Compute the Jordan canonical form of a given matrix over a finite field

Let p be a prime. Compute the Jordan canonical form over \mathbb{F}_p of the n \times n matrix whose diagonal entries are 0 and whose off-diagonal entries are 1. (n \geq 2)

Let A = [1 - \delta_{i,j}]. The computation we carried out here shows that (A - (n-1)I)(A+I) = 0 over \mathbb{F}_p (indeed, over any ring with 1). So the minimal polynomial of A divides (x-(n-1))(x+1). Note that A-(n-1)I and A+I are nonzero (any off-diagonal entry is 1), so the minimal polynomial of A is precisely m(x) = (x-(n-1))(x+1).

Let V = \mathbb{F}_p^n be an \mathbb{F}_p[x]-module via A as usual. Now the invariant factors m_k(x) of V divide m(x), and we have V = \bigoplus_k \mathbb{F}[x]/(m_k(x)).

We claim that (A+I)V has dimension 1 over \mathbb{F}_p. Indeed, if v = [v_1\ \ldots\ v_n]^\mathsf{T}, then (A+I)v = (\sum v_i)[1\ \ldots\ 1]^\mathsf{T}.

Using Lemmas 2 and 3 from this previous exercise, we have (x+1)V = \bigoplus_k (x+1)\mathbb{F}_p[x]/(m_k(x)) \cong_{\mathbb{F}_p[x]} \bigoplus_{(x+1)|m_k} \mathbb{F}_p[x]/(m_k/(x+1)) \oplus \bigoplus_{(x+1,m_k) = 1} \mathbb{F}_p[x]/(m_k(x)). Note that any mapping which realizes an isomorphism as \mathbb{F}_p[x]-modules is necessarily also an \mathbb{F}_p-vector space isomorphism; since (A+I)V has dimension 1, the right hand side of this isomorphism also has dimension 1 over \mathbb{F}_p. Now one of the left summands is \mathbb{F}_p[x]/(x-(n-1)), corresponding to the minimal polynomial. Thus the remaining summands must be trivial; that is, there are no invariant factors which are not divisible by x+1, and every invariant factor other than the minimal polynomial is exactly x+1.

So the elementary divisors of A are x+1 (n-1 times) and x-(n-1). The corresponding Jordan canonical form is J = [b_{i,j}] where b_{i,j} = n-1 if i=j=n, -1 if i=j \neq n, and 0 otherwise.

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