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\parbox[b]{2in}{\tiny
  \href{http://www.math.toronto.edu/~drorbn}{Dror Bar-Natan}: \href{http://www.math.toronto.edu/~drorbn/classes}{Classes}:
  \newline \href{http://www.math.toronto.edu/~drorbn/classes/\#2526}{2025-26}: \href{http://drorbn.net/25-347}{MAT 347 Groups, Rings, Fields}:
}~{\LARGE\bf The Cayley-Hamilton Theorem}
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\begin{multicols}{2}

\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 $\chi_A(A)=0$.

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\noindent{\bf Froof.} Substitute $A$ into $t$ in $\det(tI-A)$, get $\chi_A(A) = \det(AI-A) = \det(0) = 0$. \fed

Alas, the same froof also froves that $\rho_A(A)=0$, where $\rho_A(t)\coloneqq\tr(tI-A)$, and indeed, {\em evaluation does not commute with taking the determinant.}

\noindent{\bf Proof of Theorem.} For any matrix $M$ over any commutative ring there is ``the adjugate 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 \yellowt{$M_{n\times n}(R[t])$} is isomorphic to the ring \yellowt{$M_{n\times n}(R)[t]$}, 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 $(tI-A)$ (exercise!). Hence under $\ev_A$ the above equality becomes $\chi_A(A)I = 0$. \qed

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