Matrices with square roots

Let F be a field whose characteristic is not 2. Characterize those matrices A over F which have square roots. That is, characterize A such that A = B^2 for some matrix B.


We claim that A has a square root if and only if the following hold.

  1. Every eigenvalue of A is square in F.
  2. The Jordan blocks of A having eigenvalue 0 can be paired up in such a way that the sizes of the blocks in each pair differ by 0 or 1.

First we tackle the ‘if’ part. Suppose P^{-1}AP = \bigoplus J_i \oplus \bigoplus H_i \oplus \bigoplus K_i is in Jordan canonical form, where J_i are the Jordan blocks having nonzero eigenvalue \lambda_i and H_i and K_i are the blocks with eigenvalue 0 such that the sizes of H_i and K_i differ by 0 or 1. As we showed in this previous exercise, J_i = Q_iM_i^2Q_i^{-1} is the Jordan canonical form of the square of the Jordan block M_i with eigenvalue \sqrt{\lambda_i}, and H_i \oplus K_i = T_iN_i^2T_i^{-1} is the Jordan canonical form of the square of the Jordan block whose size is \mathsf{dim}\ H_i + \mathsf{dim}\ K_i with eigenvalue 0. So we have P^{-1}AP = C^2, and so A = (P^CP^{-1})^2 = B^2 is a square.

Conversely, suppose A = B^2 is a square, and say P^{-1}BP = J is in Jordan canonical form. Now PAP^{-1} = J^2. Now say Q^{-1}J^2Q = K is in Jordan canonical form; then Q^{-1}PAP^{-1}Q = K. By the previous exercise, the nonzero eigenvalues of K are square in F, and the Jordan blocks having eigenvalue 0 can be paired so that the sizes in each pair differ by 0 or 1.

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