Matrices with square roots over fields of characteristic 2

Let F be a field of characteristic 2. Compute the Jordan canonical form of a Jordan block J of size n and eigenvalue \lambda over F. Characterize those matrices A over F which are squares; that is, characterize A such that A = B^2 for some matrix B.


Let J = [b_{i,j}] be the Jordan block with eigenvalue \lambda and size n. That is, b_{i,j} = \lambda if j = i, 1 if j = i+1, and 0 otherwise. Now J^2 = [\sum_k b_{i,k}b_{k,j}]; if k \neq i or k \neq i+1, then b_{i,k} = 0. Evidently then we have (J^2)_{i,j} = \lambda^2 if j = i, 1 if j = i+2, and 0 otherwise, Noting that 2 = 0. So J^2 - \lambda^2 I = \begin{bmatrix} 0 & I \\ 0_2 & 0 \end{bmatrix}, where 0_2 is the 2 \times 2 zero matrix and I is the (n-2) \times (n-2) identity matrix. Now let v = \begin{bmatrix} V_1 \\ V_2 \end{bmatrix}, where V_1 has dimension 2 \times 1. Now (J^2 - \lambda^2 I)v = \begin{bmatrix} V_2 \\ 0 \end{bmatrix}. That is, J^2 - \lambda^2 I ‘shifts’ the entries of v– so e_{i+2} \mapsto e_i and e_1, e_2 \mapsto 0. In particular, the kernel of J^2 - \lambda^2 I has dimension 2, so that by this previous exercise, the Jordan canonical form of J^2 has two blocks (both with eigenvalue \lambda^2.

Now J^2 - \lambda^2 I = (J + \lambda I)(J - \lambda I) = (J - \lambda I)^2, since F has characteristic 2. Note that J-\lambda I has order n, since (evidently) we have (J - \lambda I)e_{i+1} = e_i and (J - \lambda I)e_1 = 0. So J-\lambda I has order n. If n is even, then (J^2 - \lambda^2 I)^{n/2} = 0 while (J^2 - \lambda^2 I)^{n/2-1} \neq 0, and if n is odd, then (J^2-\lambda^2 I)^{(n+1)/2} = 0 while (J^2 - \lambda^2 I)^{(n+1)/2-1} \neq 0. So the minimal polynomial of J^2 is (x-\lambda^2)^{n/2} if n is even and (x-\lambda^2)^{(n+1)/2} if n is odd.

So the Jordan canonical form of J^2 has two Jordan blocks with eigenvalue \lambda^2. If n is even, these have size n/2,n/2, and if n is odd, they have size (n+1)/2, (n-1)/2.

Now let A be an arbitrary n \times n matrix over F (with eigenvalues in F). We claim that A is a square if and only if the following hold.

  1. The eigenvalues of A are square in F
  2. For each eigenvalue \lambda of A, the Jordan blocks with eigenvalue \lambda can be paired up so that the sizes of the blocks in each pair differ by 0 or 1.

To see the ‘if’ part, suppose P^{-1}AP = \bigoplus H_i \oplus K_i is in Jordan canonical form, where H_i and K_i are Jordan blocks having the same eigenvalue \lambda_i and whose sizes differ by 0 or 1. By the first half of this exercise, H_i \oplus K_i is the Jordan canonical form of J_i^2, where J_i is a Jordan block with eigenvalue \sqrt{\lambda_i}. Now A is similar to the direct sum of these J_i^2, and so Q^{-1}AQ = J^2. Then A = (Q^{-1}JQ)^2 = B^2 is square.

Conversely, suppose A = B^2 is square, and say P^{-1}BP = J is in Jordan canonical form. So P^{-1}AP = J^2. Letting J_i denote the Jordan blocks of J, we have P^{-1}AP = \bigoplus J_i^2. The Jordan canonical form of J_i^2 has two blocks with eigenvalue \lambda_i^2 and whose sizes differ by 0 or 1, by the first half of this exercise. So the Jordan blocks of A all have eigenvalues which are square in F and can be paired so that the sizes in each pair differ by 0 or 1.

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