Show that a Jordan block is not a square of any matrix.

Question

How do you show that $$J_n(0)= \begin{pmatrix} 0 & 1 & 0 & \cdots & 0 & 0 \\ 0 & 0 & 1 & \cdots & 0 & 0 \\ 0 & 0 & 0 & \cdots & 0 & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & 0 & \cdots & 0 & 1 \\ 0 & 0 & 0 & \cdots & 0 & 0 \\ \end{pmatrix} \ne A^2$$ for any $n\times n$ matrix $A$?

I don't even know how to start proving this. Any tips?

Answer 1

Note that $J_n(0)$ is nilpotent, in fact we have $(J_n(0))^k = 0 \Leftrightarrow k \geq n$. Now suppose that $A^2 = J_n(0)$, then we have $A^{2n}=0$, so $A$ is nilpotent as well. But being nilpotent implies that the characteristic polynomial of $A$ is $T^n$, by Cayley-Hamilton, we have $A^n =0$, this means that $(J_n(0))^{\lceil \frac{n}{2} \rceil} = 0$, contradicting the fact that the smallest $k$ such that $(J_n(0))^k = 0$ is $k=n$

Answer 2

Suppose the contrary. Then what would the Jordan form of $A$ look like? With this Jordan form in mind, what is the maximum possible rank of $A^2$?