Square roots of the basic Jordan block of order $n$ associated with the eigenvalue $1$

Let $F$ be a field. The basic Jordan block of order $n$ associated with the eigenvalue $1$ will be denoted by $J_n$ where

$$J_n=\begin{pmatrix}1&1&&\\&1&1\\ &&\ddots&\ddots \\ && &1 &1\\ &&&&1\end{pmatrix}.$$

Is there an invertible matrix B such that $J_n=B^2$ ? If not, are there invertible matrices $B_1,\dots,B_k$ such that $J_n=B_1^2B_2^2\cdots B_k^2$? Moreover, can we describe the $B_i$? This is done for the case the field of characteristic different from $2$. But I do not know what to do for the case the field of characteristic $2$.

Solution 1:

On your second question: this is impossible, at least for $n=2$ and $F=GF(2)$. It can be verified that there are only six invertible $2\times2$ matrices in $M_2\left(GF(2)\right)$, namely, $$ I_2,\ R=\pmatrix{0&1\\ 1&0},\ J=\pmatrix{1&0\\ 1&1},\ J^T,\, X=\pmatrix{1&1\\ 1&0},\ X^2=\pmatrix{0&1\\ 1&1}. $$ Since $R^2=J^2=(J^T)^2=I_2$, every square of the six matrices above is a polynomial in $X$. In turn, $B_1^2B_2^2\cdots B_k^2$ is also a polynomial in $X$. Hence $B_1^2B_2^2\cdots B_k^2$ is symmetric and it cannot possibly be equal to $J$.

However, I suspect that the decomposition is possible when $n\ge3$, but I have no idea how to prove or disprove it.

Solution 2:

We can use the following fact (in all the textbooks) to give a proof that when $n\geqslant 2$ no such $B$ exists: Suppose $T:V\to V$ is a linear operator on a finite dimensional vector space. Then $(\mathop{rk}(T^i))_{i=0}^\infty$ is a non increasing sequence of natural numbers, and moreover if $\mathop{rk}(T^s)=\mathop{rk}(T^{s+1})$ for some $s$ then $\mathop{rk}(T^s)=\mathop{rk}(T^{s+t})$ for all $t$.

When $n=2$ it is easy to check by hand no such $B$ exists, so suppose $n\geqslant 3$.

Suppose that $B^2=J_n$; then $(B-I)^2=J-I$ has rank $n-1$.

Then $$n=\mathop{rk} (B-I)^{0}\geqslant \mathop{rk}(B-I) \geqslant \mathop{rk}(B-I)^2=n-1$$ and so either $\mathop{rk} (B-I)^0 =\mathop{rk}(B-I)^1=n$ or $\mathop{rk} (B-I)^1 =\mathop{rk}(B-I)^2=n-1$. In either case the sequence stabilizes and so $$n-1\leqslant \mathop{rk}(B-I)=\mathop{rk}(B-I)^4=\mathop{rk}(J-I)^2=n-2$$ a manifest contradiction. So no such $B$ exists.