## Cayley-Hamilton for commutative rings

Many people have told me that Cayley-Hamilton Theorem is their favourite theorem in linear algebra. I also like this theorem a lot (Proof: I am dedicating a blog post about it!)

Let $A$ be a linear transformation on a finite-dimensional vector space $V$ (over $F$). In other words, $A: V\to V$ satisfies $A(v_1+v_2)=A(v_1)+A(v_2)$, and $A(cv)=cA(v)$ for every scalar $c\in F$. It is clear that $k$-fold composition $A^{k}: V\to V$ is also a linear map. So we can ask when $A$ satisfies a polynomial identity. In other words, we are interested in finding some coefficients $c_{1}, c_{2}, \cdots, c_{m}$ such that $c_m A^{m}+c_{m-1} A^{m-1}+\cdots + c_0 I = 0$ where $I$ is the identity map, and $0$ is the zero map (the map that is identically zero).  Since the set of all linear transformations on $V$ forms a $n^2$ dimensional vector space, where $n = \textrm{dim}(V)$, it follows that the $n^2+1$ maps $I=A^{0}, A, A^2, \cdots, A^{n^2}$ are linearly dependent, so there exists a linear relation between the powers of $A$. As a result, we see that $A$ satisfies a polynomial of degree at most $n^2$.

Cayley-Hamilton Theorem shows that we can do significantly better. Recall the characteristic polynomial $p_{A}(x)=\det(Ix-A)$. It is clear that $p_{A}(x)$ is a polynomial of degree $n$. Cayley-Hamilton Theorem states that $p_{A}(A)=0$. I am not going to prove this theorem here. A proof can be found located in wikipedia. I am also going to assume that the reader is familiar with the notion that a linear transformation on a finite-dimensional vector space can be identified with a matrix.

Let me explain why I like this theorem. It is because the theorem is true when the field is replaced by any commutative ring! In other words, if you consider a matrix with entries from some commutative ring $R$, then define (say, using Laplace expansion) $p_{A}(x)=\det(Ix-A)$ as usual. We still have the conclusion $p_{A}(A)=0$. Of course, it doesn’t make sense to talk about vector spaces over arbitrary rings (for this purpose, we use related objects called modules), but the Cayley-Hamilton Theorem still holds in the sense I described. This is closely related with the so-called “determinant trick” (cf. Nakayama’s Lemma) which is a very convenient tool in commutative algebra.

Here is something cool I learned today (from Professor Lior Silberman). We can deduce Cayley-Hamilton Theorem for commutative rings using the result of Cayley-Hamilton Theorem for fields. This could at first seem like a hopeless task, because we are trying to prove something stronger. Here is how the proof goes. Cayley-Hamilton is true for the field of rational numbers $\mathbb{Q}$. Since $\mathbb{Z}$ is a subring of $\mathbb{Q}$, Cayley-Hamilton holds for $\mathbb{Z}$. By this, I just mean that the conclusion $p_{A}(A)=0$ holds whenever $A$ is a matrix with entries from $Z$. Now consider an arbitrary commutative ring $R$. Let $A$ be a matrix with entries from $R$, i.e. $A\in M_{n\times n}(R )$.  We view $A=(a_{ij})_{i, j}$ as an array $n^2$ indeterminates. Then $p_{A}(A)$ is a matrix with $n^2$ polynomials (each of them being multivariate in variables $a_{ij}$ with coefficients in $\mathbb{Z}$). Now, when these indeterminates $a_{ij}$ are replaced by integers, we get that $p_{A}(A)$ vanishes (precisely because Cayley-Hamilton is true for $\mathbb{Z}$). Thus, $p_{A}(A)$ vanishes on all of $M_{n\times n}(\mathbb{Z})$. But each entry of $p_{A}(A)$ is a polynomial of degree $n$, and no non-zero multivariate polynomial with coefficients in $\mathbb{Z}$ can vanish on all of $\mathbb{Z}$. It follows that each entry of $p_{A}(A)$ is the zero polynomial. Consequently, $p_{A}(A)=0$.

It is natural to wonder why $\mathbb{Z}$ plays an important role in the proof above. One of the reasons for this phenomena is that $\mathbb{Z}$ is the initial object in the category of rings.