proof of Cayley-Hamilton theorem in a commutative ring
Let $R$ be a commutative ring with identity^{} and let $A$ be an order $n$ matrix with elements from $R[x]$. For example, if $A$ is $\left(\begin{array}{cc}\hfill {x}^{2}+2x\hfill & \hfill 7{x}^{2}\hfill \\ \hfill x+1\hfill & \hfill 5\hfill \end{array}\right)$
then we can also associate with $A$ the following polynomial^{} having matrix coefficents:
$${A}^{\sigma}=[\genfrac{}{}{0pt}{}{0}{1}\mathit{\hspace{1em}}\genfrac{}{}{0pt}{}{0}{5}]+[\genfrac{}{}{0pt}{}{2}{1}\mathit{\hspace{1em}}\genfrac{}{}{0pt}{}{0}{0}]x+[\genfrac{}{}{0pt}{}{1}{0}\mathit{\hspace{1em}}\genfrac{}{}{0pt}{}{7}{0}]{x}^{2}.$$ |
In this way we have a mapping $A\u27f6{A}^{\sigma}$ which is an isomorphism^{} of the rings ${M}_{n}(R[x])$ and ${M}_{n}(R)[x]$.
Now let $A\in {M}_{n}(R)$ and consider the characteristic polynomial^{} of $A$: ${p}_{A}(x)=det(xI-A)$, which is a monic polynomial^{} of degree $n$ with coefficients in $R$. Using a property of the adjugate matrix we have
$$(xI-A)\mathrm{adj}(xI-A)={p}_{A}(x)I.$$ |
Now view this as an equation in ${M}_{n}(R)[x]$. It says that $xI-A$ is a left factor of ${p}_{A}(x)$. So by the factor theorem, the left hand value of ${p}_{A}(x)$ at $x=A$ is 0. The coefficients of ${p}_{A}(x)$ have the form $cI$, for $c\in R$, so they commute with $A$. Therefore right and left hand values are the same.
References
- 1 Malcom F. Smiley. Algebra^{} of Matrices. Allyn and Bacon, Inc., 1965. Boston, Mass.
Title | proof of Cayley-Hamilton theorem^{} in a commutative ring |
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Canonical name | ProofOfCayleyHamiltonTheoremInACommutativeRing |
Date of creation | 2013-03-22 16:03:16 |
Last modified on | 2013-03-22 16:03:16 |
Owner | Mathprof (13753) |
Last modified by | Mathprof (13753) |
Numerical id | 11 |
Author | Mathprof (13753) |
Entry type | Proof |
Classification | msc 15A18 |
Classification | msc 15A15 |