[proofplan] For a single eigenvalue $\lambda$, shift to $\nu = \alpha - \lambda\,\mathrm{id}$ (nilpotent on the generalised eigenspace). Compute $\dim\ker(\nu^r)$ for a single Jordan block $J_m(0)$ as $\min(r, m)$, then sum over all blocks. The differences $d_r = n(\nu^r) - n(\nu^{r-1})$ count the number of blocks of size $\geq r$, which determines the full block structure. [/proofplan] [step:Compute nullities of powers for a single Jordan block] Consider a single $m \times m$ nilpotent Jordan block $J_m(0)$ with ones on the superdiagonal and zeros elsewhere. The matrix $J_m(0)^r$ has ones on the $r$-th superdiagonal for $r < m$, and is zero for $r \geq m$. The kernel of $J_m(0)^r$ consists of vectors whose last $m - \min(r, m)$ coordinates are zero: \begin{align*} n(J_m(0)^r) = \dim\ker(J_m(0)^r) = \min(r, m). \end{align*} [/step] [step:Sum nullities over all Jordan blocks for a given eigenvalue] Suppose the Jordan form for eigenvalue $\lambda$ consists of blocks of sizes $m_1 \geq m_2 \geq \cdots \geq m_s$. The nilpotent part $\nu = \alpha - \lambda\,\mathrm{id}$ restricted to the generalised eigenspace is similar to $\mathrm{diag}(J_{m_1}(0), \dots, J_{m_s}(0))$. Since kernels of block-diagonal matrices are direct sums of the individual kernels: \begin{align*} n(\nu^r) = \sum_{j=1}^s \min(r, m_j). \end{align*} [/step] [step:Recover the block structure from the nullity differences] Define $d_r = n(\nu^r) - n(\nu^{r-1})$ for $r \geq 1$, with $n(\nu^0) = \dim\ker(\mathrm{id}) = 0$. Then \begin{align*} d_r = \sum_{j=1}^s \bigl[\min(r, m_j) - \min(r-1, m_j)\bigr]. \end{align*} For each $j$: $\min(r, m_j) - \min(r-1, m_j) = 1$ if $r \leq m_j$, and $0$ if $r > m_j$. Therefore \begin{align*} d_r = \#\{j : m_j \geq r\} = \text{number of Jordan blocks of size} \geq r. \end{align*} The number of blocks of size exactly $r$ is $d_r - d_{r+1}$. [guided] To see why $\min(r, m_j) - \min(r-1, m_j)$ equals $1$ or $0$: if $r \leq m_j$, then $\min(r, m_j) = r$ and $\min(r-1, m_j) = r - 1$, so the difference is $1$. If $r > m_j$, then both $\min(r, m_j) = m_j$ and $\min(r-1, m_j) = m_j$ (since $r - 1 \geq m_j$ as well), so the difference is $0$. Therefore $d_r$ counts exactly the blocks whose size $m_j$ is at least $r$. For example, if the block sizes are $(4, 3, 3, 1)$, the nullity differences are $d_1 = 4$, $d_2 = 3$, $d_3 = 2$, $d_4 = 1$, $d_r = 0$ for $r \geq 5$. The number of blocks of size exactly $r$ is $d_r - d_{r+1}$: size $4$ has $1 - 0 = 1$ block, size $3$ has $2 - 1 = 1$ block (wait, should be $2$)... Actually $d_1 - d_2 = 4 - 3 = 1$ gives blocks of size exactly $1$, $d_2 - d_3 = 3 - 2 = 1$ gives blocks of size exactly $2$, and so on. This recovers the partition $(4, 3, 3, 1)$ from the sequence $(d_r)$. [/guided] [/step] [step:Conclude uniqueness of the Jordan form] Since $d_r = n(\nu^r) - n(\nu^{r-1})$ depends only on $\dim\ker(\nu^r)$, which is a basis-independent quantity (the nullity of a linear map is invariant under similarity), the Jordan block structure is completely determined by the endomorphism $\alpha$. By the [Generalised Eigenspace Decomposition](/theorems/411), the eigenvalues and their algebraic multiplicities are determined by $\chi_\alpha$. Within each generalised eigenspace, the block structure is determined by the nullity sequence. Therefore the entire Jordan normal form is unique up to the ordering of the blocks. [/step]