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{\bf A Proof of the Cayley-Hamilton Theorem.}

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Dror Bar-Natan, December 30, 2015.

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\noindent{\bf Theorem.} Let $R$ be a commutative ring, let $A\in M_{n\times n}(R)$ be a matrix and let $\chi_A(t)\coloneqq\det(tI-A)$ be the characteristic polynomial of $A$. Then $\chi_A(A)=0$.

\noindent{\bf Proof.} For any matrix $M$ over any commutative ring there is ``the adjoint matrix $\adj(M)$ of $M$'', defined using the minors of $M$, which satisfies $\det(M)I=(\adj M)M$. Use this with $M=tI-A$, over the ring $R[t]$, and find that in the ring $M_{n\times n}(R[t])$ we have

\[ \chi_A(t)I = \det(tI-A)I = (\adj(tI-A))(tI-A). \]

Now note that the ring $M_{n\times n}(R[t])$ is isomorphic to the ring $M_{n\times n}(R)[t]$, and on the latter there is a linear ``evaluation at $t=A$'' map $\ev_A\colon M_{n\times n}(R)[t] \to M_{n\times n}(R)$, defined by ``putting $A$ to the right of all the coefficients''; namely, by $\sum B_kt^k \mapsto \sum B_kA^k$. This evaluation map $\ev_A$ is {\em not} multiplicative, but nevertheless it annihilates anything that has a right factor of $(tI-A)$. Hence under $\ev_A$ the above equality becomes

\[ \chi_A(A)I = 0. \]

\hfill$\Box$

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