The Cayley - Hamilton Theorem via Complex Integration

Let f be an analytic function on the plane and A a square matrix of order n whose eigenvalues are contained in the interior of a circle K centered at the origin. The expression f(A) = fKf(z)(z-A)-dz has been widely used to calculate (or define) the value of f at A. A special case of this formula provides us with a trivial proof of the Cayley-Hamilton theorem. We need only the most elementary notions: the integral fKf (z)dz, where f is a continuous complex valued function on K, and the formulas rS { 2iri if n=-1 (z-a)ndz 0 otherwise, where a lies inside K, and fK p(z)dz = 0, where p is any polynomial. Now let F be a matrix valued function continuous at least on K. We define fK F(z)dz to be the matrix whose entries are given by [fAF(z)dz ik [F(z) ]'jKdz, i.e., the function F is integrated entry-wise. The familiar properties of integration extend to matrix valued functions. In particular, for F: e-*n and BrC=n:fKBF(z)dz =BfKF(z)dz. The computation is easy. LEMMA. If A and K are as specified above, then fK(z-A)'-dz = 2ril. Proof. Note that (z-A)-1=a(z)-'adj(z-A), where a is the characteristic polynomial of A and adj is the classical adjoint (by adj B we mean the matrix such that B (adj B) = (det B)I). Since the entries in adj (z-A) are polynomials in z, the entries in (z-A)-' are rational functions. Let p(z)/a(z) be such an entry. We have the partial fraction decomposition p(z)/a(z) = b j(z-a) + R(z), (*) j where the aj's are simple eigenvalues and R(z) is a sum of terms b/(z-a)r, r> 1. If p(z)/a(z) is off the diagonal in (z-A)-', the degree of p is at most n-2, whereas that of a is n. Thus, to maintain the identity (*), we must have E b