31, 2010. The proof of the Cayley-Hamilton Theorem I like best (among the ones I know) is on page 21 (proof of Proposition 2.4) of Introduction to Commutative Algebra by Atiyah and MacDonald. The argument can be phrased as follows.

Let $K$ be a commutative ring; let $n$ be a positive integer; let $A=(a_{ij})\in M_n(K)$ be an $n$ by $n$ matrix with entries in $K$; let $\chi$ be its characteristic polynomial; define $B=(b_{ij})\in M_n(K[A])$ by $b_{ij}:=\delta_{ij}\,A-a_{ij}$; let $(e_i)$ be the canonical basis of $K^n$; observe $$\sum_i\ \ b_{ij}\ e_i=0,\quad\det B=\chi(A);$$ and write $(c_{ij})$ for the adjugate of $B$. Applying (a trivial case of) Fubini's Theorem to the double sum $\sum_{i,j}\ c_{jk}\ b_{ij}\ e_i$, we get $\chi(A)=0$.

Thank you very much to darij grinberg! [I'm leaving the previous edits "for the record".] END OF EDIT OF AUG. 31, 2010.

EDIT OF DEC. 11, 2010. For a nice application of the Cayley-Hamilton Theorem, see this answer by Balazs Strenner.

PREVIOUS EDITS:

Here is a proof of the Cayley-Hamilton Theorem.

Let $K$ be a commutative ring, let $n$ be a positive integer, let $X$ be an indeterminate, let $A\in M_n(K)$ be an $n$ by $n$ matrix with coefficients in $K$, and let $\chi:=\det(X-A)$ be the characteristic polynomial. Equip $K^n$ with the $K[X]$-module structure induced by $A$. We must check $\chi K^n=0$. Form the right $M_n(K[X])$-module $$H:=\mathrm{Hom}_{K[X]}(K[X]^n,K^n).$$ Let $e\in H$ be the evaluation at $A$ (note $K[X]^n=K^n[X]$). As $e$ is surjective, it suffices to show $e\chi=0$. As $X-A$ divides $\chi$ on the left, it suffices to show $e(X-A)=0$. But this is obvious.

EDIT OF AUG. 1, 2010. Here is a diagrammatic rewriting of the argument.

EDIT OF AUG. 30, 2010. Here is a coordinate version of the above argument. [Compare with the proof of Propositon 3 page 81 of Weil's Basic Number Theory, and with the proof of Propositon 2.4 page 21 of Introduction to Commutative Algebra by Atiyah and MacDonald].

Weil's formulation. Put $$B(X)=(b_{ij}(X)):=X-A\in M_n(K[X]),$$ and let $C(X)=(c_{ij}(X))$ be the adjugate of $B(X)$. We have $$\sum_j\ c_{jk}(X)\ b_{ij}(X)=\delta_{ik}\ \chi(X)\in K[X].$$ Replacing $X$ with $A$, evaluating on $e_i$ (the $i$-th vector of the canonical basis of $K^n$), and summing over $i$ gives $$\sum_j\ c_{jk}(A)\ \sum_i\ b_{ij}(A)\ e_i=\chi(A)\ e_k\in K^n.$$ But the second sum is 0 by definition of $b_{ij}(X)$.

Atiyah-MacDonald's formulation. Put $A=(a_{ij})$ and define $B=(b_{ij})\in M_n(K[A])$ by $b_{ij}:=\delta_{ij}A-a_{ij}$; observe $$\sum_i\ b_{ij}\ e_i=0,\quad\det B=\chi(A);$$ and write $(c_{ij})$ for the adjugate of $B$. Computing $\sum_{i,j}\,c_{jk}\,b_{ij}\,e_i$ in two ways we get $\chi(A)=0$.

7 Edit clearly indicated.

EDIT OF AUG. 31, 2010. The proof of the Cayley-Hamilton Theorem I like best (among the ones I know) is on page 21 (proof of Proposition 2.4) of Introduction to Commutative Algebra by Atiyah and MacDonald. The argument can be phrased as follows.

Let $K$ be a commutative ring; let $n$ be a positive integer; let $A=(a_{ij})\in M_n(K)$ be an $n$ by $n$ matrix with entries in $K$; let $\chi$ be its characteristic polynomial; define $B=(b_{ij})\in M_n(K[A])$ by $b_{ij}:=\delta_{ij}\,A-a_{ij}$; observe $$\sum_i\ \ b_{ij}\ e_i=0,\quad\det B=\chi(A);$$ and write $(c_{ij})$ for the adjugate of $B$. Applying (a trivial case of) Fubini's Theorem to the double sum $\sum_{i,j}\ c_{jk}\ b_{ij}\ e_i$, we get $\chi(A)=0$.

Thank you very much to darij grinberg! [I'm leaving the previous edits "for the record".] END OF EDIT OF AUG. 31, 2010.

6 Minor changes.

Here is a proof of the Cayley-Hamilton Theorem.

Let $K$ be a commutative ring, let $n$ be a positive integer, let $X$ be an indeterminate, let $A\in M_n(K)$ be an $n$ by $n$ matrix with coefficients in $K$, and let $\chi:=\det(X-A)$ be the characteristic polynomial. Equip $K^n$ with the $K[X]$-module structure induced by $A$. We must check $\chi K^n=0$. Form the right $M_n(K[X])$-module $$H:=\mathrm{Hom}_{K[X]}(K[X]^n,K^n).$$ Let $e\in H$ be the evaluation at $A$ (note $K[X]^n=K^n[X]$). As $e$ is surjective, it suffices to show $e\chi=0$. As $X-A$ divides $\chi$ on the left, it suffices to show $e(X-A)=0$. But this is obvious.

EDIT OF AUG. 1, 2010. Here is a diagrammatic rewriting of the argument.

EDIT OF AUG. 30, 2010. Here is a coordinate version of the above argument. [Compare with the proof of Propositon 3 page 81 of Weil's Basic Number Theory, and with the proof of Propositon 2.4 page 21 of Introduction to Commutative Algebra by Atiyah and MacDonald].

Weil's formulation. Let $(a_{ij})$ be the coefficients of $A$, put Put $$B(X)=(b_{ij}(X)):=X-A\in M_n(K[X]),$$ and let $C(X)=(c_{ij}(X))$ be the adjugate of $B(X)$. We have $$\sum_j\ c_{jk}(X)\ b_{ij}(X)=\delta_{ik}\ \chi(X)\in K[X].$$ Replacing $X$ with $A$, evaluating on $e_i$ (the $i$-th vector of the canonical basis of $K^n$), and summing over $i$ gives $$\sum_j\ c_{jk}(A)\ \sum_i\ b_{ij}(A)\ e_i=\chi(A)\ e_k\in K^n.$$ But the second sum is 0 by definition of $b_{ij}(X)$.

Atiyah-MacDonald's formulation. Define Put $A=(a_{ij})$ and define $B=(b_{ij})\in M_n(K[A])$ by $b_{ij}:=\delta_{ij}A-a_{ij}$; observe $\sum_i\,b_{ij}\,e_i=0$, $\det B=\chi(A)$; $\sum_i\ b_{ij}\ e_i=0,\quad\det B=\chi(A);$$and write$(c_{ij})$for the adjugate of$B$. Computing$\sum_{i,j}\,c_{jk}\,b_{ij}\,e_i$in two ways we get$\chi(A)=0\$.