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 $\operatorname{dim}(\operatorname{ker} X^2) \geq 2$.

Let $y$ be any $n$-dimensional vector not in Ker$(X)$, such that $X^2y$ is nonzero. [That $X^2y$ is nonzero for some $y$ is what would have to be for $X^2$ to be $J$, and furthermore, for Dim Ker$(X^2)$ to be only $1$.] Then the fact that $X$ is nilpotent gives $X^ny=0$ for some positive integer $n$. Thus, let $n$ now be the smallest integer such that $X^ny=0$. Then $n$ is at least $3$. Furthermore:

*

*Then $X^ny=X(X^{n-1}y)$ is $0$, which gives $X^{n-1}y \in$ Ker$(X)$. That $n$ is the smallest integer and at least $3$ such that $X^{n}y=0$ gives $X^{n-1}y$ a nonzero vector in Ker$(X)$.


*Furthermore, $X^{n-1}y$ a nonzero vector in Ker$(X)$ gives $z=X^{n-2}y$ a nonzero vector not in Ker$(X)$ satisfying $X^2z = X^2(X^{n-2}y) = 0$. [Indeed, $z$ cannot be in Ker$(X)$, lest $Xz=$ $X(X^{n-2}y)= X^{n-1}y$ is $0$, which would contradict that $n$ is the smallest integer such that $X^ny=0$.]
But then this gives $X^2z = X^2(X^{n-2}y) = 0$ for some $z$ not in Ker$(X)$. But also $X^2v = X(Xv)$ $= X(0) = 0$ for all nonzero $v \in$ Ker$(X)$. So then Ker$(X^2)$ is a subset of $\langle v,z \rangle$, and $z \not \in$ Ker$(X)$ and $v \in$ Ker$(X)$ gives $2$ $\le$ Dim$(\langle v,z \rangle)$ $\le$ Dim Ker$(X^2)$. Which is what you want.
