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.

Question

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

Answer 1

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.

Answer 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.