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.

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