## 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.