Power ranks determine every nilpotent Jordan-block multiplicity

On a block $J_m(0)$, $\dim \ker J_m(0{)}^k=\min (k,m)$; its contribution to $d_k-d_{k-1}$ is therefore $1$ exactly when $m\ge k$, and $0$ otherwise. Summing across the block basis from [L1] gives the first nullity formula.

Subtracting the number of blocks of size at least $k+1$ from the number of blocks of size at least $k$ gives $2d_k-d_{k-1}-d_{k+1}$ blocks of exact size $k$.

Fact [L4] gives $d_k+{\rho }_k=\dim V$ for every $k$, so replacing each $d_k$ in steps 1.1-2.1 gives the two rank formulas. For the tail, first dispose of $V=0$: there the block multiset is empty, every $d_k$ and ${\rho }_k$ is zero, and both formulas read $0=0$. So assume $V\mathrm{\ne }0$, in which case the basis of [L1] has at least one block and a largest block size $m^{\ast }$ exists; step 1.1 gives $\dim \ker J_m(0{)}^k=\min (k,m)=m$ for every block and every $k\ge m^{\ast }$, so $d_k=\dim V$ and ${\rho }_k=0$ for all $k\ge m^{\ast }$. In particular $\ker N^{m^{\ast }}=\ker N^{m^{\ast }+1}$, which is the hypothesis of [L2], and [L2] then gives the stabilised tail beyond the largest block.

These formulas recover every block multiplicity, including size one and the endpoint after the largest block; when $V=0$ each quantity and each recovered multiplicity is zero.