= Solution
A <Jordan block> $J_m(\lambda)$ has $\lambda$ on its diagonal, ones on its superdiagonal, and zeros elsewhere. A matrix is in <Jordan normal form> when it is block diagonal with Jordan blocks.
On a block $J_m(\lambda)$, $\ker(\alpha-\lambda I)^r$ has dimension $\min(r,m)$. Its increase from $r-1$ to $r$ is one exactly when $m\geq r$. Summing over blocks proves that
$$
\boxed{\dim\ker(\alpha-\lambda I)^r-\dim\ker(\alpha-\lambda I)^{r-1}}
$$
counts the $\lambda$-blocks of size at least $r$.
If $\lambda\ne0$, the polynomial $z^2$ has nonzero derivative at $\lambda$, so
$$
\boxed{J_m(\lambda)^2\sim J_m(\lambda^2).}
$$
If $\lambda=0$, squaring the nilpotent shift separates the odd and even basis chains:
$$
\boxed{J_m(0)^2\sim J_{\lceil m/2\rceil}(0)\oplus J_{\lfloor m/2\rfloor}(0),}
$$
omitting a zero-size block.
The displayed invertible matrix preserves the subspaces spanned by $(e_1,e_4)$ and $(e_2,e_3)$. Its restrictions have characteristic polynomials $t^2-a_1a_4$ and $t^2-a_2a_3$. Since invertibility makes both products nonzero, each restriction has two distinct complex eigenvalues. Its Jordan form is therefore
$$
\boxed{\operatorname{diag}\left(\sqrt{a_1a_4},-\sqrt{a_1a_4},
\sqrt{a_2a_3},-\sqrt{a_2a_3}\right),}
$$
for either choices of the square roots.
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