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

Let p be a prime. Compute the Jordan canonical form over the finite field \mathbb{F}_p of the n \times n matrix whose every entry is 1. (n \geq 2.)


Let A = [1]. Now the computation in this previous exercise holds over \mathbb{F}_p (indeed, over any ring with 1), and so we have A(A-nI) = 0.

If n is not divisible by p, then n is a unit in \mathbb{F}_p. If v = [v_1\ \ldots\ v_n]^\mathsf{T} is an eigenvector with eigenvalue n, then evidently (as we saw in the exercise referenced above) v = v_1[1\ \ldots\ 1]^\mathsf{T}. So the eigenspace of n has dimension 1, the characteristic polynomial of A is x^{n-1}(x-n), and the Jordan canonical form of A is J = [b_{i,j}] where b_{n,n} = n and b_{i,j} = 1 otherwise.

Suppose p divides n. Since A \neq 0, the minimal polynomial of A is now x^2. We claim that the remaining invariant factors of A are x. There is probably a slick linear-algebraic way to argue this, but I don’t see it. Instead I will show explicitly that A is similar to a matrix in Jordan canonical form. (If anyone can offer a simplification of this proof I’d be grateful!)

If n = 2, then the minimal polynomial of A is the only invariant factor, so the Jordan form of A is \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}. Suppose n \geq 2.

Let U = [1\ \ldots\ 1]^\mathsf{T} have dimension (n-1) \times 1, and let W be the (n-1) \times (n-1) matrix whose every entry is 1. Note that UU^\mathsf{T} = W and U^\mathsf{T}U = [n-1]. Let P = \begin{bmatrix} 1 & 0 \\ U & I \end{bmatrix} and Q = \begin{bmatrix} 1 & 0 \\ -U & I \end{bmatrix}. Evidently QP = I, so that Q = P^{-1}. Moreover, we have A = \begin{bmatrix} 1 & U^\mathsf{T} \\ U & W \end{bmatrix}, and P^{-1}AP = B = \begin{bmatrix} n & U^\mathsf{T} \\ 0 & 0 \end{bmatrix} = \begin{bmatrix} 0 & U^\mathsf{T} \\ 0 & 0 \end{bmatrix}.

Now let V = [1\ \ldots\ 1] have dimension 1 \times (n-2). Let R = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & -V \\ 0 & 0 & I \end{bmatrix} and S = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & V \\ 0 & 0 & I \end{bmatrix}. Evidently, RS = I, so that S = R^{-1}. Now B = \begin{bmatrix} 0 & 1 & V \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}, and we have R^{-1}BR = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}. If we squint just right, this matrix is in Jordan canonical form, and is similar to A.

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