## On the square of a Jordan block

Let $J$ be a Jordan block of size $n$ and eigenvalue $\lambda$ over a field $K$ whose characteristic is not 2.

1. Suppose $\lambda \neq 0$. Prove that the Jordan canonical form of $J^2$ is the Jordan block of size $n$ with eigenvalue $\lambda^2$.
2. Suppose $\lambda = 0$. Prove that the Jordan canonical form of $J^2$ has two blocks (with eigenvalue 0) of size $n/2, n/2$ if $n$ is even and of size $(n+1)/2, (n-1)/2$ if $n$ is odd.

First suppose $\lambda \neq 0$.

Lemma: Let $e_i$ denote the $i$th standard basis element (i.e. $e_i = [\delta_{i,1}\ \delta_{i,2}\ \ldots\ \delta_{i,n}]^\mathsf{T}$. If $1 \leq i < n-2$, then $(J^2-\lambda^2I)e_{i+2} = 2\lambda e_{i+1} + e_i$, $(J^2-\lambda^2I)e_{2} = 2\lambda e_1$, and $(J^2-\lambda^2I)e_1 = 0$. Proof: We have $J^2-\lambda^2I = (J+\lambda I)(J-\lambda I)$. Evidently, $(J-\lambda I)e_{i+2} = e_{i+1}$. Now $J+\lambda I = [b_{j,k}]$, where $b_{j,k} = 2\lambda$ if $j = k$ and $1$ if $k = j+1$ and 0 otherwise. So $(J+\lambda I)e_{i+1} = [\sum_k b_{j,k} \delta_{k,i+1}]$ $= [b_{j,i+1}]$ $= 2\lambda e_{i+1} + e_i$ if $i > 1$, and similarly $(J^2-\lambda^2 I)(e_2) = 2\lambda e_1$. $\square$

In particular, note that $(J^2 - \lambda^2 I)^{n-1} e_n = 2^{n-1}\lambda^{n-1} e_1$ is nonzero (here we use the noncharacteristictwoness of $K$), while $(J^2 - \lambda^2 I)^n = 0$. So the minimal polynomial of $J^2$ is $(x-\lambda^2)^n$. Thus the Jordan canonical form of $J^2$ has a single Jordan block, of size $n$, with eigenvalue $\lambda^2$.

Now suppose $\lambda = 0$. Evidently, we have $Je_{i+1} = e_i$ and $Je_1 = 0$. Now $J^n = 0$ and $J^{n-1} \neq 0$. If $n$ is even, we have $(J^2)^{n/2} = 0$ and $(J^2)^{n/2-1} = J^{n-2} \neq 0$, so that the minimal polynomial of $J^2$ is $x^{n/2}$. If $n$ is odd, we have $(J^2)^{(n+1)/2} = 0$ while $(J^2)^{(n+1)/2-1} = J^{n-1} \neq 0$, so the minimal polynomial of $J^2$ is $x^{(n+1)/2}$.

Next, we claim that $\mathsf{ker}\ J^2$ has dimension 2. To this end, note that $J^2e_{i+2} = e_i$ is nonzero, while $J^2e_2 = J^2e_1 = 0$. By this previous exercise, $J^2$ has 2 Jordan blocks; thus the blocks of $J^2$ both have eigenvalue 0 and are of size $n/2,n/2$ if $n$ is even, and $(n+1)/2, (n-1)/2$ if $n$ is odd.