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Ranks of shifted powers determine Jordan form up to block order

Statement

Let $T$ have split characteristic polynomial. For each eigenvalue $λ$ , put $ρ_{k} (λ) = rank (T - λ I)^{k} (k \geq 0) .$ Then for every $k \geq 1$ , the number of Jordan blocks for $λ$ of size exactly $k$ is $ρ_{k - 1} (λ) - 2 ρ_{k} (λ) + ρ_{k + 1} (λ) .$ Consequently the ranks of all shifted powers determine the Jordan form uniquely up to permutation of its blocks. On the zero space the rank data and block multiset are empty.

Facts & Assumptions

Given: A finite-dimensional endomorphism $T$ whose characteristic polynomial splits.

[L1]

The space is the direct sum of generalised eigenspaces $G_{μ}$ , and $N_{μ} = (T - μ I) ∣_{G_{μ}}$ is nilpotent (Split characteristic polynomials decompose into generalised eigenspaces of the algebraic multiplicities).

[L2]

For a nilpotent operator, $rank N^{k - 1} - 2 rank N^{k} + rank N^{k + 1}$ is the number of blocks of size exactly $k$ (Power ranks determine every nilpotent Jordan-block multiplicity).

[L3]

Rank is the dimension of the image (Rank and nullity of a linear map with finite-dimensional domain).

[L4]

An internal direct sum gives unique componentwise decompositions (Internal direct sum $V = ⨁_{i < n} U_{i}$ : the sum is everything and each summand meets the sum of the others only in $0_{V}$ ).

Proof

technique · direct

1.1

Fix an eigenvalue $λ$ . On $G_{λ}$ , $T - λ I = N_{λ}$ . On $G_{μ}$ with $μ \neq λ$ , it is $(μ - λ) I + N_{μ}$ , which is invertible because the finite geometric series in $- N_{μ} / (μ - λ)$ is an inverse up to the nonzero scalar $μ - λ$ .

L1algebra

2.1

By the direct sum in [L1] and uniqueness in [L4], for every $k \geq 0$ the rank on the non- $λ$ summands is their constant total dimension $c_{λ}$ , while the rank on $G_{λ}$ is $rank N_{λ}^{k}$ ; hence $ρ_{k} (λ) = c_{λ} + rank N_{λ}^{k}$ .

step 1.1L1L3L4

3.1

The constant $c_{λ}$ cancels from the second difference, and [L2] then gives the displayed exact-size block count. Varying $k$ and $λ$ recovers the entire block multiset.

step 2.1L2algebra

4.1

By the definition of Jordan form, a block multiset determines the block diagonal matrix up to block order; for $V = 0$ there are no eigenvalues or blocks.

step 3.1L1∎

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 99 results over 22 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

S. Treil, Linear Algebra Done Wrong, Chapter 9, Sections 4.3-4.4 (standard reference, not scraped)