Show that the equation $X^2=J_n$has no solution if $n\geq2$, if $J_n$ is the jordan block of size $n$ with zeros on diagonal. More clearly:
$$
J_n= 
        \begin{pmatrix}
        0 & 1 & 0 &\cdots & 0 & 0 \\
        0 & 0 & 1 &\cdots & 0 & 0 \\
        \vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\ 
        0 & 0 & 0 &\cdots & 1 & 0  \\
        0 & 0 & 0 &\cdots & 0 & 1 \\
        0 & 0 & 0 &\cdots & 0 & 0
        \end{pmatrix}
$$
The solutions manual says: $X$ is nilpotent, thus $\operatorname{dim}(\operatorname{ker} X^2) \geq 2$ and $X^2$ cannot have rank $n - 1$.
I understand that $X$ is nilpotent, but I do not understand how that implies $\operatorname{dim}(\operatorname{ker} X^2) \geq 2$.

If $X^2$ has rank $n-1$, then $n-1\ge\operatorname{rank}(X)\ge\operatorname{rank}(X^2)=n-1$. Therefore $\dim XV=\dim X^2V=n-1$, where $V=\mathbb F^n$. However, as $X^2V=X(XV)\subseteq X(V)=XV$, we must have $X^2V=XV$. It follows that $X^kV=XV\ne0$ for every integer $k>0$, which is a contradiction to the assumption that $X$ is nilpotent.
By the way, one may prove that the equation $X^2=J_n$ is not solvable without considering the rank of $X^2$: if $J_n=X^2$, then $X^{2n}=J_n^n=0$. Hence $X$ is nilpotent. Thus we must have $X^n=0$, because $X$ is $n\times n$. But then $0=X^nX^{n-2}=X^{2n-2}=J_n^{n-1}\ne0$, which is a contradiction.
