 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 $m\ge r$. Summing over blocks proves thatcounts the $\lambda$-blocks of size at least $r$.

If $\lambda \ne 0$, the polynomial $z^2$ has nonzero derivative at $\lambda$, soIf $\lambda =0$, squaring the nilpotent shift separates the odd and even basis chains: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 $t^2-a_1{a}_4$ and $t^2-a_2{a}_3$. Since invertibility makes both products nonzero, each restriction has two distinct complex eigenvalues. Its Jordan form is thereforefor either choices of the square roots.